replace matrix elements by comparing row and column names in r

replace matrix elements by comparing row and column names in r - r

I have a matrix A and vector B like this.
cname = 100*rep(1:3, each=9)+10*rep(1:3, each=3)+1:3
rname = 100*rep(2:3, each=9)+10*rep(1:3, each=3)+1:3
A = matrix(0, length(rname), length(cname), dimnames=list(rname,cname))
B = runif(length(cname))
names(B) = cname
I want to compute U(IJK) = B(IJK) - B(IJ1) - B(1JK) + B(1J1) define A such that
U = A %*% B
I did the following
A[rep(colnames(A),each=nrow(A)) == rep(rownames(A),ncol(A)) ] <- 1
A[(rep(as.integer(substr(colnames(A),1,2))*(substr(colnames(A),3,3)=="1"),each=nrow(A)) == rep(as.integer(substr(rownames(A),1,2)),ncol(A)))] <- -1
A[(rep(as.integer(substr(colnames(A),2,3))*(substr(colnames(A),1,1)=="1"),each=nrow(A)) == rep(as.integer(substr(rownames(A),2,3)),ncol(A)))] <- -1
A[(rep(as.integer(substr(colnames(A),2,2))*(substr(colnames(A),1,1)=="1"&substr(colnames(A),3,3)=="1"),each=nrow(A)) == rep(as.integer(substr(rownames(A),2,2)),ncol(A)))] <- 1
which gives
> A
111 112 113 121 122 123 131 132 133 211 212 213 221 222 223 231 232 233 311 312 313 321 322 323 331 332 333
211 1 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
212 1 -1 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
213 1 0 -1 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
221 0 0 0 1 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
222 0 0 0 1 -1 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
223 0 0 0 1 0 -1 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0 0 0 0
231 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0
232 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0
233 0 0 0 0 0 0 1 0 -1 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0 0 0 0
311 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0
312 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0
313 1 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0 0 0 0
321 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0
322 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0
323 0 0 0 1 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 1 0 0 0
331 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0
332 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0
333 0 0 0 0 0 0 1 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 1
Is there an easier way to do this?
Update
A neater code for the last part
A_row <- t(matrix(rep(rownames(A),each=ncol(A)),ncol(A),nrow(A)))
A_col <- matrix(rep(colnames(A),each=nrow(A)),nrow(A),ncol(A))
A[A_row==A_col] <- 1
A[substr(A_col,1,2)==substr(A_row,1,2) & substr(A_col,3,3)=="1"] <- -1
A[substr(A_col,2,3)==substr(A_row,2,3) & substr(A_col,1,1)=="1"] <- -1
A[substr(A_col,2,2)==substr(A_row,2,2) & substr(A_col,1,1)=="1" & substr(A_col,3,3)=="1"] <- 1

Related

Comparison of two dataframes

I have an excel table of 15200 lines, corresponding to a tree analyzed for its structures. I have all the structures in columns (48 structures), they have been counted on every tree. for example, the tree 12607 has 3 structures CV11, 1 structure IN12 and none (0) of the rest of all the structures. Thus, the table looks like a huge table with a lot of 0 and some numbers of the occurrence of structures on the trees. The very last column is the value given to the tree, according to the structures found on it (each structure giving a number of point to the tree by its presence on it).
The question is: Are there some structures, or combination of structures, which give a high value to the tree. Of course, according to the value of each structure, we can see which one has a higher value than the others (ex: structure CV11 has a value of 15, structure IN12 has a value of 4). But what I want to know is, if we take all the trees having a higher final value than 100 (we create a new dataframe "data100"), and we compare with the trees having a final value under 100 (we create another dataframe "data0"), can we find a significant difference in the number and occurrence of structures found on these trees? Because the structure with high value is maybe only found on the trees with a value under 100; because for example, this structure does not allow other structures to be found on the same tree.
Voilà, I hope I have given enough details... If you have any idea or proposition for solving this problem.. it will be great!
Below is my script.
> data100
CV11 CV12 CV13 CV14 CV15 CV21 CV22 CV23 CV24 CV25 CV26 CV31 CV32 CV33 CV41 CV42 CV43 CV44 CV51 CV52 IN11 IN12 IN13
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
13 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
IN14 IN21 IN22 IN23 IN31 IN32 IN33 IN34 BA11 BA12 BA21 DE11 DE12 DE13 DE14 DE15 GR11 GR12 GR13 GR21 GR22 GR31 GR32
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
11 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 2 0 0 0 0 0
12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 3 0 0
13 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 3 0 0
14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0
15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
EP11 EP12 EP13 EP14 EP21 EP31 EP32 EP33 EP34 EP35 NE11 NE12 NE21 OT11 OT12 OT21 OT22 ecoval
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 56
3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10
4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10
5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4
6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 24
7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 18
12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 63
13 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 77
14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 54
15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 20
[ reached getOption("max.print") -- omitted 60749 rows ]
> sortdata100<-data100[order(data100[,64],decreasing=T),]
> rsortdata100<-sortdata100[sortdata100$ecoval>100,]
> rsortdata100<-na.omit(rsortdata100)#181 lignes
> rsortdata100
CV11 CV12 CV13 CV14 CV15 CV21 CV22 CV23 CV24 CV25 CV26 CV31 CV32 CV33 CV41 CV42 CV43 CV44 CV51 CV52 IN11 IN12 IN13
1291 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1083 0 4 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3919 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0
14685 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
4021 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
5452 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
14686 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0
4022 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0
1013 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2895 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4719 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0
682 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
3444 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1299 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
2713 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0
IN14 IN21 IN22 IN23 IN31 IN32 IN33 IN34 BA11 BA12 BA21 DE11 DE12 DE13 DE14 DE15 GR11 GR12 GR13 GR21 GR22 GR31 GR32
1291 0 0 0 0 0 0 0 0 30 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1083 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3919 0 0 1 0 2 0 0 0 2 0 0 0 3 0 0 0 0 0 0 11 0 0 0
14685 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4021 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5452 0 0 1 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0
14686 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 0 0 0 2
4022 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1013 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2895 0 0 0 1 0 0 0 0 4 0 0 3 0 4 3 0 0 0 0 0 0 0 0
4719 0 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0
682 0 0 0 0 0 0 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 0
3444 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1299 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0
2713 0 0 0 2 0 3 0 0 2 0 0 0 1 5 1 0 0 0 0 0 0 0 0
EP11 EP12 EP13 EP14 EP21 EP31 EP32 EP33 EP34 EP35 NE11 NE12 NE21 OT11 OT12 OT21 OT22 ecoval
1291 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1192
1083 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 424
3919 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 380
14685 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 370
4021 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 358
5452 0 0 0 0 0 0 1 0 0 11 0 0 0 0 1 0 0 356
14686 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 354
4022 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 346
1013 0 8 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 326
2895 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 325
4719 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 324
682 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 311
3444 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 306
1299 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 302
2713 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 302
[ reached getOption("max.print") -- omitted 166 rows ]
> data0<-sortdata100[sortdata100$ecoval<100,]
> data0<-na.omit(data0)
> data0
CV11 CV12 CV13 CV14 CV15 CV21 CV22 CV23 CV24 CV25 CV26 CV31 CV32 CV33 CV41 CV42 CV43 CV44 CV51 CV52 IN11 IN12 IN13
4728 0 0 0 1 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0
5339 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
11766 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
796 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3561 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
10581 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
10618 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0
14376 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
14389 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0
790 0 0 0 1 0 0 0 0 1 0 0 2 0 0 0 0 0 0 0 0 1 0 0
3974 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
4739 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 1 0 0 0 0 0 0
156 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2740 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2950 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0
IN14 IN21 IN22 IN23 IN31 IN32 IN33 IN34 BA11 BA12 BA21 DE11 DE12 DE13 DE14 DE15 GR11 GR12 GR13 GR21 GR22 GR31 GR32
4728 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
5339 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
11766 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0
796 1 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3561 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0
10581 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
10618 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
14376 1 0 0 0 0 0 0 0 1 0 0 0 0 2 0 0 0 0 0 0 0 0 0
14389 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 1 0 0 0 0 0 0 0
790 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0
3974 0 0 0 0 0 0 0 0 1 0 0 0 4 0 0 0 1 0 0 0 0 0 0
4739 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
156 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0
2740 0 0 0 0 0 0 0 0 0 0 0 0 0 6 2 0 0 0 0 0 0 0 0
2950 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
EP11 EP12 EP13 EP14 EP21 EP31 EP32 EP33 EP34 EP35 NE11 NE12 NE21 OT11 OT12 OT21 OT22 ecoval
4728 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 99
5339 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 99
11766 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 99
796 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 98
3561 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 98
10581 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 98
10618 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 98
14376 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 98
14389 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 98
790 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 97
3974 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 97
4739 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 1 0 97
156 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 96
2740 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 96
2950 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 96
[ reached getOption("max.print") -- omitted 14984 rows ]

maybe something like this ?
library(dplyr)
data %>% group_by(ecoval > 100) %>% summarize_all(mean)
that should give you the average of each column for ecoval > and <= to 100

Select n rows after specific number

I work with a data.frame like this:
Country Date balance_of_payment business_confidence_indicator consumer_confidence_indicator CPI Crisis_IMF
1 Australia 1980-01-01 -0.87 100.215 99.780 25.4 0
2 Australia 1980-04-01 -1.62 100.061 99.746 26.2 0
3 Australia 1980-07-01 -3.70 100.599 100.049 26.6 0
4 Australia 1980-10-01 -3.13 100.597 100.735 27.2 0
5 Australia 1981-01-01 -2.73 101.149 101.016 27.8 0
6 Australia 1981-04-01 -4.11 100.936 100.150 28.4 0
I want to create a summary statistic with describe(dataset)from the HmiscPackage.
I need to differentiate between the timespans n-quarters before Crisis_IMF is 1, the time in which Crisis_IMF is 1 and the state n-quaters after Crisis_IMF is 1.
To select the time in which Crisis_IMF is 1, I did describe(dataset[dataset$Crisis_IMF==1,"balance_of_payment"]).
But I do not know how to make the command over the timespan of n-quarters (e.g. 8) after the event.
Edit:
dataset$Crisis_IMF
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[60] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[119] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[178] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[237] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[296] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[355] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[414] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[473] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[532] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
[591] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[650] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[709] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0
[768] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[827] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[886] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[945] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1004] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1063] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1122] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1181] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1240] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1299] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1358] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1417] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1476] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1
[1535] 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1
[1594] 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1653] 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1712] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1771] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0
[1830] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[1889] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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[4308] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Edit2; further information on the dataset:
Country Date balance_of_payment Crisis_IMF
1 Australia 1980-01-01 -0.87 0
2 Australia 1980-04-01 -1.62 0
3 Australia 1980-07-01 -3.70 0
4 Australia 1980-10-01 -3.13 0
5 Australia 1981-01-01 -2.73 0
6 Australia 1981-04-01 -4.11 0
7 Australia 1981-07-01 -3.98 0
8 Australia 1981-10-01 -5.27 0
9 Australia 1982-01-01 -5.31 0
10 Australia 1982-04-01 -4.67 0
11 Australia 1982-07-01 -3.30 0
12 Australia 1982-10-01 -3.24 0
13 Australia 1983-01-01 -3.45 0
14 Australia 1983-04-01 -2.86 0
15 Australia 1983-07-01 -3.58 0
...
137 Australia 2014-01-01 -2.18 0
138 Australia 2014-04-01 -3.44 0
139 Australia 2014-07-01 -3.04 0
140 Australia 2014-10-01 -2.39 0
141 Austria 1980-01-01 -3.97 0
142 Austria 1980-04-01 -3.89 0
143 Austria 1980-07-01 -1.84 0
144 Austria 1980-10-01 -1.60 0
145 Austria 1981-01-01 -2.74 0
146 Austria 1981-04-01 -2.88 0
147 Austria 1981-07-01 -2.83 0
148 Austria 1981-10-01 -2.06 0
149 Austria 1982-01-01 -0.63 0
150 Austria 1982-04-01 0.61 0
151 Austria 1982-07-01 2.42 0
152 Austria 1982-10-01 2.70 0
There can be more then one crisis period for one country. That e.g. in Australia are Crisis from 1990-01-01 to 1991-04-01 and 2002-01-01 to 2005-01-01. I want to create 3 different describe commands, which show the behaviour of the variable in the above mentioned states.

You haven't provided your full data, so I have to guess that your Crisis_IMF column has an unbroken sequence of zeroes (before the crisis), followed by an unbroken sequence of ones (during which the IMF crisis was considered to be in effect), finally followed by an unbroken sequence of zeroes (after the crisis).
Below I've synthesized my own data for testing. I only synthesized columns Crisis_IMF and balance_of_payment, because those appear to be the only columns relevant to your problem. I used 30 rows, with the first 10 before, the next 10 during, and the last 10 after the crisis. I used sort of a random parabolic arc for the balance_of_payment, but that was entirely random.
library('Hmisc');
set.seed(1);
N <- 30;
df <- data.frame(balance_of_payment=-5+2*seq(-1.5,1.5,len=N)^2+rnorm(N,0,0.2), Crisis_IMF=c(rep(0,N/3),rep(1,N/3),rep(0,N/3)) );
df;
## balance_of_payment Crisis_IMF
## 1 -0.6252908 0
## 2 -1.0625579 0
## 3 -1.8228927 0
## 4 -1.8503850 0
## 5 -2.5744076 0
## 6 -3.2324647 0
## 7 -3.3561408 0
## 8 -3.6484112 0
## 9 -3.9805631 0
## 10 -4.4136342 0
## 11 -4.2642312 1
## 12 -4.6598435 1
## 13 -4.9904788 1
## 14 -5.3947830 1
## 15 -4.7696630 1
## 16 -5.0036359 1
## 17 -4.9550811 1
## 18 -4.6774634 1
## 19 -4.5735679 1
## 20 -4.4478071 1
## 21 -4.1687610 0
## 22 -3.9392921 0
## 23 -3.7811631 0
## 24 -3.8514970 0
## 25 -2.9444058 0
## 26 -2.6515349 0
## 27 -2.2006002 0
## 28 -1.9499174 0
## 29 -1.1949166 0
## 30 -0.4164117 0
crisisRange <- range(which(df$Crisis_IMF==1));
crisisRange;
## [1] 11 20
df$Off_Crisis <- c((1-crisisRange[1]):-1,rep(0,diff(crisisRange)+1),1:(nrow(df)-crisisRange[2]));
df;
## balance_of_payment Crisis_IMF Off_Crisis
## 1 -0.6252908 0 -10
## 2 -1.0625579 0 -9
## 3 -1.8228927 0 -8
## 4 -1.8503850 0 -7
## 5 -2.5744076 0 -6
## 6 -3.2324647 0 -5
## 7 -3.3561408 0 -4
## 8 -3.6484112 0 -3
## 9 -3.9805631 0 -2
## 10 -4.4136342 0 -1
## 11 -4.2642312 1 0
## 12 -4.6598435 1 0
## 13 -4.9904788 1 0
## 14 -5.3947830 1 0
## 15 -4.7696630 1 0
## 16 -5.0036359 1 0
## 17 -4.9550811 1 0
## 18 -4.6774634 1 0
## 19 -4.5735679 1 0
## 20 -4.4478071 1 0
## 21 -4.1687610 0 1
## 22 -3.9392921 0 2
## 23 -3.7811631 0 3
## 24 -3.8514970 0 4
## 25 -2.9444058 0 5
## 26 -2.6515349 0 6
## 27 -2.2006002 0 7
## 28 -1.9499174 0 8
## 29 -1.1949166 0 9
## 30 -0.4164117 0 10
n <- 8;
describe(df[df$Off_Crisis>=-n&df$Off_Crisis<=-1,'balance_of_payment']);
## df[df$Off_Crisis >= -n & df$Off_Crisis <= -1, "balance_of_payment"]
## n missing unique Info Mean
## 8 0 8 1 -3.11
##
## -4.41363415781177 (1, 12%), -3.98056311135777 (1, 12%), -3.64841115885525 (1, 12%), -3.35614082447269 (1, 12%), -3.23246466374394 (1, 12%), -2.57440760140387 (1, 12%), -1.85038498107066 (1, 12%), -1.82289266659616 (1, 12%)
describe(df[df$Off_Crisis==0,'balance_of_payment']);
## df[df$Off_Crisis == 0, "balance_of_payment"]
## n missing unique Info Mean .05 .10 .25 .50 .75 .90 .95
## 10 0 10 1 -4.774 -5.219 -5.043 -4.982 -4.724 -4.595 -4.429 -4.347
##
## -5.39478302143074 (1, 10%), -5.00363594891363 (1, 10%), -4.99047879387293 (1, 10%), -4.95508109661503 (1, 10%), -4.76966304348196 (1, 10%), -4.67746343562751 (1, 10%), -4.65984348113626 (1, 10%), -4.57356788939893 (1, 10%), -4.44780713171369 (1, 10%), -4.26423116226702 (1, 10%)
describe(df[df$Off_Crisis>=1&df$Off_Crisis<=n,'balance_of_payment']);
## df[df$Off_Crisis >= 1 & df$Off_Crisis <= n, "balance_of_payment"]
## n missing unique Info Mean
## 8 0 8 1 -3.186
##
## -4.16876100605885 (1, 12%), -3.93929212154225 (1, 12%), -3.85149697413106 (1, 12%), -3.78116310320806 (1, 12%), -2.94440583734139 (1, 12%), -2.65153490367274 (1, 12%), -2.20060024283928 (1, 12%), -1.949917420894 (1, 12%)
The solution works by first computing the range of indexes during which the crisis was in effect as crisisRange. Then it appends to the data.frame a new column Off_Crisis which captures how many quarters offset from the crisis the row is, using negative numbers for before and positive numbers for after, and assuming each row represents exactly one quarter.
The describe() calls can then be made by subsetting on the Off_Crisis column, getting just the quarters offset from the crisis that you want for each call.
Edit: Whew! That was tough. Pretty sure I got it though:
library('Hmisc');
set.seed(1);
N <- 60;
df <- data.frame(balance_of_payment=rep(-5+2*seq(-1.5,1.5,len=N/2)^2,2)+rnorm(N,0,0.2), Crisis_IMF=c(rep(0,N/6),rep(1,N/6),rep(0,N/3),rep(1,N/6),rep(0,N/6)) );
df;
## balance_of_payment Crisis_IMF
## 1 -0.6252908 0
## 2 -1.0625579 0
## 3 -1.8228927 0
## 4 -1.8503850 0
## 5 -2.5744076 0
## 6 -3.2324647 0
## 7 -3.3561408 0
## 8 -3.6484112 0
## 9 -3.9805631 0
## 10 -4.4136342 0
## 11 -4.2642312 1
## 12 -4.6598435 1
## 13 -4.9904788 1
## 14 -5.3947830 1
## 15 -4.7696630 1
## 16 -5.0036359 1
## 17 -4.9550811 1
## 18 -4.6774634 1
## 19 -4.5735679 1
## 20 -4.4478071 1
## 21 -4.1687610 0
## 22 -3.9392921 0
## 23 -3.7811631 0
## 24 -3.8514970 0
## 25 -2.9444058 0
## 26 -2.6515349 0
## 27 -2.2006002 0
## 28 -1.9499174 0
## 29 -1.1949166 0
## 30 -0.4164117 0
## 31 -0.2282641 0
## 32 -1.1198441 0
## 33 -1.5782326 0
## 34 -2.1802021 0
## 35 -2.9157211 0
## 36 -3.1513699 0
## 37 -3.5324846 0
## 38 -3.8079388 0
## 39 -3.8757143 0
## 40 -4.1999213 0
## 41 -4.5994921 1
## 42 -4.7884845 1
## 43 -4.7268380 1
## 44 -4.8405104 1
## 45 -5.1324004 1
## 46 -5.1361483 1
## 47 -4.8789267 1
## 48 -4.7125241 1
## 49 -4.7602814 1
## 50 -4.3903659 1
## 51 -4.2729353 0
## 52 -4.2181247 0
## 53 -3.7278522 0
## 54 -3.6794993 0
## 55 -2.7817662 0
## 56 -2.2442292 0
## 57 -2.2428854 0
## 58 -1.8645939 0
## 59 -0.9853426 0
## 60 -0.5270109 0
df$Off_Crisis <- ifelse(df$Crisis_IMF==1,0,with(rle(df$Crisis_IMF),{ mids <- lengths[-c(1,length(lengths))]; c(-lengths[1]:-1,sequence(mids)-rep(rbind(0,mids+1),rbind(ceiling(mids/2),floor(mids/2))),1:lengths[length(lengths)]); }));
df;
## balance_of_payment Crisis_IMF Off_Crisis
## 1 -0.6252908 0 -10
## 2 -1.0625579 0 -9
## 3 -1.8228927 0 -8
## 4 -1.8503850 0 -7
## 5 -2.5744076 0 -6
## 6 -3.2324647 0 -5
## 7 -3.3561408 0 -4
## 8 -3.6484112 0 -3
## 9 -3.9805631 0 -2
## 10 -4.4136342 0 -1
## 11 -4.2642312 1 0
## 12 -4.6598435 1 0
## 13 -4.9904788 1 0
## 14 -5.3947830 1 0
## 15 -4.7696630 1 0
## 16 -5.0036359 1 0
## 17 -4.9550811 1 0
## 18 -4.6774634 1 0
## 19 -4.5735679 1 0
## 20 -4.4478071 1 0
## 21 -4.1687610 0 1
## 22 -3.9392921 0 2
## 23 -3.7811631 0 3
## 24 -3.8514970 0 4
## 25 -2.9444058 0 5
## 26 -2.6515349 0 6
## 27 -2.2006002 0 7
## 28 -1.9499174 0 8
## 29 -1.1949166 0 9
## 30 -0.4164117 0 10
## 31 -0.2282641 0 -10
## 32 -1.1198441 0 -9
## 33 -1.5782326 0 -8
## 34 -2.1802021 0 -7
## 35 -2.9157211 0 -6
## 36 -3.1513699 0 -5
## 37 -3.5324846 0 -4
## 38 -3.8079388 0 -3
## 39 -3.8757143 0 -2
## 40 -4.1999213 0 -1
## 41 -4.5994921 1 0
## 42 -4.7884845 1 0
## 43 -4.7268380 1 0
## 44 -4.8405104 1 0
## 45 -5.1324004 1 0
## 46 -5.1361483 1 0
## 47 -4.8789267 1 0
## 48 -4.7125241 1 0
## 49 -4.7602814 1 0
## 50 -4.3903659 1 0
## 51 -4.2729353 0 1
## 52 -4.2181247 0 2
## 53 -3.7278522 0 3
## 54 -3.6794993 0 4
## 55 -2.7817662 0 5
## 56 -2.2442292 0 6
## 57 -2.2428854 0 7
## 58 -1.8645939 0 8
## 59 -0.9853426 0 9
## 60 -0.5270109 0 10
n <- 8;
describe(df[df$Off_Crisis>=-n&df$Off_Crisis<=-1,'balance_of_payment']);
## df[df$Off_Crisis >= -n & df$Off_Crisis <= -1, "balance_of_payment"]
## n missing unique Info Mean .05 .10 .25 .50 .75 .90 .95
## 16 0 16 1 -3.133 -4.253 -4.090 -3.825 -3.294 -2.476 -1.837 -1.762
##
## -4.41363415781177 (1, 6%), -4.19992133068899 (1, 6%), -3.98056311135777 (1, 6%), -3.87571430729169 (1, 6%), -3.80793877922333 (1, 6%), -3.64841115885525 (1, 6%)
## -3.53248462570045 (1, 6%), -3.35614082447269 (1, 6%), -3.23246466374394 (1, 6%), -3.15136989958027 (1, 6%), -2.91572106713267 (1, 6%), -2.57440760140387 (1, 6%)
## -2.1802021496148 (1, 6%), -1.85038498107066 (1, 6%), -1.82289266659616 (1, 6%), -1.57823262180228 (1, 6%)
describe(df[df$Off_Crisis==0,'balance_of_payment']);
## df[df$Off_Crisis == 0, "balance_of_payment"]
## n missing unique Info Mean .05 .10 .25 .50 .75 .90 .95
## 20 0 20 1 -4.785 -5.149 -5.133 -4.964 -4.765 -4.645 -4.442 -4.384
##
## lowest : -5.395 -5.136 -5.132 -5.004 -4.990, highest: -4.599 -4.574 -4.448 -4.390 -4.264
describe(df[df$Off_Crisis>=1&df$Off_Crisis<=n,'balance_of_payment']);
## df[df$Off_Crisis >= 1 & df$Off_Crisis <= n, "balance_of_payment"]
## n missing unique Info Mean .05 .10 .25 .50 .75 .90 .95
## 16 0 16 1 -3.157 -4.232 -4.193 -3.873 -3.312 -2.244 -2.075 -1.929
##
## -4.27293530430708 (1, 6%), -4.21812466033862 (1, 6%), -4.16876100605885 (1, 6%), -3.93929212154225 (1, 6%), -3.85149697413106 (1, 6%), -3.78116310320806 (1, 6%)
## -3.72785216159621 (1, 6%), -3.67949925417454 (1, 6%), -2.94440583734139 (1, 6%), -2.78176624658013 (1, 6%), -2.65153490367274 (1, 6%), -2.24422917606577 (1, 6%)
## -2.24288543679152 (1, 6%), -2.20060024283928 (1, 6%), -1.949917420894 (1, 6%), -1.86459386937746 (1, 6%)
For this demo I synthesized five periods: 10 rows of non-crisis, 10 rows of crisis (the first), 20 rows of non-crisis, 10 rows of crisis (the second), and 10 rows of non-crisis. The algorithm is the same, namely to compute an Off_Crisis column (which was much more difficult this time!) and then use it to subset the data.frame for each describe() call. Only now, data points from different crises will be combined in the subsets.

3D Array value assignment ruins the structure of array

Here is how to reproduce my problem. I want to create a 3D array
> g=array(0,dim=c(3,31,31))
> dim(g)
[1] 3 31 31
> dim(g[1,,])
[1] 31 31
This is x with dimension 31 by 31
> dim(x)
[1] 31 31
> x
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
1 NA 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
2 0 NA 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
3 2 1 NA 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
4 0 0 0 NA 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
5 0 0 0 0 NA 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 NA 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0
7 0 0 0 0 0 1 NA 0 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 NA 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 NA 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
10 0 1 1 0 0 0 1 0 0 NA 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0
11 0 0 0 0 0 0 0 0 2 0 NA 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
12 0 0 0 0 0 0 1 1 0 0 0 NA 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
13 0 0 0 0 0 0 0 0 0 0 0 0 NA 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
14 0 0 0 0 0 0 0 0 0 0 0 0 0 NA 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
15 0 0 1 1 1 0 0 0 0 0 0 0 0 1 NA 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1
16 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 NA 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
17 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 NA 0 0 0 0 0 0 1 0 0 0 0 0 0 0
18 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 NA 0 1 1 0 0 0 0 0 0 0 0 0 0
19 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 NA 0 0 0 0 0 0 0 0 0 0 0 0
20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 NA 0 0 0 0 0 0 0 0 0 0 0
21 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 NA 0 0 0 0 0 0 0 0 0 0
22 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 NA 1 0 0 0 0 0 0 0 0
23 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 NA 0 0 0 0 0 0 0 0
24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 NA 0 0 1 0 0 0 0
25 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 NA 0 0 0 0 0 0
26 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 NA 0 0 0 0 0
27 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 NA 0 1 0 0
28 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 NA 0 0 0
29 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 NA 0 0
30 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 NA 0
31 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 NA
when I try to assign x to the first 'section' of g using
> g[1,,] = x
The array structure of g is totally changed, as now it becomes:
> dim(g)
NULL
> head(g)
[[1]]
[1] NA 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
[[2]]
[1] 0
[[3]]
[1] 0
[[4]]
[1] 0 NA 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
[[5]]
[1] 0
[[6]]
[1] 0
This is totally different from what I expected, I am just trying to put a matrix to g[1,,] and dim(g) should still be 3 by 31 by 31, am I wrong? where did I do wrong?

Thanks to Pascal's comments below I've amended my answer, though I've left the dimensionality changed to 31x31x3 for perhaps easier understanding. The problem is the way that the data are converted from your data.frame object before storing in your array. I think by converting first to a matrix you should get what you're looking for:
g <- array(0,dim=c(31,31,3))
m <- matrix(1, 31, 31)
x <- as.data.frame(m)
## Storing x as it is will result in g becoming a list...
#g[,,1] <- x
## Converting the data.frame to a matrix will result in
## g remaining an array:
g[,,1] <- as.matrix(x)

Finding values that repeat more than twice in a table

I am trying do something that in SQL might be done with "...having count(ID) > 2 ..."
I want to find those values of a col in a two column subset of a dataframe that repeat more than twice. The function table gives me sparse matrix and I am not sure how to maneuver its results into what I want. For a matrix, rowSums will find me totals, but I want the total associated with the identifier.
The dataset mtcars shows more clearly what I mean.
x <- head(table( mtcars$hp, mtcars$disp), 20)
x[,1] <- as.numeric(rownames(x))
x
71.1 75.7 78.7 79 95.1 108 120.1 120.3 121 140.8 145 146.7 160 167.6 225 258 275.8 301 304 318 350 351 360 400 440 460 472
52 52 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
62 62 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
65 65 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
66 66 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
91 91 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
93 93 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
95 95 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
97 97 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
105 105 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
109 109 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
110 110 0 0 0 0 0 0 0 0 0 0 0 2 0 0 1 0 0 0 0 0 0 0 0 0 0 0
113 113 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
123 123 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0
150 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0
175 175 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0
180 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0
205 205 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
215 215 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
230 230 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
245 245 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0
The result I would like for this dataframe of 20 rows would be:
110 3
175 3
180 3

Something like this?
df <- mtcars[ , c("hp", "disp")]
tt <- with(df, table(hp))
data.frame(count = tt[tt > 2])
# count
# 110 3
# 175 3
# 180 3

Plotting relatively large networks in R with igraph

I have two questions, the first is more complicated than the second.
I have a plot of my network below, but the plot is rather clustered how can I spread the nodes out more so I can clearly see the edges. I've tried a circular plot but I still got a rather clustered plot.
How can I rename the nodes? The current plot puts an 'X' in front of the node name, how can I remove the 'X'?
In my matrix an interacting node is marked with 1, non interacting nodes are marked with 0
Code:
library(igraph)
df <- as.matrix(read.table("sig_pairs_matrix_for_r.txt"))
g=graph.adjacency(df,mode="undirected",weighted=NULL,diag=FALSE)
plot.igraph(g,vertex.label=V(g)$name)
Matrix:
4 6 7 8 12 13 15 17 22 23 25 26 27 29 30 31 34 35 36 43 44 47 48 52 53 56 57 59 63 66 67 70 96 99 122 166 168 172 174 176 180 181 191 192 193 220 222 224 225 226 236 249 253 256 258 266 267 277 296
4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 1 1 0 1 1 1 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
8 0 0 0 0 1 1 0 1 1 0 1 1 1 1 1 0 1 1 1 1 0 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
12 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
13 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
17 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
22 0 0 1 1 0 0 0 1 0 0 0 0 1 1 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
23 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
25 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
26 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
27 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
29 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
30 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
31 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
34 0 0 1 1 0 0 0 0 1 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
35 0 0 1 1 0 0 0 0 1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
36 0 0 1 1 0 0 0 0 1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
43 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
44 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
47 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
48 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
52 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
53 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
56 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
57 0 0 1 1 0 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
59 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
63 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
66 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 1 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
67 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 1 0 0 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
70 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
96 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
99 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
122 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
166 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
168 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0
172 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
174 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
176 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
181 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0
191 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
192 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
193 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
220 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0
222 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
224 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
225 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 1 0 0 0 0
226 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 0
236 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
249 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0
253 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 1 1 0 1 0
256 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
258 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0 1 0 1 0
266 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 1 0
267 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
277 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 1 0 0 0
296 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0

For the first question, check ?igraph::layout. There's some instructions how you can alter the aligment of nodes.
For your second question, I guess the graph takes the names of the nodes from the row or column names of your matrix but read.table by default appends names with prefix X in order to make names starting with number valid variable names (see make.names). So rownames of your matrix are actually X1, X2 etc. You can circumvent this by reading your matrix like this:
df <- as.matrix(read.table("sig_pairs_matrix_for_r.txt",check.names=FALSE))
Or you could rename the rows and columns of your matrix before making the graph, i.e.
#newnames is a vector containing the new names, such as c("a","b",...)
rownames(df)<-colnames(df)<-newnames

For the second part of your question I would create a separate files for the row and column names. Once imported you can set the column and row names.
temp = read.csv("adj_matrix.csv", sep=",",head=FALSE)
adjmatrix <- as.matrix(temp)
temprow = read.csv("rownames.csv", sep=",",head=FALSE)
rowns <-as.matrix(temprow)
tempcol = read.csv("colnames.csv", sep=",",head=FALSE)
cols <-as.matrix(tempcol)
rownames(adjmatrix)<- rows
colnames(adjmatrix)<- cols
g1 <- graph.adjacency(adjmatrix, mode=c("directed"), weighted=NULL, diag=TRUE)
plot.igraph(g1)

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replace matrix elements by comparing row and column names in r - r

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Select n rows after specific number

3D Array value assignment ruins the structure of array

Finding values that repeat more than twice in a table

Plotting relatively large networks in R with igraph

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