3D Array value assignment ruins the structure of array

3D Array value assignment ruins the structure of array - r

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)

Related

R: Simulating ERGM model in R then generate adjacency matrix of that model

I use library(ergm) and library(igraph) and generate a ERGM network. But I want the adjacency matrix of that network. I am unable to find any function which can produce that.
library(ergm)
library(igraph)
g.use <- network(16,density=0.1,directed=FALSE)
#
# Starting from this network let's draw 3 realizations
# of a edges and 2-star network
#
g.sim <- simulate(~edges+kstar(2), nsim=3, coef=c(-1.8,0.03),
basis=g.use, control=control.simulate(
MCMC.burnin=1000,
MCMC.interval=100))
#g.sim[[3]]
summary(g.sim)
Is it possible to find the adjacency matrix from g.sim? and how?

EGRM package uses the network package and not the igraph package. You should maintain everythig in network and not load igraph as the two have some conflicting functions with same names.
In your case, you simulate 3 graphs thus you should have 3 adjacency matrices. The code is as below:
library(ergm)
g.use <- network(16,density=0.1,directed=FALSE)
g.sim <- simulate(~edges+kstar(2), nsim=3, coef=c(-1.8,0.03),
basis=g.use, control=control.simulate(
MCMC.burnin=1000,
MCMC.interval=100))
The code you want:
lapply(g.sim, as.matrix)
[[1]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0
3 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1
4 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0
5 1 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0
6 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0
7 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1
8 0 1 0 0 0 0 0 0 0 1 1 1 1 0 1 0
9 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1
10 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0
11 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0
12 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
13 0 0 1 0 1 0 0 1 0 1 1 0 0 0 0 1
14 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
16 0 0 1 0 0 0 1 0 1 0 0 0 1 0 0 0
[[2]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0
2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
3 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0
4 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
6 0 0 0 0 0 0 0 1 1 0 1 1 1 0 0 1
7 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
8 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0
9 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0
11 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0
12 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1
13 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0
14 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
15 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0
16 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0
[[3]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1
2 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0
3 0 0 0 0 1 0 0 0 0 1 1 0 0 0 1 0
4 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
5 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
6 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 0
7 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0
8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
10 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1
11 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1
12 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0
13 1 1 0 1 0 1 0 0 0 0 0 1 0 0 0 0
14 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0
15 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 1
16 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0

R multiple for loop

I have this loop over the file msp.chr1
for(i in names(msp.chr1[c(7:70)])){
tmp <- rle(msp.chr1[[i]])$lengths
msp.chr1$idx <- rep(1:length(tmp),tmp)
tmp2 <- unlist(by(msp.chr1[msp.chr1[[i]]==1,], list(msp.chr1$idx[msp.chr1[[i]]==1]),function(x){tail(x["epos"],1)-head(x["spos"],1)}))
assign(paste(i, ".chr1", sep=""), as.vector(tmp2))
rm(i); rm(tmp); rm(tmp2)
}
This file is a dataframe with multiple columns:
head(msp.chr1)
chm spos epos sgpos egpos nsnps PDAC1.0 PDAC1.1 PDAC10.0 PDAC10.1 PDAC100.0 PDAC100.1 PDAC101.0 PDAC101.1 PDAC102.0 PDAC102.1 PDAC103.0 PDAC103.1
1 1 123492 134160 0.12 0.13 252 0 0 0 0 1 0 0 0 0 0 0 0
2 1 134160 135025 0.13 0.14 20 0 0 0 0 1 0 0 0 0 0 0 0
3 1 135025 145600 0.14 0.15 150 0 0 0 0 1 0 0 0 0 0 0 0
4 1 145600 316603 0.15 0.32 195 0 1 0 0 1 0 0 1 0 0 0 1
5 1 316603 520140 0.32 0.52 765 0 0 0 0 0 0 0 0 0 0 0 0
6 1 520140 667054 0.52 0.67 1080 0 0 0 0 0 0 0 0 0 0 0 0
PDAC104.0 PDAC104.1 PDAC105.0 PDAC105.1 PDAC11.0 PDAC11.1 PDAC12.0 PDAC12.1 PDAC13.0 PDAC13.1 PDAC14.0 PDAC14.1 PDAC15.0 PDAC15.1 PDAC17.0 PDAC17.1
1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1
2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1
3 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1
4 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
PDAC18.0 PDAC18.1 PDAC19.0 PDAC19.1 PDAC2.0 PDAC2.1 PDAC20.0 PDAC20.1 PDAC21.0 PDAC21.1 PDAC22.0 PDAC22.1 PDAC23.0 PDAC23.1 PDAC24.0 PDAC24.1 PDAC25.0
1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0
2 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0
3 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0
4 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
5 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
PDAC25.1 PDAC3.0 PDAC3.1 PDAC4.0 PDAC4.1 PDAC5.0 PDAC5.1 PDAC6.0 PDAC6.1 PDAC7.0 PDAC7.1 PDAC8.0 PDAC8.1 PDAC807.0 PDAC807.1 PDAC810.0 PDAC810.1
1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
PDAC9.0 PDAC9.1 idx
1 0 0 1
2 0 0 1
3 0 0 1
4 0 0 1
5 1 0 1
6 1 0 1
for(i in names(msp.chr1[c(7:70)])){
tmp <- rle(msp.chr1[[i]])$lengths
msp.chr1$idx <- rep(1:length(tmp),tmp)
tmp2 <- unlist(by(msp.chr1[msp.chr1[[i]]==1,], list(msp.chr1$idx[msp.chr1[[i]]==1]),function(x){tail(x["epos"],1)-head(x["spos"],1)}))
assign(paste(i, ".chr1", sep=""), as.vector(tmp2))
rm(i); rm(tmp); rm(tmp2)
}
But I actually have 23 files, of names msp.chr1, msp.chr2, ..., msp.chr23.
I want to add another loop on the above, to do that on all files at once.
I tried several things but it is not working...
Basically, every chr1 in my loop (including in the assign) should be replaced by chr1 to chr23.
Can you help?
Thanks,

You can generate the name of the file with paste, and then get the file by its name with get. A better option would be to create these files within a list, then you'd only use the j like df=list[[j]].
for(j in 1:23){
df = get(paste("msp.chr",j,sep=""))
for(i in names(df[c(7:70)])){
tmp <- rle(df[[i]])$lengths
df$idx <- rep(1:length(tmp),tmp)
tmp2 <- unlist(by(df[df[[i]]==1,], list(df$idx[df[[i]]==1]),function(x){tail(x["epos"],1)-head(x["spos"],1)}))
assign(paste(i, ".chr1", sep=""), as.vector(tmp2))
rm(i); rm(tmp); rm(tmp2)
}
}

Error with foodweb package

I am using foodweb package to analyse a bunch of food webs. I keep getting the following error when analyzing a particular food web.
Error in [<-.data.frame(tmp, by.level == 0, value = c(NA_real_, :
'value' is the wrong length
This is the adyacency matrix for the problematic food web:
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 0 0 0 0 0 0 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 1 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0
1 0 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 1 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 0 0 0 0 0 0
0 0 1 0 1 0 0 1 0 0 0 1 0 1 0 0 1 1 0 0 1 0 0 1 1 0 0 0 0 0 1 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
1 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 0 1 0 1 0 0 1
1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 0 1 1 1 0 0 0 0 1 1 1 0 0 0 0 1 1 0
1 0 1 1 0 0 1 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 1 1 0
0 0 0 0 0 0 1 0 0 0 0 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 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 1 0 0 0 0 0 0 0 0 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 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
1 0 1 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 1 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 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 1 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0
1 0 1 1 1 0 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 1 1 0 1 0 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 0 0
1 0 1 1 0 0 0 0 1 0 0 0 1 1 1 0 1 1 1 0 0 0 0 1 1 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 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 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 1 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0
1 1 0 1 0 0 1 0 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0
0 0 1 0 1 1 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0
1 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 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 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 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0
There is something wrong with the length, but I can´t figure it out. Any help with this? Thanks in advance!

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
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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.

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
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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
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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
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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
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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
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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
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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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