I have a matrix in R. Each entry i,j is a score and the rownames and colnames are ids.
Instead of the matrix I just want a 3 column matrix that has: i,j,score
Right now I'm using nested for loops. Like:
for(i in rownames(g))
{
print(which(rownames(g)==i))
for(j in colnames(g))
{
cur.vector<-c(cur.ref, i, j, g[rownames(g) %in% i,colnames(g) %in% j])
rbind(new.file,cur.vector)->new.file
}
}
But thats very inefficient I think...I'm sure there's a better way I'm just not good enough with R yet.
Thoughts?
If I understand you correctly, you need to flatten the matrix.
You can use as.vector and rep to add the id columns e.g. :
m = cbind(c(1,2,3),c(4,5,6),c(7,8,9))
row.names(m) = c('R1','R2','R3')
colnames(m) = c('C1','C2','C3')
d <- data.frame(i=rep(row.names(m),ncol(m)),
j=rep(colnames(m),each=nrow(m)),
score=as.vector(m))
Result:
> m
C1 C2 C3
R1 1 4 7
R2 2 5 8
R3 3 6 9
> d
i j score
1 R1 C1 1
2 R2 C1 2
3 R3 C1 3
4 R1 C2 4
5 R2 C2 5
6 R3 C2 6
7 R1 C3 7
8 R2 C3 8
9 R3 C3 9
Please, note that this code converts a matrix into a data.frame, since the row and col names can be string and you can't have a matrix with different column type.
If you are sure that all row and col names are numbers, you can coerced it to a matrix.
If you convert your matrix first to a table (with as.table) then to a data frame (as.data.frame) then it will accomplish what you are asking for. A simple example:
> tmp <- matrix( 1:12, 3 )
> dimnames(tmp) <- list( letters[1:3], LETTERS[4:7] )
> as.data.frame( as.table( tmp ) )
Var1 Var2 Freq
1 a D 1
2 b D 2
3 c D 3
4 a E 4
5 b E 5
6 c E 6
7 a F 7
8 b F 8
9 c F 9
10 a G 10
11 b G 11
12 c G 12
Related
I have two sets of data frames and I am trying to create a function which takes in a data frame and row name as an argument and returns the three highest value on the row (in a descending order) and the name of the column of three highest value.
set.seed(0)
df <- data.frame(A=c(3,2,1,4,5),B=c(1,6,3,8,4),C=c(2,1,4,8,9), D=c(4,1,2,4,6))
row.names(df)<-c("R1","R2","R3","R4","R5")
df2 <- data.frame(E=c(2,5,6,1,4),F=c(2,4,2,5,1),G=c(5,6,2,7,3),H=c(8,2,7,4,1))
row.names(df2)<-c("R6","R7","R8","R9","R10")
print(df)
A B C D
R1 3 1 2 4
R2 2 6 1 1
R3 1 3 4 2
R4 4 8 8 4
R5 5 4 9 6
print(df2)
E F G H
R6 2 2 5 8
R7 5 4 6 2
R8 6 2 2 7
R9 1 5 7 4
R10 4 1 3 1
Here is an example of a result:
Let the function be maxthree. Now
maxthree(df2, "R7")
G E F
6 5 4
Here is what I have done so far:
maxthree <- function(data,row) {
if(!row %in% rownames(data)) {
print("Check value")
} else {
max_col <- which.max(data[row,])
print(max_col)
}
}
This function will now return the maximum value in that row as well as the column name. However, I don't now how to add the second and the third highest values to the function.
maxthree = function(data, row) {
data[row, order(unlist(data[row, ]), decreasing = TRUE)[1:3]]
}
maxthree(df2, "R7")
# G E F
# R7 6 5 4
The result is a 1x3 data frame.
This should work great
maxthree <- function(data,roww){
x <- data[roww,]
x[order(x, decreasing = T)][1:3]
}
> maxthree(df2, "R7")
G E F
R7 6 5 4
Try this:
df <- data.frame(A=c(3,2,1,4,5),B=c(1,6,3,8,4),C=c(2,1,4,8,9), D=c(4,1,2,4,6))
row.names(df)<-c("R1","R2","R3","R4","R5")
df2 <- data.frame(E=c(2,5,6,1,4),F=c(2,4,2,5,1),G=c(5,6,2,7,3),H=c(8,2,7,4,1))
row.names(df2)<-c("R6","R7","R8","R9","R10")
maxthree <- function(data,row) {
named_vec <- t(data)[,row]
return(sort(named_vec, decreasing = T)[1:3])
}
maxthree(df2, "R7")
# G E F
# 6 5 4
This approach transposes your data frame "t()" to allow a straightforward subset of the row as a named vector. This allows sort to be used to order the values as desired.
You can use sort and [1:3] to get the first 3 elements like:
maxthree <- function(data,row) {sort(data[row,], TRUE)[1:3]}
maxthree(df2, "R7")
# G E F
#R7 6 5 4
In case the rowname should not be shown you can add unlist:
maxthree <- function(data,row) {head(unlist(sort(data[row,], TRUE)),3)}
maxthree(df2, "R7")
#G E F
#6 5 4
You can use the order function.
maxthree <- function(data, row_name) data[row_name, order(-data[row_name,])][, 1:3]
maxthree(df2, 'R7')
G E F
R7 6 5 4
I have a matrix. The entries of the matrix are counts for the combination of the dimension levels. For example:
(m0 <- matrix(1:4, nrow=2, dimnames=list(c("A","B"),c("A","B"))))
A B
A 1 3
B 2 4
I can change it to a long format:
library("reshape")
(m1 <- melt(m0))
X1 X2 value
1 A A 1
2 B A 2
3 A B 3
4 B B 4
But I would like to have multipe entries according to value:
m2 <- m1
for (i in 1:nrow(m1)) {
j <- m1[i,"value"]
k <- 2
while ( k <= j) {
m2 <- rbind(m2,m1[i,])
k = k+1
}
}
> m2 <- subset(m2,select = - value)
> m2[order(m2$X1),]
X1 X2
1 A A
3 A B
31 A B
32 A B
2 B A
4 B B
21 B A
41 B B
42 B B
43 B B
Is there a parameter in melt which considers to multiply the entries according to value? Or any other library which can perform this issue?
We could do this with base R. We convert the dimnames of 'm0' to a 'data.frame' with two columns using expand.grid, then replicate the rows of the dataset with the values in 'm0', order the rows and change the row names to NULL (if necessary).
d1 <- expand.grid(dimnames(m0))
d2 <- d1[rep(1:nrow(d1), c(m0)),]
res <- d2[order(d2$Var1),]
row.names(res) <- NULL
res
# Var1 Var2
#1 A A
#2 A B
#3 A B
#4 A B
#5 B A
#6 B A
#7 B B
#8 B B
#9 B B
#10 B B
Or with melt, we convert the 'm0' to 'long' format and then replicate the rows as before.
library(reshape2)
dM <- melt(m0)
dM[rep(1:nrow(dM), dM$value),1:2]
As #Frank mentioned, we can also use table with as.data.frame to create 'dM'
dM <- as.data.frame(as.table(m0))
I have 8 columns of variables which I must keep column 1 to 3. For column 4 to 8 I need to keep those with only 3 levels and drop which does not qualify that condition.
I tried the following command
data3 <- data2[,sapply(data2,function(col)length(unique(col)))==3]
It managed to retain the variables with 3 levels, but deleted my first 3 columns.
You could do a two step process:
data4 <- data2[1:3]
#Your answer for the second part here:
data3 <- data2[,sapply(data2,function(col)length(unique(col)))==3]
merge(data3,data4)
Depending on what you would like your expected output to be, could try with the option all =TRUE inside the merge().
I would suggest another approach:
x = 1:3
cbind(data2[x], Filter(function(i) length(unique(i))==3, data2[-x]))
# 1 2 3 5
#1 a 1 3 b
#2 b 2 4 b
#3 c 3 5 b
#4 d 4 6 a
#5 e 5 7 c
#6 f 6 8 c
#7 g 7 9 c
#8 h 8 10 a
#9 i 9 11 c
#10 j 10 12 b
Data:
data2 = setNames(
data.frame(letters[1:10],
1:10,
3:12,
sample(letters[1:10],10, replace=T),
sample(letters[1:3],10, replace=T)),
1:5)
Assuming that the columns 4:8 are factor class, we can also use nlevels to filter the columns. We create 'toKeep' as the numeric index of columns to keep, and 'toFilter' as numeric index of columns to filter. We subset the dataset into two: 1) using the 'toKeep' as the index (data2[toKeep]), 2) using the 'toFilter', we further subset the dataset by looping with sapply to find the number of levels (nlevels), create logical index (==3) to filter the columns and cbind with the first subset.
toKeep <- 1:3
toFilter <- setdiff(seq_len(ncol(data2)), n)
cbind(data2[toKeep], data2[toFilter][sapply(data2[toFilter], nlevels)==3])
# V1 V2 V3 V4 V6
#1 B B D C B
#2 B D D A B
#3 D E B A B
#4 C B E C A
#5 D D A D E
#6 E B A A B
data
set.seed(24)
data2 <- as.data.frame(matrix(sample(LETTERS[1:5], 8*6, replace=TRUE), ncol=8))
Say I have a data frame which looks like this:
df.A
A B C
x 1 3 4
y 5 4 6
z 8 9 1
And I want to replace the column names in the first based on column values in a second:
df.B
Low High
A D
B F
C G
Such that I get:
df.A
D F G
x 1 3 4
y 5 4 6
z 8 9 1
How would I do it?
I have tried extracting the vector df.B$High from df.B and using this in names(df.A), but everything is in alphabetical order and shifted over one. Furthermore, this only works if the order of columns in df.A is conserved with respect to the elements in df.B$High, which is not always the case (and in my real example there is no numeric or alphabetical way to sort the two to the same order). So I think I need an rbind-type argument for matching elements, but I'm not sure.
Thanks!
You can use rename from plyr:
library(plyr)
dat <- read.table(text = " A B C
x 1 3 4
y 5 4 6
z 8 9 1",header = TRUE,sep = "")
> new <- read.table(text = "Low High
A D
B F
C G",header = TRUE,sep = "")
> rename(dat,replace = setNames(new$High,new$Low))
D F G
x 1 3 4
y 5 4 6
z 8 9 1
using match:
df.A <- read.table(sep=" ", header=T, text="
A B C
x 1 3 4
y 5 4 6
z 8 9 1")
df.B <- read.table(sep=" ", header=T, text="
Low High
A D
B F
C G")
df.C <- df.A
names(df.C) <- df.B$High[match(names(df.A), df.B$Low)]
df.C
# D F G
# x 1 3 4
# y 5 4 6
# z 8 9 1
You can play games with the row names of df.B to make a lookup more convenient:
rownames(df.B) <- df.B$Low
names(df.A) <- df.B[names(df.A),"High"]
df.A
## D F G
## x 1 3 4
## y 5 4 6
## z 8 9 1
Here's an approach abusing factor:
f <- factor(names(df.A), levels=df.B$Low)
levels(f) <- df.B$High
f
## [1] D F G
## Levels: D F G
names(df.A) <- f
## Desired results
I have two dataframe in R.
dataframe 1
A B C D E F G
1 2 a a a a a
2 3 b b b c c
4 1 e e f f e
dataframe 2
X Y Z
1 2 g
2 1 h
3 4 i
1 4 j
I want to match dataframe1's column A and B with dataframe2's column X and Y. It is NOT a pairwise comparsions, i.e. row 1 (A=1 B=2) are considered to be same as row 1 (X=1, Y=2) and row 2 (X=2, Y=1) of dataframe 2.
When matching can be found, I would like to add columns C, D, E, F of dataframe1 back to the matched row of dataframe2, as follows: with no matching as na.
Final dataframe
X Y Z C D E F G
1 2 g a a a a a
2 1 h a a a a a
3 4 i na na na na na
1 4 j e e f f e
I can only know how to do matching for single column, however, how to do matching for two exchangable columns and merging two dataframes based on the matching results is difficult for me. Pls kindly help to offer smart way of doing this.
For the ease of discussion (thanks for the comments by Vincent and DWin (my previous quesiton) that I should test the quote.) There are the quota for loading dataframe 1 and 2 to R.
df1 <- data.frame(A = c(1,2,4), B=c(2,3,1), C=c('a','b','e'),
D=c('a','b','e'), E=c('a','b','f'),
F=c('a','c','f'), G=c('a','c', 'e'))
df2 <- data.frame(X = c(1,2,3,1), Y=c(2,1,4,4), Z=letters[7:10])
The following works, but no doubt can be improved.
I first create a little helper function that performs a row-wise sort on A and B (and renames it to V1 and V2).
replace_index <- function(dat){
x <- as.data.frame(t(sapply(seq_len(nrow(dat)),
function(i)sort(unlist(dat[i, 1:2])))))
names(x) <- paste("V", seq_len(ncol(x)), sep="")
data.frame(x, dat[, -(1:2), drop=FALSE])
}
replace_index(df1)
V1 V2 C D E F G
1 1 2 a a a a a
2 2 3 b b b c c
3 1 4 e e f f e
This means you can use a straight-forward merge to combine the data.
merge(replace_index(df1), replace_index(df2), all.y=TRUE)
V1 V2 C D E F G Z
1 1 2 a a a a a g
2 1 2 a a a a a h
3 1 4 e e f f e j
4 3 4 <NA> <NA> <NA> <NA> <NA> i
This is slightly clunky, and has some potential collision and order issues but works with your example
df1a <- df1; df1a$A <- df1$B; df1a$B <- df1$A #reverse A and B
merge(df2, rbind(df1,df1a), by.x=c("X","Y"), by.y=c("A","B"), all.x=TRUE)
to produce
X Y Z C D E F G
1 1 2 g a a a a a
2 1 4 j e e f f e
3 2 1 h a a a a a
4 3 4 i <NA> <NA> <NA> <NA> <NA>
One approach would be to create an id key for matching that is order invariant.
# create id key to match
require(plyr)
df1 = adply(df1, 1, transform, id = paste(min(A, B), "-", max(A, B)))
df2 = adply(df2, 1, transform, id = paste(min(X, Y), "-", max(X, Y)))
# combine data frames using `match`
cbind(df2, df1[match(df2$id, df1$id),3:7])
This produces the output
X Y Z id C D E F G
1 1 2 g 1 - 2 a a a a a
1.1 2 1 h 1 - 2 a a a a a
NA 3 4 i 3 - 4 <NA> <NA> <NA> <NA> <NA>
3 1 4 j 1 - 4 e e f f e
You could also join the tables both ways (X == A and Y == B, then X == B and Y == A) and rbind them. This will produce duplicate pairs where one way yielded a match and the other yielded NA, so you would then reduce duplicates by slicing only a single row for each X-Y combination, the one without NA if one exists.
library(dplyr)
m <- left_join(df2,df1,by = c("X" = "A","Y" = "B"))
n <- left_join(df2,df1,by = c("Y" = "A","X" = "B"))
rbind(m,n) %>%
group_by(X,Y) %>%
arrange(C,D,E,F,G) %>% # sort to put NA rows on bottom of pairs
slice(1) # take top row from combination
Produces:
Source: local data frame [4 x 8]
Groups: X, Y
X Y Z C D E F G
1 1 2 g a a a a a
2 1 4 j e e f f e
3 2 1 h a a a a a
4 3 4 i NA NA NA NA NA
Here's another possible solution in base R. This solution cbind()s new key columns (K1 and K2) to both data.frames using the vectorized pmin() and pmax() functions to derive the canonical order of the key columns, and merges on those:
merge(cbind(df2,K1=pmin(df2$X,df2$Y),K2=pmax(df2$X,df2$Y)),cbind(df1,K1=pmin(df1$A,df1$B),K2=pmax(df1$A,df1$B)),all.x=T)[,-c(1:2,6:7)];
## X Y Z C D E F G
## 1 1 2 g a a a a a
## 2 2 1 h a a a a a
## 3 1 4 j e e f f e
## 4 3 4 i <NA> <NA> <NA> <NA> <NA>
Note that the use of pmin() and pmax() is only possible for this problem because you only have two key columns; if you had more, then you'd have to use some kind of apply+sort solution to achieve the canonical key order for merging, similar to what #Andrie does in his helper function, which would work for any number of key columns, but would be less performant.