I have a table
rawData <- as.data.frame(matrix(c(1,2,3,4,5,6,"a,b,c","d,e","f"),nrow=3,ncol=3))
1 4 a,b,c
2 5 d,e
3 6 f
I would like to convert to
1 2 3
4 5 6
a d f
b e
c
so far I can transpose and split the third column, however, I'm lost as to how to reconstruct a new table with the format outline above?
new = t(rawData)
for (e in 1:ncol(new)){
s<-strsplit(new[3:3,e], split=",")
print(s)
}
I tried creating new vectors for each iteration but I'm not sure how to efficiently put each one back into a dataframe. Would be grateful for any help. thanks!
You can use stri_list2matrix from the stringi package:
library(stringi)
rawData <- as.data.frame(matrix(c(1,2,3,4,5,6,"a,b,c","d,e","f"),nrow=3,ncol=3),stringsAsFactors = F)
d1 <- t(rawData[,1:2])
rownames(d1) <- NULL
d2 <- stri_list2matrix(strsplit(rawData$V3,split=','))
rbind(d1,d2)
# [,1] [,2] [,3]
# [1,] "1" "2" "3"
# [2,] "4" "5" "6"
# [3,] "a" "d" "f"
# [4,] "b" "e" NA
# [5,] "c" NA NA
You can also use cSplit from my "splitstackshape" package.
By default, it just creates additional columns after splitting the input:
library(splitstackshape)
cSplit(rawData, "V3")
# V1 V2 V3_1 V3_2 V3_3
# 1: 1 4 a b c
# 2: 2 5 d e NA
# 3: 3 6 f NA NA
You can just transpose that to get your desired output.
t(cSplit(rawData, "V3"))
# [,1] [,2] [,3]
# V1 "1" "2" "3"
# V2 "4" "5" "6"
# V3_1 "a" "d" "f"
# V3_2 "b" "e" NA
# V3_3 "c" NA NA
Related
I have a data set like that
Reproducible data
test <- list(c("1"),c("2"),c("3"),c(c("a"),c("b")),c("d"))
Desired output is:
1
2
3
a,b
d
I have tried
output <- do.call(rbind, test)
You can use toString to collapse every element in the list.
out <- data.frame(result = sapply(test, toString))
out
# result
#1 1
#2 2
#3 3
#4 a, b
#5 d
We can use tidyverse methods
Loop over the list with map
Paste the elements using str_c
Return a single column dataset by adding the suffix _dfr in map
library(dplyr)
library(purrr)
library(stringr)
map_dfr(test, ~ tibble(col1 = str_c(.x, collapse=",")))
# A tibble: 5 x 1
col1
<chr>
1 1
2 2
3 3
4 a,b
5 d
Actually you are already close to your goal. Maybe the code below can help you
> do.call(rbind, Map(toString, test))
[,1]
[1,] "1"
[2,] "2"
[3,] "3"
[4,] "a, b"
[5,] "d"
Maybe this could help too beside all the very good answers you received:
test |>
lapply(\(x) paste(x, collapse = ",")) |>
cbind()
[,1]
[1,] "1"
[2,] "2"
[3,] "3"
[4,] "a,b"
[5,] "d"
I am trying to create an edgelist out of patent data of the form:
PatentID InventorIDs CoinventorIDs
1 A ; B C,D,E ; F,G,H,C
2 J ; K ; L M,O ; N ; P, Q
What I would like is the edgelist below showing the connections between inventors and patents. (the semicolons separate the coinventors associated with each primary inventor):
1 A B
1 A C
1 A D
1 A E
1 B F
1 B G
1 B H
1 B C
2 J K
2 J L
2 J M
2 J O
2 K N
2 L P
2 L Q
Is there an easy way to do this with igraph in R?
I'm confused by the edges going between the inventorIds. But, here is a kind of brute force function that you could just apply by row. There may be a way with igraph, it being a massive library, that is better, but once you have the data in an this form it should be simple to convert to an igraph data structure.
Note that this leaves out the edges between primary inventors.
## A function to make the edges for each row
rowFunc <- function(row) {
tmp <- lapply(row[2:3], strsplit, '\\s*;\\s*')
tmp2 <- lapply(tmp[[2]], strsplit, ',')
do.call(rbind, mapply(cbind, row[[1]], unlist(tmp[[1]]), unlist(tmp2, recursive=FALSE)))
}
## Apply the function by row
do.call(rbind, apply(dat, 1, rowFunc))
# [,1] [,2] [,3]
# [1,] "1" "A" "C"
# [2,] "1" "A" "D"
# [3,] "1" "A" "E"
# [4,] "1" "B" "F"
# [5,] "1" "B" "G"
# [6,] "1" "B" "H"
# [7,] "1" "B" "C"
# [8,] "2" "J" "M"
# [9,] "2" "J" "O"
# [10,] "2" "K" "N"
# [11,] "2" "L" "P"
# [12,] "2" "L" " Q"
I am trying to populate a field in a table (or create a separate vector altogether, whichever is easier) with consecutive numbers from 1 to n, where n is the total number of records that share the same factor level, and then back to 1 for the next level, etc. That is, for a table like this
data<-matrix(c(rep('A',4),rep('B',3),rep('C',4),rep('D',2)),ncol=1)
the result should be a new column (e.g. "sample") as follows:
sample<-c(1,2,3,4,1,2,3,1,2,3,4,1,2)
You can get it as follows, using ave:
data <- data.frame(data)
new <- ave(rep(1,nrow(data)),data$data,FUN=cumsum)
all.equal(new,sample) # check if it's right.
You can use rle function together with lapply :
sample <- unlist(lapply(rle(data[,1])$lengths,FUN=function(x){1:x}))
data <- cbind(data,sample)
Or even better, you can combine rle and sequence in the following one-liner (thanks to #Arun suggestion)
data <- cbind(data,sequence(rle(data[,1])$lengths))
> data
[,1] [,2]
[1,] "A" "1"
[2,] "A" "2"
[3,] "A" "3"
[4,] "A" "4"
[5,] "B" "1"
[6,] "B" "2"
[7,] "B" "3"
[8,] "C" "1"
[9,] "C" "2"
[10,] "C" "3"
[11,] "C" "4"
[12,] "D" "1"
[13,] "D" "2"
There are lots of different ways of achieving this, but I prefer to use ddply() from plyr because the logic seems very consistent to me. I think it makes more sense to be working with a data.frame (your title talks about levels of a factor):
dat <- data.frame(ID = c(rep('A',4),rep('B',3),rep('C',4),rep('D',2)))
library(plyr)
ddply(dat, .(ID), summarise, sample = 1:length(ID))
# ID sample
# 1 A 1
# 2 A 2
# 3 A 3
# 4 A 4
# 5 B 1
# 6 B 2
# 7 B 3
# 8 C 1
# 9 C 2
# 10 C 3
# 11 C 4
# 12 D 1
# 13 D 2
My answer:
sample <- unlist(lapply(levels(factor(data)), function(x)seq_len(sum(factor(data)==x))))
factors <- unique(data)
f1 <- length(which(data == factors[1]))
...
fn <- length(which(data == factors[length(factors)]))
You can use a for loop or 'apply' family to speed that part up.
Then,
sample <- c(1:f1, 1:f2, ..., 1:fn)
Once again you can use a for loop for that part. Here is the full script you can use:
data<-matrix(c(rep('A',4),rep('B',3),rep('C',4),rep('D',2)),ncol=1)
factors <- unique(data)
f <- c()
for(i in 1:length(factors)) {
f[i] <- length(which(data == factors[i]))
}
sample <- c()
for(i in 1:length(f)) {
sample <- c(sample, 1:f[i])
}
> sample
[1] 1 2 3 4 1 2 3 1 2 3 4 1 2
I have a dataframe with one column that I would like to split into several columns, but the number of splits is dynamic throughout the rows.
Var1
====
A/B
A/B/C
C/B
A/C/D/E
I have tried using colsplit(df$Var1,split="/",names=c("Var1","Var2","Var3","Var4")), but rows with less than 4 variables will repeat.
From Hansi, the desired output would be:
Var1 Var2 Var3 Var4
[1,] "A" "B" NA NA
[2,] "A" "B" "C" NA
[3,] "C" "B" NA NA
[4,] "A" "C" "D" "E"
> read.table(text=as.character(df$Var1), sep="/", fill=TRUE)
V1 V2 V3 V4
1 A B
2 A B C
3 C B
4 A C D E
Leading zeros in digit only fields can be preserved with colClasses="character"
a <- data.frame(Var1=c("01/B","04/B/C","0098/B","8708/C/D/E"))
read.table(text=as.character(a$Var1), sep="/", fill=TRUE, colClasses="character")
V1 V2 V3 V4
1 01 B
2 04 B C
3 0098 B
4 8708 C D E
If I understood your objective correctly here is one possible solution, I'm sure there is a better way of doing it but this was the first that came to mind:
a <- data.frame(Var1=c("A/B","A/B/C","C/B","A/C/D/E"))
splitNames <- c("Var1","Var2","Var3","Var4")
# R> a
# Var1
# 1 A/B
# 2 A/B/C
# 3 C/B
# 4 A/C/D/E
b <- t(apply(a,1,function(x){
temp <- unlist(strsplit(x,"/"));
return(c(temp,rep(NA,max(0,length(splitNames)-length(temp)))))
}))
colnames(b) <- splitNames
# R> b
# Var1 Var2 Var3 Var4
# [1,] "A" "B" NA NA
# [2,] "A" "B" "C" NA
# [3,] "C" "B" NA NA
# [4,] "A" "C" "D" "E"
i do not know a function to solve your problem, but you can achieve it easily with standard R commands :
# Here are your data
df <- data.frame(Var1=c("A/B", "A/B/C", "C/B", "A/C/D/E"), stringsAsFactors=FALSE)
# Split
rows <- strsplit(df$Var1, split="/")
# Maximum amount of columns
columnCount <- max(sapply(rows, length))
# Fill with NA
rows <- lapply(rows, `length<-`, columnCount)
# Coerce to data.frame
out <- as.data.frame(rows)
# Transpose
out <- t(out)
As it relies on strsplit, you may need to make some type conversion. See type.con
I have an R data frame that looks like this:
z = as.data.frame(list(Col1=c("a","c","e","g"),Col2=c("b","d","f","h"),Col3=c("1,2,5","3,5,7","9,8","1")))
> z
Col1 Col2 Col3
1 a b 1,2,5
2 c d 3,5,7
3 e f 9,8
4 g h 1
(The third column is a text column with comma-separated values.) I would like to convert it to a data frame like this:
a b 1
a b 2
a b 5
c d 3
c d 5
c d 7
e f 9
e f 8
g h 1
Can anyone suggest a way to accomplish this using apply? I'm close using the command below but it's not quite right. Any suggestions on more efficient ways to do this would be appreciated as well...
> apply(z,1,function(a){ids=strsplit(as.character(a[3]),",")[[1]];out<-c();for(id in ids){out<-rbind(out,c(a[1:2],id))};return(out)})
[[1]]
Col1 Col2
[1,] "a" "b" "1"
[2,] "a" "b" "2"
[3,] "a" "b" "5"
[[2]]
Col1 Col2
[1,] "c" "d" "3"
[2,] "c" "d" "5"
[3,] "c" "d" "7"
[[3]]
Col1 Col2
[1,] "e" "f" "9"
[2,] "e" "f" "8"
[[4]]
Col1 Col2
[1,] "g" "h" "1"
You can use ddply.
library(plyr)
ddply(z, c("Col1", "Col2"), summarize,
Col3=strsplit(as.character(Col3),",")[[1]]
)
With reshapeor reshape2
require(reshape2)
merge(cbind(z[,-3], L1=rownames(z)), melt(strsplit(as.character(z$Col3),",")))
gives
L1 Col1 Col2 value
1 1 a b 1
2 1 a b 2
3 1 a b 5
4 2 c d 3
5 2 c d 5
6 2 c d 7
7 3 e f 9
8 3 e f 8
9 4 g h 1