I have a data frame that looks like this
set.seed(123)
test_data <- data.frame(id = 1:6,
var1 = rbinom(n = 6, size = 1, prob = .5),
var2 = rbinom(n = 6, size = 1, prob = .5),
age = sample(18:30, size = 6, replace = T))
I want to use dplyr
or purrr
to change the values that are equal to 1 in var1
and var2
, to the name of their column and keep the 0's the way they are.
The result would look like this.
id var1 var2 age
1 0 var2 26
2 var1 var2 25
3 0 var2 19
4 var1 0 29
5 var1 var2 21
6 0 0 18
I have tried using dplyr::mutate_at
mutate_at(test_data,
vars(var1, var2),
function(var_x) { ifelse(var_x == 1, colnames(var_x), var_x) })
This returns the following error. So, probably not the best way to go.
Error in evalq(sys.calls(), ) : replacement has length zero In addition: Warning message: In rep(yes, length.out = length(ans)) : 'x' is NULL so the result will be NULL
And I have tried using purrr:map_at
map_at(test_data,
c("var1", "var2"),
function(var_x) { ifelse(var_x == 1, colnames(var_x), var_x) })
And this returns this error.
Error in ans[test & ok] <- rep(yes, length.out = length(ans))[test & ok] : replacement has length zero In addition: Warning message: In rep(yes, length.out = length(ans)) : 'x' is NULL so the result will be NULL
While I'd prefer to work with dplyr
or purrr
, I'm open to suggestions using other approaches.
Here is an idea via tidyverse
. The trick here is to gather
first, replace the values and then spread
library(tidyverse)
test_data %>%
gather(var, val, -c(id, age)) %>%
mutate(val = ifelse(val == 1, var, val)) %>%
spread(var, val)
# id age var1 var2
#1 1 26 0 var2
#2 2 25 var1 var2
#3 3 19 0 var2
#4 4 29 var1 0
#5 5 21 var1 var2
#6 6 18 0 0
This doesn't have to be too messy if you use an intermediate object:
ix <- which(test_data[2:3]==1,arr.ind=TRUE)
test_data[2:3][ix] <- names(test_data[2:3])[ix[,"col"]]
# id var1 var2 age
#1 1 0 var2 26
#2 2 var1 var2 25
#3 3 0 var2 19
#4 4 var1 0 29
#5 5 var1 var2 21
#6 6 0 0 18
This should be relatively quick if you are dealing with big data as there is only one <-
assignment operation to do the replacement. The overhead of making ix
shouldn't be too large.
And a few base R solutions :
# Solution 1
test_data[, 2:3] <- sapply(2:3, function(x) ifelse(test_data[x]==1, names(test_data[x]), 0))
# Solution 2
test_data[, c("var1", "var2")] <- sapply(c("var1", "var2"), function(x) ifelse(test_data[x]==1, x, 0))
# Solution 3
for (i in 2:3) {test_data[,i] <- ifelse(test_data[,i] == 1, colnames(test_data[i]), 0)}
# Solution 4 - probably the most traightforward. Most of the job is vectorised
# works also for other values than 0 and 1
for (i in 2:3) {test_data[test_data[,i]==1,i] <- colnames(test_data[i])}
# etc...
Here is an option using data.table
library(data.table)
dcast(melt(setDT(test_data), id.var = c('id', 'age'))[,
value := as.character(value)
][value == 1, value := as.character(variable)],
id + age ~variable, value.var = "value")
# id age var1 var2
#1: 1 26 0 var2
#2: 2 25 var1 var2
#3: 3 19 0 var2
#4: 4 29 var1 0
#5: 5 21 var1 var2
#6: 6 18 0 0
Or an option suggested by @thelatemail
cols <- c("var1","var2")
test_data[, (cols) := Map(function(x,y) replace(x,x==1,y), .SD, cols), .SDcols=cols]
Or another option is set
from data.table
setDT(test_data)
for(j in seq_along(cols)){
set(test_data, i = NULL, j = cols[j], value = as.character(test_data[[cols[j]]]))
set(test_data, i = which(test_data[[cols[j]]] == 1), j = cols[j], value = cols[j])
}
Or we can use base R
methods
d1 <- `dim<-`(names(test_data)[2:3][col(test_data[, 2:3])], dim(test_data[, 2:3]))
d1[test_data[, 2:3]==0] <- 0
test_data[, 2:3] <- d1
I would use these lines to do that, not sure, being an apprentice as I am, whether they are too clumsy:
test_data[test_data$var1==1,]$var1='var1'
test_data[test_data$var2==1,]$var2='var2'
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.