Replacing values in a R dataframe given certain conditions-CodePudding

I'm trying to loop through columns 10 - 12 of the following data set df2, replacing 1 with a.2. with b. 3 with c and 4 with d

I've figured out how to perform the substitution on a single column using the code below.

df2[df2$DX == 1,]$DX <-'a'
df2[df2$DX == 2,]$DX <-'b'
df2[df2$DX == 3,]$DX <- 'c'
df2[df2$DX == 4,]$DX <- 'd'
df2$DX <- as.factor(df2$DX)

I'm not sure how to set up the for loop to loop over the other two columns. Thank you for your help.

CodePudding user response：

library(dplyr)  
df2 <- data.frame(replicate(30,sample(0:10,1000,rep=TRUE)))

df3 <- df2 %>% mutate(across(10:12, ~ case_when(
  .x == 1 ~ "a",
  .x == 2 ~ "b",
  .x == 3 ~ "c",
  .x == 4 ~ "d",
  T ~ as.character(.x)
)))


> head(df2)
  X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13
1  7  3  0  6  2  0  3  4  3   1   6   1   1
2  0  3  7  2  7  2  1  2  6   0   3   4  10
3  4  3  1  8  0  7  6  3  5   5   9   8   5
4  8  0  7  4  0 10  8  3  8   3   9   7   4
5  7  4  9  3  2  3  0  8  9   1   2   3   4
6  8  1  2  2  7  6  3  9  4   1   2  10  10
  X14 X15 X16 X17 X18 X19 X20 X21 X22 X23 X24
1   2   3   4   6   9   1   4   7   1   6   0
2   9   0   1   6   9   8   7   2   5   2   5
3   9   3   2   8   8   6  10   5   0   2   2
4   3   5   8   7   9   0   9   6   6   2   6
5   2   4   6   7   0   5   4   6   9   0   4
6   3   0   4  10   8   3  10   8   0   3   1
  X25 X26 X27 X28 X29 X30
1   8   5   3   5   3   2
2  10   1   2   5   8   2
3   3   2  10   7   9   5
4   7   6   0   7   5  10
5   6   7   7   5   2  10
6   9   8   9   6   6   5

> head(df3)
  X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13
1  7  3  0  6  2  0  3  4  3   a   6   a   1
2  0  3  7  2  7  2  1  2  6   0   c   d  10
3  4  3  1  8  0  7  6  3  5   5   9   8   5
4  8  0  7  4  0 10  8  3  8   c   9   7   4
5  7  4  9  3  2  3  0  8  9   a   b   c   4
6  8  1  2  2  7  6  3  9  4   a   b  10  10
  X14 X15 X16 X17 X18 X19 X20 X21 X22 X23 X24
1   2   3   4   6   9   1   4   7   1   6   0
2   9   0   1   6   9   8   7   2   5   2   5
3   9   3   2   8   8   6  10   5   0   2   2
4   3   5   8   7   9   0   9   6   6   2   6
5   2   4   6   7   0   5   4   6   9   0   4
6   3   0   4  10   8   3  10   8   0   3   1
  X25 X26 X27 X28 X29 X30
1   8   5   3   5   3   2
2  10   1   2   5   8   2
3   3   2  10   7   9   5
4   7   6   0   7   5  10
5   6   7   7   5   2  10
6   9   8   9   6   6   5

CodePudding user response：

library(dplyr)
mutate(df2, across(c(col10, col11, col12), ~factor(case_when(
    . == 1 ~ "a",
    . == 2 ~ "b",
    . == 3 ~ "c",
    . == 4 ~ "d",
    TRUE ~ as.character(.),
))))

CodePudding user response：

Here's a base R solution:

df2[10:12] <- lapply( df2[10:12], function( x) { ifelse(x %in% 1:4, letters[x], x) })

For each column run ifelse on the column vector of numerics and return "a":"d" ( = letters[1:4] )when x is in the range 1:4, else return the original value. Necessarily the values outside 1:4 will still get converted to character values rather than staying numeric class..