changing the values in many r variables - r

Question

I want to do the equivalent of find and replace 1=0;2=0;3=0;4=1;5=2;6=3 for many different variables in my data set.
Things I've tried:
making 1=0;2=0;3=0;4=1;5=2;6=3 into a function and using sapply. I changed the ; to , and changed the = to <- and no combination of these were recognized as a function. I tried creating a function with that definition and putting it into sapply and it didn't work.
I tried using recode and it did not work:
wdata[ ,cols2] = recode(wdata[ ,cols2], 1=0;2=0;3=0;4=1;5=2;6=3)

Answer 1

Assuming you are working with a data.frame or matrix you can use direct indexing:
# Sample data
set.seed(2017);
df <- as.data.frame(matrix(sample(1:6, 20, replace = T), ncol = 4));
df;
#V1 V2 V3 V4
#1 6 5 5 3
#2 4 1 1 3
#3 3 3 1 5
#4 2 3 3 6
#5 5 2 3 5
df[df == 1 | df == 2 | df == 3] <- 0;
df[df == 4] <- 1;
df[df == 5] <- 2;
df[df == 6] <- 3;
df;
# V1 V2 V3 V4
#1 3 2 2 0
#2 1 0 0 0
#3 0 0 0 2
#4 0 0 0 3
#5 2 0 0 2
Note that the order of the substitutions matters. For example, df[df == 4] = 1; df[df == 1] <- 0; will give a different output from df[df == 1] <- 0; df[df == 4] <- 1;

Answer 2

Alternative solution using recode from dplyr with sapply or mutate_all:
set.seed(2017);
df <- as.data.frame(matrix(sample(1:6, 20, replace = T), ncol = 4));
df
library(dplyr)
f = function(x) recode(x, `1`=0, `2`=0, `3`=0, `4`=1, `5`=2, `6`=3)
sapply(df, f)
# V1 V2 V3 V4
# [1,] 3 2 2 0
# [2,] 1 0 0 0
# [3,] 0 0 0 2
# [4,] 0 0 0 3
# [5,] 2 0 0 2
df %>% mutate_all(f)
# V1 V2 V3 V4
# 1 3 2 2 0
# 2 1 0 0 0
# 3 0 0 0 2
# 4 0 0 0 3
# 5 2 0 0 2

Answer 3

A looping alternative with lapply and match is as follows:
dat[] <- lapply(dat, function(x) c(0, 0, 0, 1, 2, 3)[match(x, 1:6)])
This uses a lookup table on the vector c(0,0,0,1,2,3) with match selecting the indices. Using the data.frame created by Maurits Evers, we get
dat
V1 V2 V3 V4
1 3 2 2 0
2 1 0 0 0
3 0 0 0 2
4 0 0 0 3
5 2 0 0 2
To do this for a subset of the columns, just select them on each side, like
dat[, cols2] <-
lapply(dat[, cols2], function(x) c(0, 0, 0, 1, 2, 3)[match(x, 1:6)])
or
dat[cols2] <- lapply(dat[cols2], function(x) c(0, 0, 0, 1, 2, 3)[match(x, 1:6)])