Iterating through data frame and changing values on condition [R]

Question

Had to make an account because this sequence of for loops has been annoying me for quite some time.

I have a data frame in R with 1000 rows and 10 columns, with each value ranging from 1:3. I would like to re-code EVERY entry so that: 1==3, 2==2, 3==1. I understand that there are easier ways to do this, such as sub-setting each column and hard coding the condition, but this isn't always ideal as many of the data sets that I work with have up to 100 columns.

I would like to use a nested loop in order to accomplish this task -- this is what I have thus far:

for(i in 1:nrow(dat_trans)){
  for(j in length(dat_trans)){
    if(dat_trans[i,j] == 1){
      dat_trans[i,j] <- 3
    } else if(dat_trans[i,j] == 2){
      dat_trans[i,j] <- 2
    } else{
      dat_trans[i,j] <- 1
    }
  }
}

So I iterate through the first column, grab every value and change it based on the if/else's condition, I am still learning R so if you have any pointers in my code, feel free to point it out.

edit: code

Answer 1

R is a vectorized language, so you really don't need the inner loop.
Also if you notice that 4-"old value" = "new value", you can eliminate the if statements.

for(i in 1:ncol(dat_trans)){
        dat_trans[,i] <- 4-dat_trans[,i]
}

The outer loop is now iterating across the columns for only 10 iterations as opposed to 1000 for all of rows. This will greatly improve performance.

Answer 2

This type of operation is a swap operation. The ways to swap values without for loops are numerous.

To set up a simple dataframe:

df <- data.frame(
  col1 = c(1,2,3),
  col2 = c(2,3,1),
  col3 = c(3,1,2)
)

Using a dummy value:

df[df==1] <- 4
df[df==3] <- 1
df[df==4] <- 3

Using a temporary variable:

dftemp <- df
df[dftemp==1] <- 3
df[dftemp==3] <- 1

Using multiplication/division and addition/subtraction:

df <- 4 - df

Using Boolean operations:

df <- (df==1) * 3 + (df==2) * 2 + (df==3) * 1

Using a bitwise xor (in case you really have a need for speed):

df[df!=2] <- sapply(df, function(x){bitwXor(2,x)})[df!=2]

If a nested for loop is required the switch function is a good option.

for(i in seq(ncol(df))){
  for(j in seq(nrow(df))){
    df[j,i] <- switch(df[j,i],3,2,1)
  }
}

Text can be used if the values are not as nicely indexed as 1, 2, and 3.

for(i in seq(ncol(df))){
  for(j in seq(nrow(df))){
    df[j,i] <- switch(as.character(df[j,i]),
                      "1" = 3,
                      "2" = 2,
                      "3" = 1)
  }
}

Answer 3

This sounds like a merge / join operation.

set.seed(42)
dat_trans <- as.data.frame(
  setNames(lapply(1:3, function(ign) sample(1:3, size=10, replace=TRUE)),
           c("V1", "V2", "V3"))
)
dat_trans
#    V1 V2 V3
# 1   3  2  3
# 2   3  3  1
# 3   1  3  3
# 4   3  1  3
# 5   2  2  1
# 6   2  3  2
# 7   3  3  2
# 8   1  1  3
# 9   2  2  2
# 10  3  2  3

newvals <- data.frame(old = c(1, 3), new = c(3, 1))
newvals
#   old new
# 1   1   3
# 2   3   1

Using dplyr and tidyr :

library(dplyr)
library(tidyr) # gather, spread
dat_trans %>%
  mutate(rn = row_number()) %>%
  gather(k, v, -rn) %>%
  left_join(newvals, by = c("v" = "old")) %>%
  mutate(v = if_else(is.na(new), v, new)) %>%
  select(-new) %>%
  spread(k, v) %>%
  select(-rn)
#    V1 V2 V3
# 1   1  2  1
# 2   1  1  3
# 3   3  1  1
# 4   1  3  1
# 5   2  2  3
# 6   2  1  2
# 7   1  1  2
# 8   3  3  1
# 9   2  2  2
# 10  1  2  1

(The need for rn is likely due to my use of an older version of tidyr : I'm at 0.8.2, though 1.0.0 has recently been released. That release did a lot of enhancement/work on spread / gather and introduced the pivot_* functions which are likely much smoother at this. If you have a more recent version, try this without the rn portions.)

---

Or a much-more-direct approach using a "recode" mindset:

dat_trans[,c("V1", "V2", "V3")] <- lapply(dat_trans[,c("V1", "V2", "V3")], car::recode, "1=3; 3=1")
# or
dat_trans[,c("V1", "V2", "V3")] <- lapply(dat_trans[,c("V1", "V2", "V3")], dplyr::recode, '1' = 3L, '3' = 1L)

Answer 4

You could use an assignment matrix am . match() each value of an attribute of df1 with column 1 of am but select column 2, then assign it to df1 . In a lapply() of course.

df1
#   V1 V2 V3
# 1  1  2  1
# 2  1  2  1
# 3  1  1  2
# 4  1  3  2
# 5  2  3  2

am <- matrix(c(1, 2, 3, 3, 2, 1), 3)
am
#      [,1] [,2]
# [1,]    1    3
# [2,]    2    2
# [3,]    3    1

df1[] <- lapply(df1, function(x) am[match(x, am[,1]), 2])
df1
#   V1 V2 V3
# 1  3  2  3
# 2  3  2  3
# 3  3  3  2
# 4  3  1  2
# 5  2  1  2

Data

df1 <- structure(list(V1 = c(1L, 1L, 1L, 1L, 2L), V2 = c(2L, 2L, 1L, 
3L, 3L), V3 = c(1L, 1L, 2L, 2L, 2L)), class = "data.frame", row.names = c(NA, 
-5L))