Looping same function over data frames in R

Question

I'm new to R. I want to recode the same variable through multiple data frames. But I keep getting errors. Please see the following example of 3 small data frames. I want to create a new variable in each of the 3 dfs called Q2_nom that is recoded as "1" or "0" if Q2 is greater or less than the mean(Q2). Please see my code below.

df1:

Q1 <- c('ABC','DEF','GHI', 'DEF','JKL','XYZ')
Q2 <- c(21000, 23400, 26800, 26000, 20400, 30800)
df1 <- data.frame(Q1, Q2)

df2:

Q1 <- c('DEF','JKL','XYZ', 'ABC', 'MNO', 'PQR')
Q2 <- c(30100, 20200, 15800, 21000, 23400, 26800)
df2 <- data.frame(Q1, Q2)

df3:

Q1 <- c('ABC','DEF','GHI', 'XYZ', 'MNO', 'PQR')
Q2 <- c(17800, 23060, 13080, 27000, 22400, 26500)
df3 <- data.frame(Q1, Q2)

a <- c("Q1", "Q2", "Q3")

for (i in a) {
  newname <- paste(i)
  newname$Q2_mean_nom <- ifelse(newmame$Q2 > mean(newname$Q2, na.rm = TRUE), "1", "0")
}

I noticed that in doing the above, newname is not a df and so the mean won't run. Is there a way to make the loop recognise newname as a df?

I tried using a list but it didn't work either.

newlist <- c(df1, df2, df3)

for (i in 1:length(newlist)) {
  newlist[[i]]$Q2_mean_nom <- ifelse(newlist[[i]]$Q2 > mean(newlist[[i]]$Q2, na.rm = TRUE),
"1", "0")
 }

Please help. Thank you very much!

Answer 1

Using your nice base R approach with slight modification it works:

# list instead of vector:
newlist <- list(df1, df2, df3)
# someone told me once `seq_along`is more stable
for (i in seq_along(newlist)) {
  newlist[[i]]$Q2_mean_nom <- ifelse(newlist[[i]]$Q2 > mean(newlist[[i]]$Q2, na.rm = TRUE), "1", "0")
}

EDIT (see comments):

# df1:
Q1 <- c('ABC','DEF','GHI', 'DEF','JKL','XYZ')
Q2 <- c(21000, 23400, 26800, 26000, 20400, 30800)
df1 <- data.frame(Q1, Q2)
# df2:
Q1 <- c('DEF','JKL','XYZ', 'ABC', 'MNO', 'PQR')
Q2 <- c(30100, 20200, 15800, 21000, 23400, 26800)
df2 <- data.frame(Q1, Q2)
# df3:
Q1 <- c('ABC','DEF','GHI', 'XYZ', 'MNO', 'PQR')
Q2 <- c(17800, 23060, 13080, 27000, 22400, 26500)
df3 <- data.frame(Q1, Q2)

# list instead of vector:
newlist <- list(df1, df2, df3)
# someone told me once `seq_along`is more stable
for (i in seq_along(newlist)) {
  newlist[[i]]$Q2_mean_nom <- ifelse(newlist[[i]]$Q2 > mean(newlist[[i]]$Q2, na.rm = TRUE), "1", "0")
}

newlist # displays output, see next chunk.

# output:
#> [[1]]
#>    Q1    Q2 Q2_mean_nom
#> 1 ABC 21000           0
#> 2 DEF 23400           0
#> 3 GHI 26800           1
#> 4 DEF 26000           1
#> 5 JKL 20400           0
#> 6 XYZ 30800           1
#> 
#> [[2]]
#>    Q1    Q2 Q2_mean_nom
#> 1 DEF 30100           1
#> 2 JKL 20200           0
#> 3 XYZ 15800           0
#> 4 ABC 21000           0
#> 5 MNO 23400           1
#> 6 PQR 26800           1
#> 
#> [[3]]
#>    Q1    Q2 Q2_mean_nom
#> 1 ABC 17800           0
#> 2 DEF 23060           1
#> 3 GHI 13080           0
#> 4 XYZ 27000           1
#> 5 MNO 22400           1
#> 6 PQR 26500           1

Created on 2021-09-11 by the reprex package (v2.0.1)