How to keep non-numeric columns when using apply() function on a data frame in R?

Question

I have a data frame where one column has non-numeric values. I need to apply a function on the numeric columns and create new data-frame, but I want to keep the column with non-numeric values in this new data frame.

#Data frame (exemple)
df <- data.frame(col1 = c(1:5), col2 = c(6:10), col3 = c("x","y","z","w","x"))

#Function (example)
sum1 <- function(x){x + 1}

df_aplly <- as.data.frame(apply(df, MARGIN = 2, FUN = sum1))

This example return an error:

Error in x + 1 : non-numeric argument to binary operator

I tried to apply only on the numeric columns:

#Function (exemple)
sum1 <- function(x){x + 1}

df_aplly <- as.data.frame(apply(df[1:2], MARGIN = 2, FUN = sum1))

> df_aplly
  col1 col2
1    2    7
2    3    8
3    4    9
4    5   10
5    6   11

It works, but return a new data frame with just 2 columns, but I need to keep the columns non-numeric in this new data frame.

Is there a way to do it using only the 'apply()' function?

#data frame I need:
> df_aplly
  col1 col2 df$col3
1    2    7       x
2    3    8       y
3    4    9       z
4    5   10       w
5    6   11       x

You can cbind your col3 to as.data.frame(apply(df[1:2], MARGIN = 2, FUN = sum1)), i.e. cbind(as.data.frame(apply(df[1:2], 2, sum1)), df$col3) — benson23
– benson23, Commented Feb 14, 2024 at 15:21

Jilber Urbina · Accepted Answer · 2024-02-14 15:27:32Z

0

You can use dplyr

> df %>% 
    mutate_if(is.numeric, sum1)
  col1 col2 col3
1    2    7    x
2    3    8    y
3    4    9    z
4    5   10    w
5    6   11    x

In base R

> ind <- sapply(df, is.numeric)
> as.data.frame(c(lapply(as.list(df[ind]), sum1), as.list(df[!ind])))
  col1 col2 col3
1    2    7    x
2    3    8    y
3    4    9    z
4    5   10    w
5    6   11    x

edited Feb 14, 2024 at 15:27

answered Feb 14, 2024 at 15:19

Jilber Urbina

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Comments

Jake · Accepted Answer · 2024-02-14 16:09:40Z

0

You can nest a conditional statement within your function:

sum1 <- function(x){   
     ifelse(!is.na(as.numeric(x)),
     as.numeric(x) + 1,
     x) 
}

Note that you need to coerce x to numeric for this to work, otherwise apply will apparently parse it as character class. Which is odd... but this should work.

answered Feb 14, 2024 at 16:09

Jake

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