Use functional programming and update each iteration the underlying dataframe R?

Question

Given a dataframe I want to run on multiple column names, calculate something and add the output as new column. The next calculation will be added as a new column to the updated dataframe.

For example:

Given a simple df:

df <- structure(list(a = c(1, 2, 3), b = c(4, 5, 6), c = c(7, 8, 9), 
    d = c(10, 11, 12)), .Names = c("a", "b", "c", "d"), row.names = c(NA, 
-3L), class = c("tbl_df", "tbl", "data.frame"))

For each column a, b, c, d I want to calculate, say a square:

a2 = a^2
b2 = b^2 ...

For technical reasons I can't publish the whole dataframe but I am going to pass a column name each time and expect the function to mutate a new column (for example a2) next time when I will add b2, a2 will be already there:

If I would use for loop it would look like:

for (x in column_names) {

     df <- df %>% mutate("x2" = x^2)

}

So each time my df updates with new calculated column.

Please advise how can I do this without for loop with functional programming.

I am trying to do this with map, lapply but I have the problem that my df doesn't get updated each iteration.

s_baldur · Accepted Answer · 2018-12-06 14:06:52Z

2

Is this the function you are looking for?

add_x2 <- function(df, x) {
  df[paste0(x, "2")] <- df[x]^2
  df
}

df %>%
  add_x2(c("a", "b"))

# A tibble: 3 x 6
      a     b     c     d    a2    b2
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     4     7    10     1    16
2     2     5     8    11     4    25
3     3     6     9    12     9    36

edited Dec 6, 2018 at 14:06

answered Dec 6, 2018 at 13:58

s_baldur

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3 Comments

SteveS Over a year ago

Yes but I want to to be applied on 100 columns at once. It seems the for loop is the best simple solution here aren't it. I want to build and expand my dataframe with new columns each iteration.

IceCreamToucan Over a year ago

couldn't you just use names(df)[1:100] (or whatever columns you want) as the second argument to add_x2?

s_baldur Over a year ago

@steves Updated a bit so you can use it as suggested by IceCreamToucan

jyjek · Accepted Answer · 2018-12-06 13:55:00Z

0

With tidyverse:

df %>% 
+   mutate_if(is.numeric,funs(.^2))
# A tibble: 3 x 4
      a     b     c     d
  <dbl> <dbl> <dbl> <dbl>
1     1    16    49   100
2     4    25    64   121
3     9    36    81   144

or

df %>% mutate_all(funs(.^2))
# A tibble: 3 x 4
      a     b     c     d
  <dbl> <dbl> <dbl> <dbl>
1     1    16    49   100
2     4    25    64   121
3     9    36    81   144

answered Dec 6, 2018 at 13:55

jyjek

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