CODEWITHSUNDEEP
rdplyr
Konrad Rudolph
Konrad Rudolph

Reputation: 545885

How do I sweep specific columns with dplyr?

An incredibly common operation for my type of data is applying a normalisation factor to all columns. This can be done efficiently using sweep or scale:

normalized = scale(data, center = FALSE, scale = factors)
# or
normalized = sweep(data, 2, factors, `/`)

Where

data = structure(list(A = c(3L, 174L, 6L, 1377L, 537L, 173L),
    B = c(1L, 128L, 2L, 1019L, 424L, 139L),
    C = c(3L, 66L, 2L, 250L, 129L, 40L),
    D = c(4L, 57L, 4L, 251L, 124L, 38L)),
    .Names = c("A", "B", "C", "D"),
    class = c("tbl_df", "data.frame"), row.names = c(NA, -6L))

factors = c(A = 1, B = 1.2, C = 0.8, D = 0.75)

However, how do I do this with dplyr, when my data has additional columns in front? I can do it in separate statements, but I’d like doing it in one pipeline. This is my data:

data = structure(list(ID = c(1, 2, 3, 4, 5, 6),
    Type = c("X", "X", "X", "Y", "Y", "Y"),
    A = c(3L, 174L, 6L, 1377L, 537L, 173L),
    B = c(1L, 128L, 2L, 1019L, 424L, 139L),
    C = c(3L, 66L, 2L, 250L, 129L, 40L),
    D = c(4L, 57L, 4L, 251L, 124L, 38L)),
    .Names = c("ID", "Type", "A", "B", "C", "D"),
    class = c("tbl_df", "data.frame"), row.names = c(NA, -6L))

And I’d like to mutate the data columns without touching the first two columns. Normally I can do this with mutate_each; however, how I cannot pass my normalisation factors to that function:

data %>% mutate_each(funs(. / factors), A:D)

This, unsurprisingly, assumes that I want to divide each column by factors, rather than each column by its matching factor.

Upvotes: 14

Views: 2797

Answers (2)

tmfmnk
tmfmnk

Reputation: 40141

From dplyr 1.0.0, you can do:

data %>%
 rowwise() %>%
 mutate(across(A:D)/factors)

     ID Type      A       B      C      D
  <dbl> <chr> <dbl>   <dbl>  <dbl>  <dbl>
1     1 X         3   0.833   3.75   5.33
2     2 X       174 107.     82.5   76   
3     3 X         6   1.67    2.5    5.33
4     4 Y      1377 849.    312.   335.  
5     5 Y       537 353.    161.   165.  
6     6 Y       173 116.     50     50.7

Upvotes: 3

jazzurro
jazzurro

Reputation: 23574

Given akrun's encouragement, let me post what I did as an answer here. I just intuitively thought that you might want to ask R to indicate columns with a same name to do this mutate_each. For instance, if . indicates the column, A, I thought another column named A from another data.frame might be something dplyr might like. So, I created a data frame for factors then used mutate_each. It seems that the outcome is right. Since I have no technical background, I am afraid that I cannot really provide any explanation. I hope you do not mind that.

factors <- data.frame(A = 1, B = 1.2, C = 0.8, D = 0.75)

mutate_at(data, vars(A:D), funs(. / foo$.))

# By the time I answered this question, the following was working.
# But mutate_each() is now deprecated.

# mutate_each(data, funs(. / factors$.), A:D)

#  ID Type    A           B      C          D
#1  1    X    3   0.8333333   3.75   5.333333
#2  2    X  174 106.6666667  82.50  76.000000
#3  3    X    6   1.6666667   2.50   5.333333
#4  4    Y 1377 849.1666667 312.50 334.666667
#5  5    Y  537 353.3333333 161.25 165.333333
#6  6    Y  173 115.8333333  50.00  50.666667

EDIT

This also works. Given data frame is a special case of list, this is not perhaps surprising.

# Experiment
foo <- list(A = 1, B = 1.2, C = 0.8, D = 0.75)

mutate_at(data, vars(A:D), funs(. / foo$.))

# mutate_each(data, funs(. / foo$.), A:D)

#  ID Type    A           B      C          D
#1  1    X    3   0.8333333   3.75   5.333333
#2  2    X  174 106.6666667  82.50  76.000000
#3  3    X    6   1.6666667   2.50   5.333333
#4  4    Y 1377 849.1666667 312.50 334.666667
#5  5    Y  537 353.3333333 161.25 165.333333
#6  6    Y  173 115.8333333  50.00  50.666667

Upvotes: 12

Related Questions