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I am tracking the body weights of individuals over time, and the function below allow me to calculate the % body weight of the individual on a particular day, relative to the initial value (essentially dividing the body weight on a particular day by the body weight observed on day 1).

variability <- function(df, column_number) {
  variable_name <- paste0("var_BW", column_number)

   df %>% 
  mutate(!!variable_name := round(100*(df[,column_number]/df[1,column_number]), 1))

}

This function works fine if I use it on one column, but since I have a number of individuals, I would like to use the apply() family to use the function on multiple columns of one dataframe (for instance on columns 1:8 of the dataframe below):

 BW1  BW2  BW3  BW4  BW5  BW6  BW7  BW8
1 18.4 19.6 20.7 17.4 18.7 18.9 19.0 17.8
2 18.1 19.3 20.0 17.5 18.3 19.4 19.5 18.0
3 17.7 18.9 20.4 17.3 18.3 19.2 19.3 17.9

My initial guess is to store the column numbers in a list, and then pass that list as an argument in the lapply() function, as such: 

l <- list(1:8)
lapply(working_df, variability, l)

However, when I do that, I get the following error:

Error in UseMethod("mutate_") : 
  no applicable method for 'mutate_' applied to an object of class "c('double', 'numeric')"

Any thoughts?

DTYK
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ThomasC
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  • I think you are interested in `apply()` function. Check `apply(working_df, 2, variability) ` – Carles Dec 29 '18 at 00:11
  • `sweep` function might be the easiest here - I feel like its often overlooked. ```100*sweep(BW,2,unlist(BW[1,]),`/`)``` – smarchese Dec 29 '18 at 00:16
  • To answer your question about the `mutate()` error, that's because you're calling mutate on an object that is not a data frame. – thus__ Dec 29 '18 at 00:23
  • It would be worthwhile posting the output of your original function, or stating explicitly what it does, since I can see that some of the responses have assumed that you are indexing row-wise rather than column-wise. I'm assuming that they weren't in front of a computer to run the code to check. – g_t_m Dec 29 '18 at 05:45
  • @g_t_m you're right, I was indexing column-wise rather than row-wise. The output that both you and AkselA posted below is what I was aiming for. Thanks – ThomasC Dec 30 '18 at 00:58

3 Answers3

0

Does this fit?
As it's possible to vectorize the relative percentage part we can simplify things greatly.

bw <- read.table(text="
 BW1  BW2  BW3  BW4  BW5  BW6  BW7  BW8
1 18.4 19.6 20.7 17.4 18.7 18.9 19.0 17.8
2 18.1 19.3 20.0 17.5 18.3 19.4 19.5 18.0
3 17.7 18.9 20.4 17.3 18.3 19.2 19.3 17.9", header=TRUE)

apply(bw, 2, function(x) round(100*x/x[1], 1))
#     BW1   BW2   BW3   BW4   BW5   BW6   BW7   BW8
# 1 100.0 100.0 100.0 100.0 100.0 100.0 100.0 100.0
# 2  98.4  98.5  96.6 100.6  97.9 102.6 102.6 101.1
# 3  96.2  96.4  98.6  99.4  97.9 101.6 101.6 100.6

Or using sweep()

round(sweep(bw, 2, unlist(bw[1,]), "/")*100, 1)
#     BW1   BW2   BW3   BW4   BW5   BW6   BW7   BW8
# 1 100.0 100.0 100.0 100.0 100.0 100.0 100.0 100.0
# 2  98.4  98.5  96.6 100.6  97.9 102.6 102.6 101.1
# 3  96.2  96.4  98.6  99.4  97.9 101.6 101.6 100.6

Or even simpler

round(100 * t(t(bw) / as.matrix(bw)[1,]), 1)
#     BW1   BW2   BW3   BW4   BW5   BW6   BW7   BW8
# 1 100.0 100.0 100.0 100.0 100.0 100.0 100.0 100.0
# 2  98.4  98.5  96.6 100.6  97.9 102.6 102.6 101.1
# 3  96.2  96.4  98.6  99.4  97.9 101.6 101.6 100.6
AkselA
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0

You don't really need apply in this case.

pctvals <- round(100.0 * bw[,1:ncol(bw)] / bw[,1], 2)

yields

  BW1    BW2    BW3   BW4    BW5    BW6    BW7    BW8
1 100 106.52 112.50 94.57 101.63 102.72 103.26  96.74
2 100 106.63 110.50 96.69 101.10 107.18 107.73  99.45
3 100 106.78 115.25 97.74 103.39 108.47 109.04 101.13
Edward Carney
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0

There's a super simple option in using mutate_at from the dplyr package:

library(dplyr)

working_df <-
  data.frame(BW1 = c(18.4, 18.1, 17.7),
             BW2 = c(19.6, 19.3, 18.9),
             BW3 = c(20.7, 20.0, 20.4))

variability_v2 <- function(df, column_numbers) {

  df %>% 
    mutate_at(vars(column_numbers), funs(var = round(100*(./first(.)), 1)))

}

variability_v2(working_df, 1:3)
#>    BW1  BW2  BW3 BW1_var BW2_var BW3_var
#> 1 18.4 19.6 20.7   100.0   100.0   100.0
#> 2 18.1 19.3 20.0    98.4    98.5    96.6
#> 3 17.7 18.9 20.4    96.2    96.4    98.6

The only 2 (very minor issues, in my opinion) with this method are:

  • If you only feed a single column number into the function, then the new column is simply called "var"
  • The "var" is appended after the column name, not before it

The former could be dealt with by a simple "if" statement within the function, carving out the situation where there is only one column specified. Hopefully you just don't care about the latter!

g_t_m
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