CODEWITHSUNDEEP
rdplyrnanna
ricke
ricke

Reputation: 175

Dplyr produces NaN while base R produces NA

Consider the following toy data and computations:

library(dplyr)

df <-  tibble(x = 1)

stats::sd(df$x)

dplyr::summarise(df, sd_x = sd(x))

The first calculation results in NA whereas the second, when the calculation is included in the dplyr function summarise produces NaN. I would expect both calculations to generate the same result and I wonder why they differ?

Upvotes: 9

Views: 722

Answers (1)

James
James

Reputation: 66834

It is calling a different function. I'm not clear what the function is, but it is not the stats one.

dplyr::summarise(df, sd_x = stats::sd(x))
# A tibble: 1 x 1
   sd_x
  <dbl>
1    NA

debugonce(sd) # debug to see when sd is called

Not called here:

dplyr::summarise(df, sd_x = sd(x))
# A tibble: 1 x 1
   sd_x
  <dbl>
1   NaN

But called here:

dplyr::summarise(df, sd_x = stats::sd(x))
debugging in: stats::sd(1)
debug: sqrt(var(if (is.vector(x) || is.factor(x)) x else as.double(x), 
    na.rm = na.rm))
...

Update

It appears that the sd within summarise gets calculated outside of R, hinted at in this header file: https://github.com/tidyverse/dplyr/blob/master/inst/include/dplyr/Result/Sd.h

A number of functions seem to be redefined by dplyr. Given that var gives the same result in both cases, I think the sd behaviour is a bug.

Upvotes: 6

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