[英]Error in using custom vectorized function in mutate/case_when
Below is a simple code to reproduce the error.下面是重现错误的简单代码。 I define a simple function, vectorize it with another function using purrr::map
, and then try to use it in a mutate case_when where the condition should normally ensure that the arguments are valid.我定义了一个简单的 function,使用purrr::map
将其与另一个 function 向量化,然后尝试在 mutate case_when 中使用它,其中条件通常应确保 arguments 有效。 The error occurs in the condition if(arg1 > 0)
when arg1 = NA
, but I don't understand why that even happens.错误发生在条件if(arg1 > 0)
when arg1 = NA
时,但我不明白为什么会发生这种情况。 If I apply the filter, the error disappears.如果我应用过滤器,错误就会消失。 Does anyone have an idea what I'm doing wrong?有谁知道我做错了什么? My feeling is that it should work.我的感觉是它应该工作。
require(tidyverse)
f_single <- function(arg1, arg2) {
if (arg1 > 0) {
return(arg1 * arg2)
}
}
f_vector <- function(arg1, arg2) {
result <- map2_dbl(arg1, arg2, f_single)
return(result)
}
x <- tribble(~ arg1, ~ arg2,
NA, 1,
2, 3,
4, 5,)
x %>%
# filter(!is.na(arg1)) %>%
mutate(y = case_when(arg1 > 0 ~ f_vector(arg1, arg2)))
The error is the following:错误如下:
Error in `mutate()`:
! Problem while computing `y = case_when(arg1 > 0 ~ f_vector(arg1, arg2))`.
Caused by error in `if (arg1 > 0) ...`:
! missing value where TRUE/FALSE needed
Two issues:两个问题:
NA
to an if
statement will throw an error.将NA
传递给if
语句将引发错误。 You can avoid this by wrapping the condition with isTRUE
.您可以通过使用isTRUE
包装条件来避免这种情况。f_single
returns NULL
when arg1
is missing or <= 0, but map_*
expects a return value for every input.您的代码仍会引发错误,因为当arg1
缺失或 <= 0 时f_single
返回NULL
,但map_*
期望每个输入都有一个返回值。 Changing f_single
as follows will solve both problems:如下更改f_single
将解决这两个问题:
f_single <- function(arg1, arg2) {
if (isTRUE(arg1 > 0)) {
arg1 * arg2
} else {
NA_real_
}
}
# rest of code unchanged from original
# # A tibble: 3 x 3
# arg1 arg2 y
# <dbl> <dbl> <dbl>
# 1 NA 1 NA
# 2 2 3 6
# 3 4 5 20
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