[英]Using mutate_at with mutate_if
我正在我的包中創建一個通用函數。 目標是找到百分比列的列,然后在它們是character
列時使用parse_number
。 我一直無法使用mutate_at
和ifelse
找出解決方案。 我在下面粘貼了一個reprex。
library(tidyverse)
df <- tibble::tribble(
~name, ~pass_percent, ~attendance_percent, ~grade,
"Jon", "90%", 0.85, "B",
"Jim", "100%", 1, "A"
)
percent_names <- df %>% select(ends_with("percent"))%>% names()
# Error due to attendance_percent already being in numeric value
if (percent_names %>% length() > 0) {
df <-
df %>%
dplyr::mutate_at(percent_names, readr::parse_number)
}
#> Error in parse_vector(x, col_number(), na = na, locale = locale, trim_ws = trim_ws): is.character(x) is not TRUE
您的attendance_percent
變量是數字,而不是字符,而parse_number
只需要字符變量,請參見此處。 所以一個解決方案是:
edited_parse_number <- function(x, ...) {
if (mode(x) == 'numeric') {
x
} else {
parse_number(x, ...)
}
}
df %>%
dplyr::mutate_at(vars(percent_names), edited_parse_number)
# name pass_percent attendance_percent grade
# <chr> <dbl> <dbl> <chr>
#1 Jon 90 0.85 B
#2 Jim 100 1 A
或者
如果您不想使用該額外函數,請在開始時提取字符變量:
percent_names <- df %>%
select(ends_with("percent")) %>%
select_if(is.character) %>%
names()
percent_names
# [1] "pass_percent"
df %>%
dplyr::mutate_at(vars(percent_names), parse_number)
# name pass_percent attendance_percent grade
# <chr> <dbl> <dbl> <chr>
# 1 Jon 90 0.85 B
# 2 Jim 100 1 A
或者,無需創建函數,您只需將ifelse
語句添加到mutate_at
例如:
if (percent_names %>% length() > 0) {
df <-
df %>% rowwise() %>%
dplyr::mutate_at(vars(percent_names), ~ifelse(is.character(.),
parse_number(.),
.))
}
Source: local data frame [2 x 4]
Groups: <by row>
# A tibble: 2 x 4
name pass_percent attendance_percent grade
<chr> <dbl> <dbl> <chr>
1 Jon 90 0.85 B
2 Jim 100 1 A
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