从 %>% 管道运算符获取倒数第二个输入
Get second to last input from %>% pipe operator
我想从我刚刚在管道中创建的列中提取一个因子。
我可以使用以下代码执行此操作:
library("dplyr")
library("magrittr")
library("janitor")
iris <- iris %>% janitor::clean_names()
iris %>% filter(species %in% c("setosa","versicolor")) %>% group_by(species) %>%
summarise(mean_sepal_length = mean(sepal_length)) %>% ungroup() %>%
mutate(species = factor(species, levels = (iris %>% group_by(species) %>% #<- works but messy
summarise(mean_sepal_length = mean(sepal_width)) %>%
ungroup() %>% arrange(mean_sepal_length) %$% species))) %>%
arrange(species)
我想知道是否有“更清洁”的方法来做到这一点。像这样的东西:
iris %>% filter(species %in% c("setosa","versicolor")) %>% group_by(species) %>%
summarise(mean_sepal_length = mean(sepal_length)) %>% ungroup() %>%
mutate(species = factor(species, levels = (. %>% arrange(mean_sepal_length) %$% species))) %>%
arrange(species)
其中 .
是倒数第二个参数,而不是给管道的最后一个参数?
这会引发错误,因为管道的最后一个参数是 mutate 语句:
Error: Problem with `mutate()` input `species`. x 'match' requires vector arguments i Input `species` is `factor(...)`.
我认为这根本不是管道运算符的工作方式,所以这可能是不可能的。
为了使第二个选项起作用,我们可以将 .
包裹在 {}
中
library(dplyr)
library(magrittr)
iris %>%
filter(species %in% c("setosa","versicolor")) %>%
group_by(species) %>%
summarise(mean_sepal_length = mean(sepal_length)) %>%
ungroup() %>%
mutate(species = factor(species,
levels = ({.} %>%
arrange(mean_sepal_length) %$% species))) %>%
arrange(species)
# A tibble: 2 x 2
# species mean_sepal_length
# <fct> <dbl>
#1 setosa 5.01
#2 versicolor 5.94
您可以 arrange
基于 mean_sepal_length
的数据,然后使用 unique
.
根据它们的出现分配 factor
级别
library(dplyr)
iris %>%
filter(species %in% c("setosa","versicolor")) %>%
group_by(species) %>%
summarise(mean_sepal_length = mean(sepal_length)) %>%
arrange(mean_sepal_length) %>%
mutate(species = factor(species, levels = unique(species)))
# species mean_sepal_length
# <fct> <dbl>
#1 setosa 5.01
#2 versicolor 5.94
我想从我刚刚在管道中创建的列中提取一个因子。 我可以使用以下代码执行此操作:
library("dplyr")
library("magrittr")
library("janitor")
iris <- iris %>% janitor::clean_names()
iris %>% filter(species %in% c("setosa","versicolor")) %>% group_by(species) %>%
summarise(mean_sepal_length = mean(sepal_length)) %>% ungroup() %>%
mutate(species = factor(species, levels = (iris %>% group_by(species) %>% #<- works but messy
summarise(mean_sepal_length = mean(sepal_width)) %>%
ungroup() %>% arrange(mean_sepal_length) %$% species))) %>%
arrange(species)
我想知道是否有“更清洁”的方法来做到这一点。像这样的东西:
iris %>% filter(species %in% c("setosa","versicolor")) %>% group_by(species) %>%
summarise(mean_sepal_length = mean(sepal_length)) %>% ungroup() %>%
mutate(species = factor(species, levels = (. %>% arrange(mean_sepal_length) %$% species))) %>%
arrange(species)
其中 .
是倒数第二个参数,而不是给管道的最后一个参数?
这会引发错误,因为管道的最后一个参数是 mutate 语句:
Error: Problem with `mutate()` input `species`. x 'match' requires vector arguments i Input `species` is `factor(...)`.
我认为这根本不是管道运算符的工作方式,所以这可能是不可能的。
为了使第二个选项起作用,我们可以将 .
包裹在 {}
library(dplyr)
library(magrittr)
iris %>%
filter(species %in% c("setosa","versicolor")) %>%
group_by(species) %>%
summarise(mean_sepal_length = mean(sepal_length)) %>%
ungroup() %>%
mutate(species = factor(species,
levels = ({.} %>%
arrange(mean_sepal_length) %$% species))) %>%
arrange(species)
# A tibble: 2 x 2
# species mean_sepal_length
# <fct> <dbl>
#1 setosa 5.01
#2 versicolor 5.94
您可以 arrange
基于 mean_sepal_length
的数据,然后使用 unique
.
factor
级别
library(dplyr)
iris %>%
filter(species %in% c("setosa","versicolor")) %>%
group_by(species) %>%
summarise(mean_sepal_length = mean(sepal_length)) %>%
arrange(mean_sepal_length) %>%
mutate(species = factor(species, levels = unique(species)))
# species mean_sepal_length
# <fct> <dbl>
#1 setosa 5.01
#2 versicolor 5.94