dplyr mutate_each_标准评估 [英] dplyr mutate_each_ standard evaluation
问题描述
这是我正在寻找的最小工作示例要完成,使用虹膜数据集(我删除'物种'列,使其全部为数字)。我从每一列减去Petal.Width列。但是我需要列名作为变量,例如My.Petal.Width
#删除Species列,因此我们只有数字数据
iris_numeric< - iris%>%select(-Species)
#这是所需的结果,使用NSE
result_NSE< - iris_numeric %>%mutate_each(funs(。 - 'Petal.Width`))
#这是我尝试使用SE
SubtractCol< - Petal.Width
result_SE< - iris_numeric%>%mutate_each_(funs(。 - as.name(SubtractCol)))
#第二次尝试
SubtractCol< - Petal.Width
列< - colnames(iris_numeric)
mutate_call = lazyeval :: interp(〜。-a,a = as.name(SubtractCol))
result_SE< - iris_numeric%>%mutate_each_点= setNames(列表(mutate_call),列))
我收到各种错误:
colwise_(tbl,funs_(funs),vars)中的错误:
参数vars缺少,没有默认
mutate_each_(。,.do中的错误) ts = setNames(list(mutate_call),Columns)):
unused参数(.dots = setNames(list(mutate_call),Columns))
请帮忙,谢谢提前。
你正在寻找什么是SE版本的 funs
,即 funs _
:
库(lazyeval); library(dplyr)
SubtractCol< - Petal.Width
iris%>%mutate_each(funs_(interp(〜。-x,x = as.name(SubtractCol))), - )%>%head
#Sepal.Length Sepal.Width Petal.Length Petal.Width物种
#1 4.9 3.3 1.2 0 setosa
#2 4.7 2.8 1.2 0 setosa
#3 4.5 3.0 1.1 0 setosa
#4 4.4 2.9 1.3 0 setosa
#5 4.8 3.4 1.2 0 setosa
#6 5.0 3.5 1.3 0 setosa
如果您想将我所写的-Species作为-Species提供,您将使用 mutate_each _
一个字符串/变量。
请注意, mutate_each _
.dot
code>和 summarise_each _
。 I'm racking my brain over the SE implementation of mutate_each_ in dplyr. What I want to do is to subtract the value in one column in a DF from every column in the DF.
Here is a minimum working example of what I'm looking to accomplish, using the iris dataset (I remove the 'Species' column so that it is all numeric). I subtract the Petal.Width column from every column. But I need the column name to be a variable, such as "My.Petal.Width"
# Remove Species column, so that we have only numeric data
iris_numeric <- iris %>% select(-Species)
# This is the desired result, using NSE
result_NSE <- iris_numeric %>% mutate_each(funs(. - `Petal.Width`))
# This is my attempt at using SE
SubtractCol <- "Petal.Width"
result_SE <- iris_numeric %>% mutate_each_(funs(. - as.name(SubtractCol)))
# Second attempt
SubtractCol <- "Petal.Width"
Columns <- colnames(iris_numeric)
mutate_call = lazyeval::interp(~.-a, a = as.name(SubtractCol))
result_SE <- iris_numeric %>% mutate_each_(.dots = setNames(list(mutate_call), Columns))
I get various errors:
Error in colwise_(tbl, funs_(funs), vars) :
argument "vars" is missing, with no default
Error in mutate_each_(., .dots = setNames(list(mutate_call), Columns)) :
unused argument (.dots = setNames(list(mutate_call), Columns))
Please help and many thanks in advance.
What you are looking for is the SE version of funs
, i.e. funs_
:
library(lazyeval); library(dplyr)
SubtractCol <- "Petal.Width"
iris %>% mutate_each(funs_(interp(~.-x, x = as.name(SubtractCol))), -Species) %>% head
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#1 4.9 3.3 1.2 0 setosa
#2 4.7 2.8 1.2 0 setosa
#3 4.5 3.0 1.1 0 setosa
#4 4.4 2.9 1.3 0 setosa
#5 4.8 3.4 1.2 0 setosa
#6 5.0 3.5 1.3 0 setosa
You would use mutate_each_
if you wanted to supply what I wrote as "-Species" as a string/variable.
Note that there's no .dots
argument in mutate_each_
and summarise_each_
.
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