英文:
Passing a string as argument for data column (whilst having another column identifier)
问题
I'm writing a function that finds the mean at 4 different categorical variables and returns a size 4 vector with all of these values. In doing so, I want to pass different columns of a dataframe such that I can find these 4 values for any column of the dataframe. (I would like for the dataframe to be variable as well but thats not important)
makeMean <- function(df, colName){
x <- c()
levels <- levels(beetleData$Herbicide.dosage)
for(i in levels){
tempdata <- summarize(df, mean.temp = mean(Species.richness[Herbicide.dosage == i]))
x <- append(x, tempdata)
}
x <- unlist(x)
return(x)
}
This is the function. Species.richness is the column name I'd like to replace with "colName" but I can't find a way to do that. I can pass it by doing df[colName] (I think) but I'd like to have the other filter too (Herbicide.dosage == i).
R doesn't let me do mean(colName[Herbicide.dosage == i]). I've tried a couple other ways as well and none work as R doesn't convert the string to the dataframe column I want, it just reads it as a string.
英文:
I'm writing a function that finds the mean at 4 different categorical variables and returns a size 4 vector with all of these values. In doing so, I want to pass different columns of a dataframe such that I can find these 4 values for any column of the dataframe. (I would like for the dataframe to be variable as well but thats not important)
makeMean <- function(df, colName){
x <- c()
levels <- levels(beetleData$Herbicide.dosage)
for(i in levels){
tempdata <- summarize(df, mean.temp = mean(Species.richness[Herbicide.dosage == i]))
x <- append(x, tempdata)
}
x <- unlist(x)
return(x)
}
This is the function. Species.richness is the column name I'd like to replace with "colName" but I can't find a way to do that. I can pass it by doing df[colName] (I think) but I'd like to have the other filter too (Herbicide.dosage == i).
R doesn't let me do mean(colName[Herbicide.dosage == i]). I've tried a couple other ways as well and none work as R doesn't convert the string to the dataframe column I want, it just reads it as a string
答案1
得分: 1
将列名传递给函数有点棘手。但dplyr编程概念会有所帮助。链接
无法在您的数据上工作，但可以通过以下方式传递一个不是字符串的列名。
library(dplyr)
makeMeans <- function(df, colName) {
df %>%
summarise(Mean = mean({{colName}}, na.rm = TRUE))
}
makeMeans(mtcars, mpg)
返回结果如下：
Mean
1 20.09062
这样，您可以使用双花括号{{}}在函数中传递列并进行管理。
可能还有其他方法，但以这种方式传递列很容易。您可以尝试链接中提到的方法。
英文:
Passing column names to a function is tricky. But dplyr programming concepts will help. Link
Cannot work on ur data but passing a column name which is not read a sring is possible this way .
library(dplyr)
makeMeans <- function(df,colName){
df%>%
summarise(Mean= mean({{colName}},na.rm=T))
}
makeMeans(mtcars,mpg)
returns
Mean
1 20.09062
This way you can just pass columns and manage them in your function with double curly brackets {{
there might be other ways but passing columns this way is easy. You can try approaches in the link
答案2
得分: 1
"Like @anuanand's answer, but passing a string."
"类似于 @anuanand 的回答，但是传递一个字符串。"
library(tidyverse)
makeMeans <- function(df, colName){
df %>%
summarise(Mean = mean(.data[[colName]], na.rm = TRUE))
}
"See: https://www.tidyverse.org/blog/2019/06/rlang-0-4-0/ regarding passing column names as part of non-standard evaluation (NSE) in R."
"参考：https://www.tidyverse.org/blog/2019/06/rlang-0-4-0/，了解如何在 R 中作为非标准评估（NSE）的一部分传递列名。"
英文:
Like @anuanand's answer, but passing a string.
library(tidyverse)
makeMeans <- function(df, colName){
df %>%
summarise(Mean = mean(.data[[colName]],na.rm=T))
}
See: https://www.tidyverse.org/blog/2019/06/rlang-0-4-0/ regarding passing column names as part of non-standard evaulation (NSE) in R.
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