如何使用管道传递数据框列作为函数中的参数? [英] How to pass a dataframe column as an argument in a function using piping?
问题描述
我正在弄乱R中的内置数据集economics，并且我试图将数据框列作为参数传递给使用管道的函数( top_n()在我的自定义函数中.以下是我如何将人口最多的5个国家(没有自定义功能)划分为子集的方法:
I'm messing around with the built-in dataset economics in R, and I'm trying to pass a dataframe column as an argument in a function that uses piping (dplyr, %>%). But I'm experiencing some seemingly strange problems. Somehow I can't successfully pass a column name as an argument to the function top_n() within my custom function. Here's how I would subset the 5 countries with the biggest population without a custom functon:
代码1:
library(dplyr)
df_econ <- economics
df_top_5 <- df_econ %>% top_n(5, pop)
df_top_5
输出1:
2014-12-01 12122.0 320201 5.0 12.6 8688
2015-01-01 12080.8 320367 5.5 13.4 8979
2015-02-01 12095.9 320534 5.7 13.1 8705
2015-03-01 12161.5 320707 5.2 12.2 8575
2015-04-01 12158.9 320887 5.6 11.7 8549
包装到自定义函数中后，它看起来可能像这样:
Wrapped into a custom function, it could look like this:
代码2:
library(dplyr)
# data
data(economics)
df_econ <- economics
# custom function
fxtop <- function(df, number, column){
tops <- df %>% top_n(number, column)
return(tops)
}
# build a df using custom function
df_top_5 <- fxtop(df=df_econ, number=5, column='pop')
df_top_5
输出2:
1967-07-01 507.4 198712 12.5 4.5 2944
1967-08-01 510.5 198911 12.5 4.7 2945
1967-09-01 516.3 199113 11.7 4.6 2958
1967-10-01 512.9 199311 12.5 4.9 3143
1967-11-01 518.1 199498 12.5 4.7 3066
1967-12-01 525.8 199657 12.1 4.8 3018
1968-01-01 531.5 199808 11.7 5.1 2878
1968-02-01 534.2 199920 12.2 4.5 3001
1968-03-01 544.9 200056 11.6 4.1 2877
1968-04-01 544.6 200208 12.2 4.6 2709
此输出有10行，而不是预期的5行.我怀疑参数number=5只是被忽略了，而实际使用的数字默认为10.数据似乎也没有按'pop'排序.
This output has 10 rows and not 5 as expected. I suspect that the argument number=5 is simply ignored and that the number that is actually used is defaulted to 10. The data does not seem to be sorted by 'pop' either.
到目前为止，我已经尝试过:
尝试1:自定义功能中的pop和number硬代码:
Attempt 1: hard-code pop and number within the custom function:
library(dplyr)
# data
data(economics)
df_econ <- economics
# custom function
fxtop <- function(df, number, column){
tops <- df %>% top_n(5, pop)
return(tops)
}
# build a df using custom function
df_top_5 <- fxtop(df=df_econ, number=5, column='pop')
df_top_5
尝试1:输出:
2014-12-01 12122.0 320201 5.0 12.6 8688
2015-01-01 12080.8 320367 5.5 13.4 8979
2015-02-01 12095.9 320534 5.7 13.1 8705
2015-03-01 12161.5 320707 5.2 12.2 8575
2015-04-01 12158.9 320887 5.6 11.7 8549
尝试1:评论
这是所需的输出！
让我们看看当我通过函数传递变量时会发生什么情况
Let's see what happens when I'm passing the variables through the function
尝试2:将变量作为对象而不是字符串:
Attempt 2: pass variables as object instead of string:
library(dplyr)
# data
data(economics)
df_econ <- economics
# custom function
fxtop <- function(df, number, column){
tops <- df %>% top_n(5, column)
return(tops)
}
# build a df using custom function
df_top_5 <- fxtop(df=df_econ, number=5, column='pop')
df_top_5
尝试2:输出:
现在输出与第一个示例相同.这两个变量似乎都被忽略了.
Now the output is the same as in the first example. Both variables are seemingly ignored.
那么，有什么建议吗?
推荐答案
我们可以对curl-curly({{}})使用非标准评估
We can use non-standard evaluation with curly-curly ({{}})
library(dplyr)
library(rlang)
fxtop <- function(df, number, column){
tops <- df %>% top_n(number, {{column}})
return(tops)
}
并传递未加引号的变量名
and pass unquoted variable names
fxtop(df=df_econ, number=5, pop)
# date pce pop psavert uempmed unemploy
# <date> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 2014-12-01 12062 319746. 7.6 12.9 8717
#2 2015-01-01 12046 319929. 7.7 13.2 8903
#3 2015-02-01 12082. 320075. 7.9 12.9 8610
#4 2015-03-01 12158. 320231. 7.4 12 8504
#5 2015-04-01 12194. 320402. 7.6 11.5 8526
如果您想将列名作为字符串(用引号引起来)传递，我们可以将sym与!!
fxtop <- function(df, number, column){
tops <- df %>% top_n(number, !!sym(column))
return(tops)
}
fxtop(df=df_econ, number=5, 'pop')
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