用外部变量总结
Summarise with an external variable
我正在尝试使此代码适用于外部变量:
#This is the original code
library(dplyr)
library(tibble)
nFinal <- 5
graphData <- tibble(y = rep(1:nFinal,2),
ponde = runif(10, min = 0, max = 1),
)
totalData <- graphData %>%
**summarise**(val1 = prop.table(questionr::wtd.table(y, weights = ponde))[1]*100,
val2 = prop.table(questionr::wtd.table(y, weights = ponde))[2]*100,
val3 = prop.table(questionr::wtd.table(y, weights = ponde))[3]*100,
val4 = prop.table(questionr::wtd.table(y, weights = ponde))[4]*100,
val5 = prop.table(questionr::wtd.table(y, weights = ponde))[5]*100)
我想要的是能够针对提供的值数量完成此操作,因此最后一个 val
应该取决于外部整数 nFinal
中提供的数字,一些像这样:
nFinal <- 10 #or any value
totalData <- graphData %>%
summarise(val1 = prop.table(questionr::wtd.table(y, weights = ponde))[1]*100,
val2 = prop.table(questionr::wtd.table(y, weights = ponde))[2]*100,
val3 = prop.table(questionr::wtd.table(y, weights = ponde))[3]*100,
val4 = prop.table(questionr::wtd.table(y, weights = ponde))[4]*100,
val5 = prop.table(questionr::wtd.table(y, weights = ponde))[5]*100,
...
valnFinal = prop.table(questionr::wtd.table(y, weights = ponde))[nFinal]*100)
我正在为此使用 dplyr
,但我会接受任何其他解决方案。
(编辑,使示例数据具有 nFinal 元素,正如在对 arkun 的评论中指出的那样。)
您可以将输出保存在列表中并使用 unnest_wider
创建新列。
library(dplyr)
library(tidyr)
graphData %>%
summarise(val = list(prop.table(questionr::wtd.table(y, weights = ponde)) * 100)) %>%
unnest_wider(val)
# `1` `2` `3` `4` `5`
# <dbl> <dbl> <dbl> <dbl> <dbl>
#1 37.2 20.0 21.8 12.2 8.92
我正在尝试使此代码适用于外部变量:
#This is the original code
library(dplyr)
library(tibble)
nFinal <- 5
graphData <- tibble(y = rep(1:nFinal,2),
ponde = runif(10, min = 0, max = 1),
)
totalData <- graphData %>%
**summarise**(val1 = prop.table(questionr::wtd.table(y, weights = ponde))[1]*100,
val2 = prop.table(questionr::wtd.table(y, weights = ponde))[2]*100,
val3 = prop.table(questionr::wtd.table(y, weights = ponde))[3]*100,
val4 = prop.table(questionr::wtd.table(y, weights = ponde))[4]*100,
val5 = prop.table(questionr::wtd.table(y, weights = ponde))[5]*100)
我想要的是能够针对提供的值数量完成此操作,因此最后一个 val
应该取决于外部整数 nFinal
中提供的数字,一些像这样:
nFinal <- 10 #or any value
totalData <- graphData %>%
summarise(val1 = prop.table(questionr::wtd.table(y, weights = ponde))[1]*100,
val2 = prop.table(questionr::wtd.table(y, weights = ponde))[2]*100,
val3 = prop.table(questionr::wtd.table(y, weights = ponde))[3]*100,
val4 = prop.table(questionr::wtd.table(y, weights = ponde))[4]*100,
val5 = prop.table(questionr::wtd.table(y, weights = ponde))[5]*100,
...
valnFinal = prop.table(questionr::wtd.table(y, weights = ponde))[nFinal]*100)
我正在为此使用 dplyr
,但我会接受任何其他解决方案。
(编辑,使示例数据具有 nFinal 元素,正如在对 arkun 的评论中指出的那样。)
您可以将输出保存在列表中并使用 unnest_wider
创建新列。
library(dplyr)
library(tidyr)
graphData %>%
summarise(val = list(prop.table(questionr::wtd.table(y, weights = ponde)) * 100)) %>%
unnest_wider(val)
# `1` `2` `3` `4` `5`
# <dbl> <dbl> <dbl> <dbl> <dbl>
#1 37.2 20.0 21.8 12.2 8.92