4

以下是我们使用两个名称更广泛地透视的一些愚蠢数据：

library(tidyr)

df <- data.frame(
    food = c('banana','banana','banana','banana','cheese','cheese','cheese','cheese'),
    binary = c(rep(c('yes','no'), 4)),
    car = c('toyota','subaru','mazda','skoda','toyota','subaru','mazda','skoda'),
    fun = c(2,4,3,6,2,4,2,3))

df %>%
    pivot_wider(
        id_cols = food,
        names_from = c(car, binary),
        values_from = fun)

如果我们想改变新变量名的格式，例如，从toyota_yesto yes_toyota，我们使用names_glue参数：

df %>%
    pivot_wider(
        id_cols = food,
        names_from = c(car, binary),
        names_glue = "{binary}_{car}",
        values_from = fun)

我面临的问题是找到将变量名传递给names_glue参数的正确语法。将变量传递给 很容易names_from，例如：

var1 <- 'car'
var2 <- 'binary'
df %>%
    pivot_wider(
        id_cols = food,
        names_from = c(var1, var2),
        values_from = fun)

但我们不能直接这样做names_glue：

df %>%
    pivot_wider(
        id_cols = food,
        names_from = c(var1, var2),
        names_glue = "{var1}_{var2}",
        values_from = fun)

错误：列名car_binary、car_binary和car_binary不得重复。

大概它正在评估变量并将结果字符串（即“汽车”或“二进制”）传递给胶水函数。我玩过一些我通常用于整洁评估的东西（!!sym(...)等），但没有任何工作。所需的输出如下，使用names_glue参数的变量：

# A tibble: 2 x 5
  food   yes_toyota no_subaru yes_mazda no_skoda
  <fct>       <dbl>     <dbl>     <dbl>    <dbl>
1 banana          2         4         3        6
2 cheese          2         4         2        3

这是将 names_glue 参数添加到 pivot_wider 的提交——尽管我真的不知道如何解决这个问题。

4

2 回答 2

4

您可以使用sprtinf/paste0来构造字符串：

library(tidyr)
df %>%
  pivot_wider(
    id_cols = food,
    names_from = c(var1, var2),
    names_glue = sprintf('{%s}_{%s}', var2, var1),
    values_from = fun)

#  food   yes_toyota no_subaru yes_mazda no_skoda
#  <chr>       <dbl>     <dbl>     <dbl>    <dbl>
#1 banana          2         4         3        6
#2 cheese          2         4         2        3
于 2020-09-29T01:19:53.743 回答
2

我们可以用glue

library(dplyr)
library(tidyr)
df %>%
     pivot_wider(
         id_cols = food,
         names_from = c(var1, var2),
         names_glue = glue::glue("{[var1]}_{[var2]}", .open = '[', .close = ']'),
         values_from = fun)
# A tibble: 2 x 5
#  food   toyota_yes subaru_no mazda_yes skoda_no
#  <chr>       <dbl>     <dbl>     <dbl>    <dbl>
#1 banana          2         4         3        6
#2 cheese          2         4         2        3
于 2020-09-30T00:06:06.207 回答