带有 names_pattern 和列对的 pivot_longer答案

【问题标题】：pivot_longer with names_pattern and pairs of columns带有 names_pattern 和列对的 pivot_longer
【发布时间】：2022-09-26 05:41:43
【问题描述】：

我试图弄清楚如何在以下示例中使用来自tidyr 的pivot_longer。这就是名为dat_plot 的原始表的结构如下：

   year organizational_based action_based ideological_based share_org_based share_ideo_based share_act_based
  <dbl>                <dbl>        <dbl>             <dbl>           <dbl>            <dbl>           <dbl>
1  1956                    1            0                 0               2               95              95
2  2000                    0            0                 0              92               87              91

也在这里：

dat_plot <- structure(list(year = c(1956, 2000), organizational_based = c(1, 
0), action_based = c(0, 0), ideological_based = c(0, 0), share_org_based = c(2, 
92), share_ideo_based = c(95, 87), share_act_based = c(95, 91
)), row.names = c(NA, -2L), class = c(\"tbl_df\", \"tbl\", \"data.frame\"
))

我想通过以下方式将其转换为长格式：

  year          based based_value      share share_value
1 1956 organizational           1  org_based           2
2 1956         action           0 ideo_based          95
3 1956    ideological           0  act_based          95
4 2000 organizational           0  org_based          92
5 2000         action           0 ideo_based          87
6 2000    ideological           0  act_based          91

或者，使用dput：

solution <- structure(list(year = c(1956, 1956, 1956, 2000, 2000, 2000), 
    based = c(\"organizational\", \"action\", \"ideological\", \"organizational\", 
    \"action\", \"ideological\"), based_value = c(1, 0, 0, 0, 0, 
    0), share = c(\"org_based\", \"ideo_based\", \"act_based\", \"org_based\", 
    \"ideo_based\", \"act_based\"), share_value = c(2, 95, 95, 92, 
    87, 91)), class = \"data.frame\", row.names = c(NA, -6L))

我以为我必须与names_pattern 合作，我尝试过的是这样的，但是如果您尝试一下，您会发现这不是我想要的：

pivot_longer(data=dat_plot, cols=c(\"share_org_based\", \"share_ideo_based\", \"share_act_based\",
                    \"organizational_based\", \"action_based\", \"ideological_based\"),
             names_pattern = c(\"(share_[A-Za-z]+)([A-Za-z]+_based)\"),
             names_to = c(\"share\", \".value\"),
             values_to = \"value\")

我很感激任何关于names_pattern 工作方式的线索，或者我错过了什么。

您可能遇到麻烦的一个地方是您的列名不完全匹配，例如您希望 \"organizational\" 和 \"org\" 匹配。您可能还希望将列类型标记为共享或基于：现在您标记了共享，但没有标记其他。为此，您可能需要先重命名列。为什么你有行动与思想相结合，思想与行为相结合？

标签： r dataframe pivot-table tidyr data-manipulation

【解决方案1】：

您可以使用两个pivot_longers：

dat_plot %>% 
  pivot_longer(cols = starts_with("share"), names_to = "share", names_prefix = "share_", values_to = "share_value") %>%
  pivot_longer(cols = ends_with("based"), names_to = "based", names_pattern = "(.*)_based", values_to = "based_value") %>% 
  filter(substr(share, 1, 3) == substr(based, 1, 3))

输出

# A tibble: 6 x 5
   year share      share_value based          based_value
  <dbl> <chr>            <dbl> <chr>                <dbl>
1  1956 org_based            2 organizational           1
2  1956 ideo_based          95 ideological              0
3  1956 act_based           95 action                   0
4  2000 org_based           92 organizational           0
5  2000 ideo_based          87 ideological              0
6  2000 act_based           91 action                   0

【讨论】：

【解决方案2】：

或者另一个选项是来自sjmisc 的to_long

library(sjmisc)
dat_plot %>%
   to_long(keys = c("based", "share"), 
   values = c("value_based", "value_share"), names(.)[2:4], names(.)[5:7] ) 
  year                based value_based            share value_share
1 1956 organizational_based           1  share_org_based           2
2 2000 organizational_based           0  share_org_based          92
3 1956         action_based           0 share_ideo_based          95
4 2000         action_based           0 share_ideo_based          87
5 1956    ideological_based           0  share_act_based          95
6 2000    ideological_based           0  share_act_based          91

【讨论】：

【解决方案3】：

这是另一种tidyverse 方法：数据按based 的字母顺序排列：

library(dplyr)
library(tidyr)

dat_plot %>%
  pivot_longer(-year) %>% 
  arrange(name) %>% 
  mutate(name = str_remove_all(name, "share_")) %>% 
  group_by(group = rep(row_number(), each=6, length.out = n())) %>% 
  mutate(name = ifelse(group==1, str_remove_all(name, "_based"), name)) %>% 
  pivot_wider(names_from = group, values_from = c(name, value), values_fn = list) %>% 
  unnest(cols = c(name_1, name_2, value_1, value_2)) %>% 
  select(year, based=name_1, based_value=value_1, share=name_2, share_value=value_2)

 year based          based_value share      share_value
  <dbl> <chr>                <dbl> <chr>            <dbl>
1  1956 action                   0 act_based           95
2  1956 ideological              0 ideo_based          95
3  1956 organizational           1 org_based            2
4  2000 action                   0 act_based           91
5  2000 ideological              0 ideo_based          87
6  2000 organizational           0 org_based           92

【讨论】：