pivot_longer 与 names_pattern 和成对的列

Question

我试图弄清楚如何在以下示例中使用 tidyr 中的 pivot_longer。原来的 table 称为 dat_plot 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"
))

我想把它变成长格式，方法如下：

  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，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))

我想我必须和 names_pattern 一起工作，我试过的是这样的，但是如果你尝试你会发现，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")

对于 names_pattern 工作原理或我遗漏了什么的任何线索，我深表感谢。

Answer 1

或者另一个选项是 to_long 来自 sjmisc

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

Answer 2

这是 tidyr 的另一种方式：

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

Answer 3

这是另一种 tidyverse 方法：数据按字母顺序排列 based:

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

pivot_longer 与 names_pattern 和成对的列

pivot_longer with names_pattern and pairs of columns

pivot-table

r

data-manipulation

dataframe

tidyr