r - 如何将列表值列拆分为多列？

Question

我有以下情况，该列power_dbm0的值作为列表。所有元素都是长度为 11 的列表。

# A tibble: 10 x 2
   real_pat power_dbm0
   <chr>    <list>    
 1 am       <dbl [11]>
 2 fax      <dbl [11]>
 3 fp       <dbl [11]>
 4 fpw      <dbl [11]>

我想知道如何拆分这些值，因为每个订单都是一个新列。最好，我想要一个类似 dplyr 的解决方案。我尝试了一些带有unnest或separate功能的解决方案tidyr，但没有成功。

提前致谢，

关注数据：

structure(list(real_pat = c("am", "fax", "fp", "fpw"), power_dbm0 = list(
    structure(c(0.0142857142857143, 0.0742857142857143, 0.111428571428571, 
    0.138571428571429, 0.208571428571429, 0.278571428571429, 
    0.368571428571429, 0.508571428571429, 0.648571428571429, 
    0.771428571428571, 0.871428571428571), .Names = c("0%", "10%", 
    "20%", "30%", "40%", "50%", "60%", "70%", "80%", "90%", "100%"
    )), structure(c(0.342857142857143, 0.342857142857143, 0.342857142857143, 
    0.342857142857143, 0.342857142857143, 0.342857142857143, 
    0.342857142857143, 0.342857142857143, 0.342857142857143, 
    0.342857142857143, 0.342857142857143), .Names = c("0%", "10%", 
    "20%", "30%", "40%", "50%", "60%", "70%", "80%", "90%", "100%"
    )), structure(c(0.0142857142857143, 0.622857142857143, 0.808571428571429, 
    0.851428571428571, 0.857142857142857, 0.871428571428571, 
    0.874285714285714, 0.885714285714286, 0.894285714285714, 
    0.911428571428571, 0.914285714285714), .Names = c("0%", "10%", 
    "20%", "30%", "40%", "50%", "60%", "70%", "80%", "90%", "100%"
    )), structure(c(0.514285714285714, 0.514285714285714, 0.514285714285714, 
    0.514285714285714, 0.514285714285714, 0.514285714285714, 
    0.514285714285714, 0.514285714285714, 0.514285714285714, 
    0.514285714285714, 0.514285714285714), .Names = c("0%", "10%", 
    "20%", "30%", "40%", "50%", "60%", "70%", "80%", "90%", "100%"
    )))), .Names = c("real_pat", "power_dbm0"), row.names = c(NA, 
-4L), class = c("tbl_df", "tbl", "data.frame"))

标签： rlistsplitdplyr

Answer 1

1）这是一个单行基础解决方案：

with(dd, do.call("rbind", setNames(power_dbm0, real_pat)))

给予：

            0%        10%       20%       30%       40%       50%       60%
am  0.01428571 0.07428571 0.1114286 0.1385714 0.2085714 0.2785714 0.3685714
fax [phone].34285714 0.3428571 0.3428571 0.3428571 0.3428571 0.3428571
fp  0.01428571 0.62285714 0.8085714 0.8514286 0.8571429 0.8714286 0.8742857
fpw 0.51428571 0.51428571 0.5142857 0.5142857 0.5142857 0.5142857 0.5142857
          70%       80%       90%      100%
am  0.5085714 0.6485714 0.7714286 0.8714286
fax [phone].3428571 0.3428571 0.3428571
fp  0.8857143 0.8942857 0.9114286 0.9142857
fpw 0.5142857 0.5142857 0.5142857 0.5142857

2）或获取real_pat作为列而不是名称：

with(dd, data.frame(real_pat, do.call("rbind", power_dbm0), check.names = FALSE))

给予：

  real_pat         0%        10%       20%       30%       40%       50%
1       am 0.01428571 0.07428571 0.1114286 0.1385714 0.2085714 0.2785714
2      fax [phone].34285714 0.3428571 0.3428571 0.3428571 0.3428571
3       fp 0.01428571 0.62285714 0.8085714 0.8514286 0.8571429 0.8714286
4      fpw 0.51428571 0.51428571 0.5142857 0.5142857 0.5142857 0.5142857
        60%       70%       80%       90%      100%
1 0.3685714 0.5085714 0.6485714 0.7714286 0.8714286
2 0.3428571 0.3428571 0.3428571 0.3428571 0.3428571
3 0.8742857 0.8857143 0.8942857 0.9114286 0.9142857
4 0.5142857 0.5142857 0.5142857 0.5142857 0.5142857

3）使用 dplyr 我们可以这样写：

library(dplyr)
dd %>% { bind_cols(select(., real_pat), bind_rows(!!!.$power_dbm0)) }

给予：

# A tibble: 4 x 12
  real_pat   `0%`  `10%` `20%` `30%` `40%` `50%` `60%` `70%` `80%` `90%` `100%`
  <chr>     <dbl>  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>
1 am       0.0143 0.0743 0.111 0.139 0.209 0.279 0.369 0.509 0.649 0.771  0.871
2 fax      0.343  0.343  0.343 0.343 0.343 0.343 0.343 0.343 0.343 0.343  0.343
3 fp       0.0143 0.623  0.809 0.851 0.857 0.871 0.874 0.886 0.894 0.911  0.914
4 fpw      0.514  0.514  0.514 0.514 0.514 0.514 0.514 0.514 0.514 0.514  0.514

3a)或使用magrittr的.id=参数：bind_rows%$%

library(dplyr)
library(magrittr)

dd %$%
   setNames(power_dbm0, real_pat) %$%
   bind_rows(!!!., .id = "real_pat")

给予：

# A tibble: 4 x 12
  real_pat   `0%`  `10%` `20%` `30%` `40%` `50%` `60%` `70%` `80%` `90%` `100%`
  <chr>     <dbl>  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>
1 am       0.0143 0.0743 0.111 0.139 0.209 0.279 0.369 0.509 0.649 0.771  0.871
2 fax      0.343  0.343  0.343 0.343 0.343 0.343 0.343 0.343 0.343 0.343  0.343
3 fp       0.0143 0.623  0.809 0.851 0.857 0.871 0.874 0.886 0.894 0.911  0.914
4 fpw      0.514  0.514  0.514 0.514 0.514 0.514 0.514 0.514 0.514 0.514  0.514

3b)或没有%$%：

library(dplyr)

dd %>%
   { setNames(.$power_dbm0, .$real_pat) } %>%
   { bind_rows(!!!., .id = "real_pat") }

给予：

# A tibble: 4 x 12
  real_pat   `0%`  `10%` `20%` `30%` `40%` `50%` `60%` `70%` `80%` `90%` `100%`
  <chr>     <dbl>  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>
1 am       0.0143 0.0743 0.111 0.139 0.209 0.279 0.369 0.509 0.649 0.771  0.871
2 fax      0.343  0.343  0.343 0.343 0.343 0.343 0.343 0.343 0.343 0.343  0.343
3 fp       0.0143 0.623  0.809 0.851 0.857 0.871 0.874 0.886 0.894 0.911  0.914
4 fpw      0.514  0.514  0.514 0.514 0.514 0.514 0.514 0.514 0.514 0.514  0.514