将从文本列中提取的列表分离为多个数字列

Question

我有一个文本列，我想将其数字分成列：

df <- structure(list(text = c("((i: DH=0, SZ=0, RM=0, FO=0, GP=0, BC=0))", 
                              "((i: DH=0, SZ=0, RM=0, FO=0, GP=0, BC=0))", "((i: DH=1, SZ=0, RM=0, FO=NA, GP=0, BC=0))", 
                              "((i: DH=1, SZ=0, RM=NA, FO=0, GP=0, BC=0))", "((i: DH=1, SZ=1, RM=0, FO=0, GP=0, BC=0))", 
                              "((i: DH=0, SZ=1, RM=0, FO=0, GP=0, BC=0))", "((i: DH=0, SZ=0, RM=1, FO=0, GP=0, BC=1))", 
                              "((i: DH=0, SZ=1, RM=0, FO=0, GP=0, BC=0))", "((i: DH=0, SZ=1, RM=0, FO=0, GP=0, BC=0))", 
                              "((i: DH=0, SZ=1, RM=0, FO=1, GP=0, BC=0))", "((i: DH=0, SZ=1, RM=0, FO=0, GP=0, BC=1))", 
                              "((i: DH=0, SZ=NA, RM=0, FO=0, GP=0, BC=0))", "((i: DH=0, SZ=0, RM=0, FO=0, GP=0, BC=0))"
)), row.names = c(NA, -13L), class = "data.frame")

我可以 str_extractthe 数字，但无法使用 tidyr 的函数 separate 将数字分成数字列：

library(stringr)
library(tidyr)
library(dplyr)
df %>%
  mutate(text = str_extract_all(text,"(?<==)\\d|NA(?=,)")) %>%
  separate(col = text,
           into = c("DH", "SZ", "RM", "FO", "GP", "BC"),
           sep = ", ",
           convert = TRUE)
      DH   SZ   RM   FO  GP   BC
1  c("0"  "0"  "0"  "0" "0" "0")
2  c("0"  "0"  "0"  "0" "0" "0")
3  c("1"  "0"  "0" "NA" "0" "0")
4  c("1"  "0" "NA"  "0" "0" "0")
5  c("1"  "1"  "0"  "0" "0" "0")
6  c("0"  "1"  "0"  "0" "0" "0")
7  c("0"  "0"  "1"  "0" "0" "1")
8  c("0"  "1"  "0"  "0" "0" "0")
9  c("0"  "1"  "0"  "0" "0" "0")
10 c("0"  "1"  "0"  "1" "0" "0")
11 c("0"  "1"  "0"  "0" "0" "1")
12 c("0" "NA"  "0"  "0" "0" "0")
13 c("0"  "0"  "0"  "0" "0" "0")

我的直觉是，这是因为str_extract_all 返回 列表 中的数字，而 separate 不适用于列表。这里有什么解决办法？

Answer 1

您可以使用unnest_wider 代替separate，然后做一些额外的 tidyverse 魔术。注意：您会收到警告，因为 NA 是由强制引入的，但在您的情况下这是预期的。

df %>%
  mutate(text = str_extract_all(text,"(?<==)\\d|NA(?=,)")) %>%
  unnest_wider(text) %>%
  mutate(across(starts_with("..."), ~as.numeric(.))) %>%
  rename_with(.cols = starts_with("..."),
              .fn = ~c("DH", "SZ", "RM", "FO", "GP", "BC"))

给出：

# A tibble: 13 x 6
      DH    SZ    RM    FO    GP    BC
   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
 1     0     0     0     0     0     0
 2     0     0     0     0     0     0
 3     1     0     0    NA     0     0
 4     1     0    NA     0     0     0
 5     1     1     0     0     0     0
 6     0     1     0     0     0     0
 7     0     0     1     0     0     1
 8     0     1     0     0     0     0
 9     0     1     0     0     0     0
10     0     1     0     1     0     0
11     0     1     0     0     0     1
12     0    NA     0     0     0     0
13     0     0     0     0     0     0

---

另一种方法（在重命名列方面）是直接在列表中设置名称属性，这解决了所有重命名问题：

library(tidyverse)
df %>%
  mutate(text = str_extract_all(text,"(?<==)\\d|NA(?=,)"),
         text = map(text, setNames, c("DH", "SZ", "RM", "FO", "GP", "BC"))) %>%
  unnest_wider(text) %>%
  mutate(across(c("DH", "SZ", "RM", "FO", "GP", "BC"), ~as.numeric(.)))

Answer 2

这是simplify=T 参数的目的，因为它将输出强制转换为矩阵。

df %>%
  mutate(text = str_extract_all(text,"(?<==)\\d|NA(?=,)", simplify=T))

   text.1 text.2 text.3 text.4 text.5 text.6
1       0      0      0      0      0      0
2       0      0      0      0      0      0
3       1      0      0     NA      0      0
4       1      0     NA      0      0      0
5       1      1      0      0      0      0
6       0      1      0      0      0      0
7       0      0      1      0      0      1
8       0      1      0      0      0      0
9       0      1      0      0      0      0
10      0      1      0      1      0      0
11      0      1      0      0      0      1
12      0     NA      0      0      0      0
13      0      0      0      0      0      0