無法將串列列轉換為R中的因子

我正在嘗試使用資料來預測一些樂譜。其中一列是流派，它看起來像這樣：

流派專欄

c("['rock-and-roll', 'space age pop', 'surf music']", "['dance pop', 'pop', 'post-teen pop']", 
"['pop', 'post-teen pop']", "['country', 'country dawn', 'nashville sound']", 
"['australian country', 'contemporary country', 'country', 'country road']", 
"['blues rock', 'garage rock', 'modern blues rock', 'neo-psychedelic', 'nu gaze', 'punk blues']", 
"['pop', 'post-teen pop']", "['adult standards', 'brill building pop', 'folk', 'folk rock', 'mellow gold', 'singer-songwriter', 'soft rock', 'yacht rock']", 
"['adult standards', 'brill building pop', 'bubblegum pop', 'folk rock', 'lounge', 'mellow gold', 'rock-and-roll', 'rockabilly', 'sunshine pop']", 
"['adult standards', 'brill building pop', 'canadian pop', 'easy listening', 'lounge', 'rock-and-roll']", 
"[]", "['boston rock', 'dance rock', 'new romantic', 'new wave', 'new wave pop']", 
"['classic soul']", "['classic country pop', 'country', 'nashville sound', 'outlaw country', 'singer-songwriter', 'texas country']", 
"['adult standards', 'brill building pop', 'bubblegum pop', 'doo-wop', 'rock-and-roll', 'rockabilly']", 
"['brill building pop', 'doo-wop', 'rhythm and blues']", "[]", 
"['album rock', 'art rock', 'blues rock', 'classic rock', 'hard rock', 'metal', 'psychedelic rock', 'rock', 'soft rock']", 
"['blues', 'blues rock', 'classic rock', 'electric blues', 'folk rock', 'funk', 'jazz blues', 'louisiana blues', 'new orleans blues', 'piano blues', 'psychedelic rock', 'roots rock', 'soul']", 
"['album rock', 'canadian pop', 'canadian singer-songwriter', 'classic canadian rock', 'heartland rock', 'mellow gold', 'rock', 'soft rock']", 
"['art rock', 'dance rock', 'new romantic', 'new wave', 'new wave pop', 'permanent wave', 'rock', 'synthpop']", 
"['album rock', 'blues rock', 'classic rock', 'country rock', 'hard rock', 'mellow gold', 'rock', 'soft rock', 'southern rock']", 
"['adult standards', 'brill building pop', 'easy listening', 'lounge', 'rock-and-roll', 'rockabilly']", 
"['christmas instrumental']", "['adult standards', 'brill building pop', 'bubblegum pop', 'classic country pop', 'country rock', 'folk', 'folk rock', 'mellow gold', 'soft rock']", 
"['adult standards', 'brill building pop', 'chicago soul', 'classic soul', 'motown', 'quiet storm', 'rhythm and blues', 'rock-and-roll', 'rockabilly', 'soul']")

我想將其用作預測的因子變數（或虛擬因子變數）。如何從串列中提取流派名稱并將它們轉換為虛擬變數列？

當我將流派轉換為虛擬列時會發生什么：

我想要的是：

uj5u.com熱心網友回復：

一個整潔的解決方案

tidyr::separate_rows()以和為中心tidyr::pivot_longer()：

library(tidyr)
library(dplyr)
library(stringr)

gdata <- gdata %>% 
  mutate(
    id = row_number(),
    genre = na_if(str_remove_all(genre, "\\[|\\]|'"), ""),
    value = 1
  ) %>% 
  separate_rows(genre, sep = ", ") %>% 
  pivot_wider(names_from = genre, values_fill = 0) %>% 
  select(!`NA`)

gdata

# A tibble: 26 × 71
      id rock-and-r…1 space…2 surf …3 dance…?   pop post-…? country count…? nashv…?
   <int>        <dbl>   <dbl>   <dbl>   <dbl> <dbl>   <dbl>   <dbl>   <dbl>   <dbl>
 1     1            1       1       1       0     0       0       0       0       0
 2     2            0       0       0       1     1       1       0       0       0
 3     3            0       0       0       0     1       1       0       0       0
 4     4            0       0       0       0     0       0       1       1       1
 5     5            0       0       0       0     0       0       1       0       0
 6     6            0       0       0       0     0       0       0       0       0
 7     7            0       0       0       0     1       1       0       0       0
 8     8            0       0       0       0     0       0       0       0       0
 9     9            1       0       0       0     0       0       0       0       0
10    10            1       0       0       0     0       0       0       0       0
# … with 16 more rows, 61 more variables: `australian country` <dbl>,
#   `contemporary country` <dbl>, `country road` <dbl>, `blues rock` <dbl>,
#   `garage rock` <dbl>, `modern blues rock` <dbl>, `neo-psychedelic` <dbl>,
#   `nu gaze` <dbl>, `punk blues` <dbl>, `adult standards` <dbl>,
#   `brill building pop` <dbl>, folk <dbl>, `folk rock` <dbl>,
#   `mellow gold` <dbl>, `singer-songwriter` <dbl>, `soft rock` <dbl>,
#   `yacht rock` <dbl>, `bubblegum pop` <dbl>, lounge <dbl>, rockabilly <dbl>, …

基礎 R 解決方案

1. 對于每個值，洗掉無關字符并以逗號分隔。這為您提供了一個串列，其中每行包含一個字符向量。
2. 用于unique(unlist())獲取所有獨特流派的向量。
3. 回圈播放獨特的流派；對于每個，在您的資料框中添加一列，測驗該型別是否出現在每一行中。如果您更喜歡 0 和 1，可以在as.integer()此處添加。

genre_list <- sapply(
  gdata$genre, 
  \(x) strsplit(gsub("\\[|\\]|'", "", x), ", ")
)

all_genres <- unique(unlist(genre_list))

for (g in all_genres) {
  gdata[[g]] <- sapply(genre_list, \(x) g %in% x)
}

gdata[1:10, 2:8]

   rock-and-roll space age pop surf music dance pop   pop post-teen pop country
1           TRUE          TRUE       TRUE     FALSE FALSE         FALSE   FALSE
2          FALSE         FALSE      FALSE      TRUE  TRUE          TRUE   FALSE
3          FALSE         FALSE      FALSE     FALSE  TRUE          TRUE   FALSE
4          FALSE         FALSE      FALSE     FALSE FALSE         FALSE    TRUE
5          FALSE         FALSE      FALSE     FALSE FALSE         FALSE    TRUE
6          FALSE         FALSE      FALSE     FALSE FALSE         FALSE   FALSE
7          FALSE         FALSE      FALSE     FALSE  TRUE          TRUE   FALSE
8          FALSE         FALSE      FALSE     FALSE FALSE         FALSE   FALSE
9           TRUE         FALSE      FALSE     FALSE FALSE         FALSE   FALSE
10          TRUE         FALSE      FALSE     FALSE FALSE         FALSE   FALSE

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