如何將包含“0/0”的字符列轉換為數字

Question

我有一個數據框，所有列都歸類為一個字符。 每列包含一個分數。 我想將列轉換為整數，但有些列有分數“0/0”，R 似乎不喜歡。 我嘗試了以下但得到了

df2 <- as.numeric(df)

並得到以下

Error: (list) object cannot be coerced to type 'double'

我找不到解釋如何將包含“0/0”小數部分的字符轉換為數值 0 的帖子。我意識到 R 給我帶來麻煩是有原因的由零。 我只是在處理遺傳數據，將數據轉換為數字並將所有內容相加比進行某種替換 function 容易得多。 實際的數據框是 10 幾百萬行和 500+ 行。

這是一個示例數據框

df <- structure(list(`GEN[5].GT` = c("0/1", "0/0", "0/0", "0/0", 
 "0/1", "0/0", "0/0", "1/1", "0/0", "0/0"), `GEN[1].GT` = c("0/0", 
 "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0"
 ), `GEN[6].GT` = c("1/1", "0/0", "0/0", "0/0", "0/0", "0/0", 
 "0/1", "0/0", "0/0", "0/0"), `GEN[9].GT` = c("0/0", "0/0", 
 "0/0", "0/0", "0/1", "0/0", "0/0", "0/1", "0/0", "0/0"), `GEN[89].GT` = c("0/0", 
 "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0"
 ), `GEN[453].GT` = c("0/0", "0/0", "0/1", "0/0", "0/0", "0/0", 
 "0/0", "0/0", "0/0", "0/0"), `GEN[554].GT` = c("0/0", "0/0", 
 "0/0", "0/0", "0/0", "0/0", "1/1", "0/0", "0/0", "0/0"), `GEN[9864].GT` = c("0/0", 
 "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0", "0/0"
 ), `GEN[1234].GT` = c("1/1", "0/0", "0/0", "0/0", "0/0", "0/0", 
 "0/0", "0/0", "0/0", "0/0"), `GEN[3333].GT` = c("0/0", "0/0", 
 "0/0", "0/0", "0/0", "1/1", "0/0", "0/1", "0/0", "0/0")), row.names = c(NA, 
 10L), class = "data.frame")

# Expected output
 df2 <- structure(list(`GEN[5].GT` = c("0.5", "0", "0", "0", 
 "0.5", "0", "0", "1", "0", "0"), `GEN[1].GT` = c("0", 
 "0", "0", "0", "0", "0", "0", "0", "0", "0"
 ), `GEN[6].GT` = c("1", "0", "0", "0", "0", "0", 
 "0.5", "0", "0", "0"), `GEN[9].GT` = c("0", "0", 
 "0", "0", "0.5", "0", "0", "0.5", "0", "0"), `GEN[89].GT` = c("0", 
 "0", "0", "0", "0", "0", "0", "0", "0", "0"
 ), `GEN[453].GT` = c("0", "0", "0.5", "0", "0", "0", 
 "0", "0", "0", "0"), `GEN[554].GT` = c("0", "0", 
 "0", "0", "0", "0", "1", "0", "0", "0"), `GEN[9864].GT` = c("0", 
 "0", "0", "0", "0", "0", "0", "0", "0", "0"
 ), `GEN[1234].GT` = c("1", "0", "0", "0", "0", "0", 
 "0", "0", "0", "0"), `GEN[3333].GT` = c("0", "0", 
 "0", "0", "0", "1", "0", "0.5", "0", "0")), row.names = c(NA, 
 10L), class = "data.frame")

Answer 1

我們可以創建一個行名列（來自rownames_to_column的tibble ），然后在分隔符（ / ）處使用separate_rows的行分割每一列，同時自動convert類型，按 'rn' 分組，得到每一列的mean

library(dplyr)
library(tibble)
library(tidyr)
df %>%
   rownames_to_column('rn') %>%
   separate_rows(-1, convert = TRUE) %>%
   group_by(rn) %>% 
   summarise_all(mean) %>%
   select(-rn)
# A tibble: 10 x 10
#   `GEN[5].GT` `GEN[1].GT` `GEN[6].GT` `GEN[9].GT` `GEN[89].GT` `GEN[453].GT` `GEN[554].GT` `GEN[9864].GT` `GEN[1234].GT` `GEN[3333].GT`
#         <dbl>       <dbl>       <dbl>       <dbl>        <dbl>         <dbl>         <dbl>          <dbl>          <dbl>          <dbl>
# 1         0.5           0         1           0              0           0               0              0              1            0  
# 2         0             0         0           0              0           0               0              0              0            0  
# 3         0             0         0           0              0           0               0              0              0            0  
# 4         0             0         0           0              0           0.5             0              0              0            0  
# 5         0             0         0           0              0           0               0              0              0            0  
# 6         0.5           0         0           0.5            0           0               0              0              0            0  
# 7         0             0         0           0              0           0               0              0              0            1  
# 8         0             0         0.5         0              0           0               1              0              0            0  
# 9         1             0         0           0.5            0           0               0              0              0            0.5
#10         0             0         0           0              0           0               0              0              0            0  

或者@IceCreamToucan 在評論中提到的另一個選項，它用strsplit拆分單元格並取mean

library(purrr)    
df %>% 
   mutate_all(~ map_dbl(strsplit(., '/'), ~ mean(as.numeric(.))))

---

或者另一個可能更有效的選擇（在base R中）是傳遞一個命名向量，通過匹配替換值

nm1 <- setNames(c(0, 0.5, 0.5, 1), c("0/0", "1/0", "0/1", "1/1"))
df[] <- lapply(df, function(x) nm1[x])
df
#   GEN[5].GT GEN[1].GT GEN[6].GT GEN[9].GT GEN[89].GT GEN[453].GT GEN[554].GT GEN[9864].GT GEN[1234].GT GEN[3333].GT
#1        0.5         0       1.0       0.0          0         0.0           0            0            1          0.0
#2        0.0         0       0.0       0.0          0         0.0           0            0            0          0.0
#3        0.0         0       0.0       0.0          0         0.5           0            0            0          0.0
#4        0.0         0       0.0       0.0          0         0.0           0            0            0          0.0
#5        0.5         0       0.0       0.5          0         0.0           0            0            0          0.0
#6        0.0         0       0.0       0.0          0         0.0           0            0            0          1.0
#7        0.0         0       0.5       0.0          0         0.0           1            0            0          0.0
#8        1.0         0       0.0       0.5          0         0.0           0            0            0          0.5
#9        0.0         0       0.0       0.0          0         0.0           0            0            0          0.0
#10       0.0         0       0.0       0.0          0         0.0           0            0            0          0.0

Answer 2

我們可以使用gsub來截取兩邊並將它們放在括號內，將/替換為+ ，然后除以 2。

library(dplyr)
df %>% 
   rowwise() %>% 
   #try mutate_all(~gsub('(.*)/(.*)','(\\1+\\2)/2',.)) to see the underlying formula 
   mutate_all(~eval(parse(text=gsub('(.*)/(.*)','(\\1+\\2)/2',.))))  %>% 
   ungroup()

# A tibble: 10 x 10
   `GEN[5].GT` `GEN[1].GT` `GEN[6].GT` `GEN[9].GT` `GEN[89].GT` `GEN[453].GT` `GEN[554].GT` `GEN[9864].GT` `GEN[1234].GT`
         <dbl>       <dbl>       <dbl>       <dbl>        <dbl>         <dbl>         <dbl>          <dbl>          <dbl>
 1         0.5           0         1           0              0           0               0              0              1
 2         0             0         0           0              0           0               0              0              0
 3         0             0         0           0              0           0.5             0              0              0
 4         0             0         0           0              0           0               0              0              0
 5         0.5           0         0           0.5            0           0               0              0              0
 6         0             0         0           0              0           0               0              0              0
 7         0             0         0.5         0              0           0               1              0              0
 8         1             0         0           0.5            0           0               0              0              0
 9         0             0         0           0              0           0               0              0              0
10         0             0         0           0              0           0               0              0              0
# ... with 1 more variable: `GEN[3333].GT` <dbl>