如何通过匹配来自另一个数据帧的整个列中的字符串来检索一个数据帧中的值？

Question

假设我有一个数据框df1，如下所示：

> df1
          probe                         OMIM
1  1565034_s_at                       601464
2     201000_at 601065 /// 613287 /// 616339
3     204565_at                       615652
4     205355_at            600301 /// 610006
5   205734_s_at                       601464
6   205735_s_at                       601464
7     206527_at            137150 /// 613163
8     209173_at                       606358
9   209459_s_at            137150 /// 613163
10    209460_at            137150 /// 613163
11    215465_at                             
12    223864_at                       610856
13    224742_at            612674 /// 613599

还有第二个数据框，df2：

> df2
                                         platprobe   symbol
1   1565034_s_at,205734_s_at,242078_at,205735_s_at     AFF3
2                                        201000_at     AARS
3                                        201884_at   DNALI1
4                                      202779_s_at     PLK1
5                                        204565_at   ACOT13
6                              205355_at,226030_at   ACADSB
7      205808_at,207284_s_at,209135_at,210896_s_at   LIMCH1
8      206164_at,206165_s_at,206166_s_at,217528_at   SLC7A8
9                  206527_at,209459_s_at,209460_at     ABAT
10                             209173_at,228969_at     AGR2
11                                       215465_at   ABCA12
12                                     221024_s_at  TMEM144
13                                       223864_at ANKRD30A
14                 224742_at,228123_s_at,228124_at   ABHD12
15                           225421_at,225431_x_at   GALNT7
16                                       226120_at    PSAT1
17                                       228241_at     AGR3

我想在df1$probe 值与df2$platprobe 匹配的基础上，向df1、df1$symbol 添加一个新列。结果应该是这样的：

> df1
          probe                         OMIM    symbol
1  1565034_s_at                       601464      AFF3
2     201000_at 601065 /// 613287 /// 616339      AARS
3     204565_at                       615652    ACOT13
4     205355_at            600301 /// 610006    ACADSB
5   205734_s_at                       601464      AFF3
6   205735_s_at                       601464      AFF3
7     206527_at            137150 /// 613163      ABAT
8     209173_at                       606358      AGR2
9   209459_s_at            137150 /// 613163      ABAT
10    209460_at            137150 /// 613163      ABAT
11    215465_at                                 ABCA12
12    223864_at                       610856  ANKRD30A
13    224742_at            612674 /// 613599    ABHD12

对我来说具有挑战性的部分是df2$platprobe 在许多情况下包含除了in df1$probe 之外的各种注释。所以，如果我尝试：

#This will retrieve only perfect matches (where df2$platprobe contains only one possible value, such as ABCA12):
df1$symbol <- df2$symbol[df2$probe %in% df1$platprobe]

#And if I use 'grepl', that won't work:
#(The reason for using 'unlist' and 'strsplit' is because I thought that maybe breaking all possible
#values from the entire df2$platprobe into a object that would work. But it doesn't)

df1$symbol <- df2$symbol[grepl(df1$probe, unlist(strsplit(paste(df2$platprobe, sep=",", collapse=","), ",")))]

非常感谢任何帮助。

PS：另外，如果您对更多主题的标题有更好的想法，非常欢迎。

更新 谢谢@Anoushiravan R。很抱歉之前没有放可重现的df。现在，他们在这里：

df1 <- data.frame(probe=c("1565034_s_at", "201000_at", "204565_at", 
"205355_at", "205734_s_at", "205735_s_at", "206527_at", "209173_at", 
"209459_s_at", "209460_at", "215465_at", "223864_at", "224742_at"
), OMIM = c("601464", "601065 /// 613287 /// 616339", "615652", 
"600301 /// 610006", "601464", "601464", "137150 /// 613163", 
"606358", "137150 /// 613163", "137150 /// 613163", "", "610856", 
"612674 /// 613599"))

df2 <- data.frame(platprobe = c("1565034_s_at, 205734_s_at, 205735_s_at, 
227198_at, 242078_at, 243967_at", "201000_at", "201884_at", "202779_s_at",
"204565_at", "205355_at,226030_at", "205808_at, 207284_s_at, 209135_at, 
210896_s_at, 224996_at, 225008_at, 242037_at", "206164_at, 206165_s_at, 
206166_s_at, 217528_at", "206527_at, 209459_s_at,209460_at", "209173_at, 
228969_at", "215465_at", "221024_s_at", "223864_at","224742_at, 228123_s_at, 
228124_at", "225421_at,225431_x_at", "226120_at", "228241_at"), symbol=c("AFF3", 
"AARS", "DNALI1", "PLK1", "ACOT13", "ACADSB", "LIMCH1", "SLC7A8", "ABAT", "AGR2", 
"ABCA12", "TMEM144", "ANKRD30A", "ABHD12", "GALNT7", "PSAT1", "AGR3"))

Answer 1

虽然上面的答案可以达到目的，但还表明没有purrr也可以完成

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

df1 %>% left_join(df2 %>% separate_rows(platprobe, sep = ',') %>%
                    mutate(platprobe = str_trim(platprobe)), by = c('probe' = 'platprobe'))

          probe                         OMIM   symbol
1  1565034_s_at                       601464     AFF3
2     201000_at 601065 /// 613287 /// 616339     AARS
3     204565_at                       615652   ACOT13
4     205355_at            600301 /// 610006   ACADSB
5   205734_s_at                       601464     AFF3
6   205735_s_at                       601464     AFF3
7     206527_at            137150 /// 613163     ABAT
8     209173_at                       606358     AGR2
9   209459_s_at            137150 /// 613163     ABAT
10    209460_at            137150 /// 613163     ABAT
11    215465_at                                ABCA12
12    223864_at                       610856 ANKRD30A
13    224742_at            612674 /// 613599   ABHD12

Answer 2

您可以使用以下解决方案：

library(dplyr)
library(stringr)
library(purrr)

df1 %>%
  mutate(symbol = map_chr(probe, ~ df2$symbol[which(str_detect(df2$platprobe, .x))]))


          probe                         OMIM   symbol
1  1565034_s_at                       601464     AFF3
2     201000_at 601065 /// 613287 /// 616339     AARS
3     204565_at                       615652   ACOT13
4     205355_at            600301 /// 610006   ACADSB
5   205734_s_at                       601464     AFF3
6   205735_s_at                       601464     AFF3
7     206527_at            137150 /// 613163     ABAT
8     209173_at                       606358     AGR2
9   209459_s_at            137150 /// 613163     ABAT
10    209460_at            137150 /// 613163     ABAT
11    215465_at                                ABCA12
12    223864_at                       610856 ANKRD30A
13    224742_at            612674 /// 613599   ABHD12

Answer 3

处理您的问题的另一种方法是基于您的观点/观察，即您的匹配键在第二个数据帧中“折叠”。

{tidyr} 具有将嵌套值拆分为新行的强大功能，即tidyr()::separate_rows()。这会将您的第二个 df 转换为长格式。

注意：separate_rows() 允许在需要时拆分多个列。 但是这里我们只使用你的密钥platprobe。

library(dplyr)   # data crunching 
library(tidyr)   # data manipulation for generating tidy df

# how to separate the nested column values to rows
df2 %>% separate_rows(platprobe, sep = ",")

检查行分布：

# A tibble: 33 x 2
   platprobe    symbol
   <chr>        <chr> 
 1 1565034_s_at AFF3  
 2 205734_s_at  AFF3  
 3 242078_at    AFF3  
 4 205735_s_at  AFF3  
 5 201000_at    AARS  
...

您现在已正确对齐匹配键并执行left_join() 以合并两个数据框。

# merging the "long" lookup df2 with df1
df1 %>% left_join(
     df2 %>% separate_rows(platprobe, sep = ",")
   , by = c("probe" = "platprobe")    # define matching keys in df1 and df2
)

这交付

          probe   symbol
1  1565034_s_at     AFF3
2     201000_at     AARS
3     204565_at   ACOT13
4     205355_at   ACADSB
...

Answer 4

如果您想使用grep 进行匹配，您可以通过sapply 或lapply 执行此操作。

df1$symbol <- df2$symbol[sapply(df1$probe, grep, df2$platprobe)]

df1
#          probe                         OMIM   symbol
#1  1565034_s_at                       601464     AFF3
#2     201000_at 601065 /// 613287 /// 616339     AARS
#3     204565_at                       615652   ACOT13
#4     205355_at            600301 /// 610006   ACADSB
#5   205734_s_at                       601464     AFF3
#6   205735_s_at                       601464     AFF3
#7     206527_at            137150 /// 613163     ABAT
#8     209173_at                       606358     AGR2
#9   209459_s_at            137150 /// 613163     ABAT
#10    209460_at            137150 /// 613163     ABAT
#11    215465_at                                ABCA12
#12    223864_at                       610856 ANKRD30A
#13    224742_at            612674 /// 613599   ABHD12