假设我有一个如下所示的数据框 df1:

> 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

我想向 df1 添加一个新列，df1$symbol，基于 df1$probe 值与 的匹配df2$platprobe。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$platprobe 在许多情况下包含各种注释，而不是在 in df1$probe 中找到的注释。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，如果您对更多主题标题有更好的想法，非常欢迎。

更新 谢谢@Anoushiravan R。很抱歉没有把可重现的 df 放在前面。现在，它们在这里:

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"))

最佳答案

虽然上面的答案达到了目的，但还表明它可以在没有 purrr 的情况下完成

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

https://stackoverflow.com/questions/67397619/