Question

我有一个具有这种结构的地图表：

    structure(list(REF_ID = structure(1:10, .Label = c("202533_s_at",  
"202534_x_at", "202551_s_at", "202552_s_at", "202555_s_at", "202565_s_at",  
"202566_s_at", "202580_x_at", "202581_at", "202589_at"), class = "factor"),  
GeneSymbol = structure(c(2L, 2L, 1L, 1L, 5L, 6L, 6L, 3L, 4L, 7L), .Label =  
c("CRIM1 /// LOC101929500", "DHFR", "FOXM1", "HSPA1A /// HSPA1B", "MYLK",  
"SVIL", "TYMS"), class = "factor")), .Names = c("REF_ID", "GeneSymbol"),  
class = "data.frame", row.names = c(NA, -10L))

在第3,4和9行中，有多个GeneSymbol与单个REF_ID匹配。（这里///是分隔符）。因此，在第3行中，两个基因符号与单个REF_ID匹配。

我想要一个修改过的表（包含所有现有的映射），这样REF_ID将重复多次，因为它与单独的基因符号匹配。
因此，我希望第3行包含两个单独的行：一行REF_ID == 202551_s_at和GeneSymbol == CRIM1，另一行REF_ID == 202551_s_at和GeneSymbol == LOC101929500。

请你帮帮我。

Answer 1

只是为了补充Rui Barradas的答案，一个整齐的方法可能是使用separate_rows()包中包含的tidyr：

library(tidyverse)
df %>% separate_rows(GeneSymbol, sep = " /// ")
#>         REF_ID   GeneSymbol
#> 1  202533_s_at         DHFR
#> 2  202534_x_at         DHFR
#> 3  202551_s_at        CRIM1
#> 4  202551_s_at LOC101929500
#> 5  202552_s_at        CRIM1
#> 6  202552_s_at LOC101929500
#> 7  202555_s_at         MYLK
#> 8  202565_s_at         SVIL
#> 9  202566_s_at         SVIL
#> 10 202580_x_at        FOXM1
#> 11   202581_at       HSPA1A
#> 12   202581_at       HSPA1B
#> 13   202589_at         TYMS

数据

df <- structure(list(REF_ID = structure(1:10, .Label = c("202533_s_at", "202534_x_at", "202551_s_at", "202552_s_at", "202555_s_at", "202565_s_at", "202566_s_at", "202580_x_at", "202581_at", "202589_at"), class = "factor"), GeneSymbol = structure(c(2L, 2L, 1L, 1L, 5L, 6L, 6L, 3L, 4L, 7L), .Label = c("CRIM1 /// LOC101929500", "DHFR", "FOXM1", "HSPA1A /// HSPA1B", "MYLK", "SVIL", "TYMS"), class = "factor")), .Names = c("REF_ID", "GeneSymbol"), class = "data.frame", row.names = c(NA, -10L))

Answer 2

以下是您想要的。它仅使用基数R，可能在tidyverse中有更简单的解决方案。

map$GeneSymbol <- as.character(map$GeneSymbol)

out <- lapply(seq_along(map$GeneSymbol), function(i){
    g <- map$GeneSymbol[i]
    if(grepl("///", g)){
        g <- trimws(unlist(strsplit(g, "///")))
        data.frame(REF_ID = rep(map$REF_ID[i], length(g)), GeneSymbol = g)
    } else {
        data.frame(REF_ID = map$REF_ID[i], GeneSymbol = g)
    }

})

map$GeneSymbol <- as.factor(map$GeneSymbol)

out <- do.call(rbind, out)
out
#        REF_ID   GeneSymbol
#1  202533_s_at         DHFR
#2  202534_x_at         DHFR
#3  202551_s_at        CRIM1
#4  202551_s_at LOC101929500
#5  202552_s_at        CRIM1
#6  202552_s_at LOC101929500
#7  202555_s_at         MYLK
#8  202565_s_at         SVIL
#9  202566_s_at         SVIL
#10 202580_x_at        FOXM1
#11   202581_at       HSPA1A
#12   202581_at       HSPA1B
#13   202589_at         TYMS

根据多个匹配修改映射表

2 个答案: