Question

我有一个包含多个向量的列表，如下所示：

$`56`
[1] "OTU2998"             "UniRef90_A0A1Z9FS94" "UniRef90_A0A257ESC3"
[4] "UniRef90_A0A293NAV3" "UniRef90_A0A2E1NMU8" "UniRef90_A0A2E1NPX9"
[7] "UniRef90_A0A2E1NQL1" "UniRef90_A0A2E1NRD2" "UniRef90_X0UC66"    

$`57`
[1] "OTU3820"             "UniRef90_A0A1Z9H3N2" "UniRef90_A0A2D5I161"
[4] "UniRef90_A0A2E6PRN5"

$`58`
[1] "OTU4452"                "UniRef90_A0A1Z9KBI8"    "UniRef90_A0A2E1VTI6"   
[4] "UniRef90_A0A2G2KCN6"    "UniRef90_UPI000BFEC744"

$`59`
[1] "OTU0245"             "UniRef90_A0A1Z9MPM9" "UniRef90_A0A2E2ME98"
[4] "UniRef90_A0A2E8X9N7"

是否可以仅提取“ OTUXXX”信息？我的意思是，我想得到这样的东西：

$`56`
[1] "OTU2998"       

$`57`
[1] "OTU3820"  

$`58`
[1] "OTU4452"   

$`59`
[1] "OTU0245"

Answer 1

我喜欢purrr::map函数家族，因为它们易于传递函数和参数。提取这些元素的两个快速选择是使用grep使用value = T返回匹配的字符串，而不仅仅是返回它们的索引，或者使用stringr::str_subset返回相同的字符串。

此处的正则表达式匹配以“ OTU”开头，后跟1个或多个数字的字符串。

这两种方法一次都可以缩放多个匹配项：我在最后一个列表元素中添加了一个项目“ OTU1234”来说明这一点。

dl <- list(
  `56` = c("OTU2998", "UniRef90_A0A1Z9FS94", "UniRef90_A0A257ESC3", "UniRef90_A0A293NAV3", "UniRef90_A0A2E1NMU8", "UniRef90_A0A2E1NPX9", "UniRef90_A0A2E1NQL1", "UniRef90_A0A2E1NRD2", "UniRef90_X0UC66"),
  `57` = c("OTU3820", "UniRef90_A0A1Z9H3N2", "UniRef90_A0A2D5I161", "UniRef90_A0A2E6PRN5"),
  `58` = c("OTU4452", "UniRef90_A0A1Z9KBI8", "UniRef90_A0A2E1VTI6", "UniRef90_A0A2G2KCN6", "UniRef90_UPI000BFEC744"),
  `59` = c("OTU0245", "UniRef90_A0A1Z9MPM9", "UniRef90_A0A2E2ME98", "UniRef90_A0A2E8X9N7", "OTU1234")
)

purrr::map(dl, ~grep("^OTU\\d+$", ., value = T))
#> $`56`
#> [1] "OTU2998"
#> 
#> $`57`
#> [1] "OTU3820"
#> 
#> $`58`
#> [1] "OTU4452"
#> 
#> $`59`
#> [1] "OTU0245" "OTU1234"
purrr::map(dl, stringr::str_subset, "^OTU\\d+$")
# same output as above

Answer 2

我们可以遍历list并提取与字符串开头（^）的子字符串'OTU'相匹配的元素，后跟四位数字（\\d{4}），直到带有$

的字符串的结尾（grepl）

lapply(lst1, function(x) x[grepl("^OTU\\d{4}$", x)])
#$`56`
#[1] "OTU2998"

#$`57`
#[1] "OTU3820"

#$`58`
#[1] "OTU4452"

#$`59`
#[1] "OTU0245" "OTU1234"

注意：仅使用base R方法

或者，如果我们是整洁的迷，请使用keep

library(tidyverse)
map(lst1, keep, str_detect, '^OTU\\d{4}$')

数据

lst1 <-  list(
  `56` = c("OTU2998", "UniRef90_A0A1Z9FS94", "UniRef90_A0A257ESC3", "UniRef90_A0A293NAV3", "UniRef90_A0A2E1NMU8", "UniRef90_A0A2E1NPX9", "UniRef90_A0A2E1NQL1", "UniRef90_A0A2E1NRD2", "UniRef90_X0UC66"),
  `57` = c("OTU3820", "UniRef90_A0A1Z9H3N2", "UniRef90_A0A2D5I161", "UniRef90_A0A2E6PRN5"),
  `58` = c("OTU4452", "UniRef90_A0A1Z9KBI8", "UniRef90_A0A2E1VTI6", "UniRef90_A0A2G2KCN6", "UniRef90_UPI000BFEC744"),
  `59` = c("OTU0245", "UniRef90_A0A1Z9MPM9", "UniRef90_A0A2E2ME98", "UniRef90_A0A2E8X9N7", "OTU1234")
)

如何在R中提取列表的元素？

2 个答案:

数据