r - grepl 与比赛

Question

我在使用 dplyr 包中的 matches 函数时遇到了一些小问题。

从这个数据集中，我想提取以 enj

开头的列名

    enj1 enj2 Enjm
bbc    1    1    2
bca    1    1    2

有了grepl，我可以做到

dt[, grepl('enj', colnames(dt))]

并获得

    enj1 enj2
bbc    1    1
bca    1    1

但是函数 matches 没有给我正确答案

library(dplyr) 

dt %>% select(matches('enj') )
# or 
dt %>% select(matches('^enj') )

知道为什么吗？

dt = structure(list(enj1 = structure(c(1L, 1L), .Names = c("bbc", 
"bca"), .Label = "1", class = "factor"), enj2 = structure(c(1L, 
1L), .Names = c("bbc", "bca"), .Label = "1", class = "factor"), 
    Enjm = structure(c(1L, 1L), .Names = c("bbc", "bca"), .Label = "2", class = "factor")), .Names = c("enj1", 
"enj2", "Enjm"), row.names = c("bbc", "bca"), class = "data.frame")

Answer 1

因为你没有设置ignore.case = F。

> dt %>% select(matches('^enj',  ignore.case = F) )
    enj1 enj2
bbc    1    1
bca    1    1
>

r - grepl 与比赛

r - grepl vs matches

r

dplyr

grepl