我有以下几点:
library(tidyverse)
df <- structure(list(`Input paired-end (STAR)` = c(9394981, 100), `Multi-mapped pair (STAR)` = c(1493691,
400), `Uniquely mapped paired-end (STAR)` = c(6826405, 200),
`Unmapped pair (STAR-appx)` = c(1074885, 300)), class = c("tbl_df",
"tbl", "data.frame"), row.names = 1:2, .Names = c("Input paired-end (STAR)",
"Multi-mapped pair (STAR)", "Uniquely mapped paired-end (STAR)",
"Unmapped pair (STAR-appx)"))
看起来像这样:
# A tibble: 2 × 4
`Input paired-end (STAR)` `Multi-mapped pair (STAR)` `Uniquely mapped paired-end (STAR)` `Unmapped pair (STAR-appx)`
* <dbl> <dbl> <dbl> <dbl>
1 9394981 1493691 6826405 1074885
2 100 400 200 300
如何重命名列:
foo bar qux gop
9394981 6826405 1074885 1493691
100 200 300 400
答案 0 :(得分:2)
一个简单的选项是setNames
,也可以合并到链中
library(dplyr)
df %>%
setNames(., c('foo', 'bar', 'qux', 'gop'))
答案 1 :(得分:2)
如果你真的想使用`dplyr函数,你可以使用重命名
df %>% rename(foo=`Input paired-end (STAR)`,
bar=`Multi-mapped pair (STAR)`,
qux=`Uniquely mapped paired-end (STAR)`,
qop=`Unmapped pair (STAR-appx)`)
但使用基本功能
会更容易names(df) <- c("foo","bar","qux","qop")