使用 dplyr 有条件地替换 tibble 中的列名
Conditional replacement of column name in tibble using dplyr
我有以下问题：
df <- structure(list(gene_symbol = c("0610005C13Rik", "0610007P14Rik",
"0610009B22Rik", "0610009L18Rik", "0610009O20Rik", "0610010B08Rik"
), foo.control.cv = c(1.16204038288333, 0.120508045270669, 0.205712615954009,
0.504508040948641, 0.333956330117591, 0.543693011377001), foo.control.mean = c(2.66407458486012,
187.137728870855, 142.111269303428, 16.7278587043453, 69.8602872478098,
4.77769028710622), foo.treated.cv = c(0.905769898934564, 0.186441944401973,
0.158552512842753, 0.551955061149896, 0.15743983656006, 0.290447431974039
), foo.treated.mean = c(2.40658723367692, 180.846795140269, 139.054032348287,
11.8584348984435, 76.8141734599118, 2.24088124240385)), .Names = c("gene_symbol",
"foo.control.cv", "foo.control.mean", "foo.treated.cv", "foo.treated.mean"
), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
6L))
看起来像这样：
# A tibble: 6 × 5
gene_symbol foo.control.cv foo.control.mean foo.treated.cv foo.treated.mean
* <chr> <dbl> <dbl> <dbl> <dbl>
1 0610005C13Rik 1.1620404 2.664075 0.9057699 2.406587
2 0610007P14Rik 0.1205080 187.137729 0.1864419 180.846795
3 0610009B22Rik 0.2057126 142.111269 0.1585525 139.054032
4 0610009L18Rik 0.5045080 16.727859 0.5519551 11.858435
5 0610009O20Rik 0.3339563 69.860287 0.1574398 76.814173
6 0610010B08Rik 0.5436930 4.777690 0.2904474 2.240881
我要做的是把里面所有有mean的列名都替换成mean_expr。结果
gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv foo.treated.mean_expr
1 0610005C13Rik 1.1620404 2.664075 0.9057699 2.406587
2 0610007P14Rik 0.1205080 187.137729 0.1864419 180.846795
3 0610009B22Rik 0.2057126 142.111269 0.1585525 139.054032
4 0610009L18Rik 0.5045080 16.727859 0.5519551 11.858435
5 0610009O20Rik 0.3339563 69.860287 0.1574398 76.814173
6 0610010B08Rik 0.5436930 4.777690 0.2904474 2.240881
我怎样才能做到这一点？
这是一个非dplyr基础的R方法：
names(df) <- sub("mean$", "mean_expr", names(df))
# or names(df) <- sub("mean", "mean_expr", names(df)) if the mean doesn't have to be at the
# end of the string
names(df)
#[1] "gene_symbol" "foo.control.cv" "foo.control.mean_expr"
#[4] "foo.treated.cv" "foo.treated.mean_expr"
如果你想让它成为pipe的一部分，你可以使用setNames函数：
df %>% setNames(sub("mean", "mean_expr", names(.))) %>% names(.)
#[1] "gene_symbol" "foo.control.cv" "foo.control.mean_expr"
#[4] "foo.treated.cv" "foo.treated.mean_expr"
对于当前版本的 dplyr，您可以使用 rename_at:
library(dplyr)
df %>% rename_at(vars(contains('mean')), funs(sub('mean', 'mean_expr', .)))
#> # A tibble: 6 × 5
#> gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv
#> * <chr> <dbl> <dbl> <dbl>
#> 1 0610005C13Rik 1.1620404 2.664075 0.9057699
#> 2 0610007P14Rik 0.1205080 187.137729 0.1864419
#> 3 0610009B22Rik 0.2057126 142.111269 0.1585525
#> 4 0610009L18Rik 0.5045080 16.727859 0.5519551
#> 5 0610009O20Rik 0.3339563 69.860287 0.1574398
#> 6 0610010B08Rik 0.5436930 4.777690 0.2904474
#> # ... with 1 more variables: foo.treated.mean_expr <dbl>
实际上，您也可以使用 rename_all，因为不匹配的名称无论如何都不会受到影响。此外，您可以使用 quosure 或任何可以通过 rlang::as_function 强制转换为函数的任何东西 .funs，因此您可以使用 purrr 样式的表示法：
df %>% rename_all(~sub('mean', 'mean_expr', .x))
由于数据框是一个列表，purrr 的 set_names 可以做同样的事情：
library(purrr) # or library(tidyverse)
df %>% set_names(~sub('mean', 'mean_expr', .x))
都是return一样的东西。
另一种选择是在rename_at中paste（使用dplyr的devel版本）
library(dplyr)
df %>%
rename_at(vars(matches('mean')), funs(sprintf('%s_expr', .)))
# A tibble: 6 × 5
# gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv foo.treated.mean_expr
#* <chr> <dbl> <dbl> <dbl> <dbl>
#1 0610005C13Rik 1.1620404 2.664075 0.9057699 2.406587
#2 0610007P14Rik 0.1205080 187.137729 0.1864419 180.846795
#3 0610009B22Rik 0.2057126 142.111269 0.1585525 139.054032
#4 0610009L18Rik 0.5045080 16.727859 0.5519551 11.858435
#5 0610009O20Rik 0.3339563 69.860287 0.1574398 76.814173
#6 0610010B08Rik 0.5436930 4.777690 0.2904474 2.240881
或使用rename_if
df %>%
rename_if(grepl("mean", names(.)), funs(sprintf("%s_expr", .)))
另一种选择是dplyr::select_all():
df %>% select_all(~gsub("mean", "mean_expr", .))
并且使用 magritrr 你可以
library(magrittr)
names(df)[df %>% names %>% grep(pattern = "mean")] %<>% paste0("_expr")
df
# A tibble: 6 x 5
gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv foo.treated.mean_expr
* <chr> <dbl> <dbl> <dbl> <dbl>
1 0610005C13Rik 1.16 2.66 0.906 2.41
2 0610007P14Rik 0.121 187. 0.186 181.
3 0610009B22Rik 0.206 142. 0.159 139.
4 0610009L18Rik 0.505 16.7 0.552 11.9
5 0610009O20Rik 0.334 69.9 0.157 76.8
6 0610010B08Rik 0.544 4.78 0.290 2.24
我有以下问题：
df <- structure(list(gene_symbol = c("0610005C13Rik", "0610007P14Rik",
"0610009B22Rik", "0610009L18Rik", "0610009O20Rik", "0610010B08Rik"
), foo.control.cv = c(1.16204038288333, 0.120508045270669, 0.205712615954009,
0.504508040948641, 0.333956330117591, 0.543693011377001), foo.control.mean = c(2.66407458486012,
187.137728870855, 142.111269303428, 16.7278587043453, 69.8602872478098,
4.77769028710622), foo.treated.cv = c(0.905769898934564, 0.186441944401973,
0.158552512842753, 0.551955061149896, 0.15743983656006, 0.290447431974039
), foo.treated.mean = c(2.40658723367692, 180.846795140269, 139.054032348287,
11.8584348984435, 76.8141734599118, 2.24088124240385)), .Names = c("gene_symbol",
"foo.control.cv", "foo.control.mean", "foo.treated.cv", "foo.treated.mean"
), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
6L))
看起来像这样：
# A tibble: 6 × 5
gene_symbol foo.control.cv foo.control.mean foo.treated.cv foo.treated.mean
* <chr> <dbl> <dbl> <dbl> <dbl>
1 0610005C13Rik 1.1620404 2.664075 0.9057699 2.406587
2 0610007P14Rik 0.1205080 187.137729 0.1864419 180.846795
3 0610009B22Rik 0.2057126 142.111269 0.1585525 139.054032
4 0610009L18Rik 0.5045080 16.727859 0.5519551 11.858435
5 0610009O20Rik 0.3339563 69.860287 0.1574398 76.814173
6 0610010B08Rik 0.5436930 4.777690 0.2904474 2.240881
我要做的是把里面所有有mean的列名都替换成mean_expr。结果
gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv foo.treated.mean_expr
1 0610005C13Rik 1.1620404 2.664075 0.9057699 2.406587
2 0610007P14Rik 0.1205080 187.137729 0.1864419 180.846795
3 0610009B22Rik 0.2057126 142.111269 0.1585525 139.054032
4 0610009L18Rik 0.5045080 16.727859 0.5519551 11.858435
5 0610009O20Rik 0.3339563 69.860287 0.1574398 76.814173
6 0610010B08Rik 0.5436930 4.777690 0.2904474 2.240881
我怎样才能做到这一点？
这是一个非dplyr基础的R方法：
names(df) <- sub("mean$", "mean_expr", names(df))
# or names(df) <- sub("mean", "mean_expr", names(df)) if the mean doesn't have to be at the
# end of the string
names(df)
#[1] "gene_symbol" "foo.control.cv" "foo.control.mean_expr"
#[4] "foo.treated.cv" "foo.treated.mean_expr"
如果你想让它成为pipe的一部分，你可以使用setNames函数：
df %>% setNames(sub("mean", "mean_expr", names(.))) %>% names(.)
#[1] "gene_symbol" "foo.control.cv" "foo.control.mean_expr"
#[4] "foo.treated.cv" "foo.treated.mean_expr"
对于当前版本的 dplyr，您可以使用 rename_at:
library(dplyr)
df %>% rename_at(vars(contains('mean')), funs(sub('mean', 'mean_expr', .)))
#> # A tibble: 6 × 5
#> gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv
#> * <chr> <dbl> <dbl> <dbl>
#> 1 0610005C13Rik 1.1620404 2.664075 0.9057699
#> 2 0610007P14Rik 0.1205080 187.137729 0.1864419
#> 3 0610009B22Rik 0.2057126 142.111269 0.1585525
#> 4 0610009L18Rik 0.5045080 16.727859 0.5519551
#> 5 0610009O20Rik 0.3339563 69.860287 0.1574398
#> 6 0610010B08Rik 0.5436930 4.777690 0.2904474
#> # ... with 1 more variables: foo.treated.mean_expr <dbl>
实际上，您也可以使用 rename_all，因为不匹配的名称无论如何都不会受到影响。此外，您可以使用 quosure 或任何可以通过 rlang::as_function 强制转换为函数的任何东西 .funs，因此您可以使用 purrr 样式的表示法：
df %>% rename_all(~sub('mean', 'mean_expr', .x))
由于数据框是一个列表，purrr 的 set_names 可以做同样的事情：
library(purrr) # or library(tidyverse)
df %>% set_names(~sub('mean', 'mean_expr', .x))
都是return一样的东西。
另一种选择是在rename_at中paste（使用dplyr的devel版本）
library(dplyr)
df %>%
rename_at(vars(matches('mean')), funs(sprintf('%s_expr', .)))
# A tibble: 6 × 5
# gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv foo.treated.mean_expr
#* <chr> <dbl> <dbl> <dbl> <dbl>
#1 0610005C13Rik 1.1620404 2.664075 0.9057699 2.406587
#2 0610007P14Rik 0.1205080 187.137729 0.1864419 180.846795
#3 0610009B22Rik 0.2057126 142.111269 0.1585525 139.054032
#4 0610009L18Rik 0.5045080 16.727859 0.5519551 11.858435
#5 0610009O20Rik 0.3339563 69.860287 0.1574398 76.814173
#6 0610010B08Rik 0.5436930 4.777690 0.2904474 2.240881
或使用rename_if
df %>%
rename_if(grepl("mean", names(.)), funs(sprintf("%s_expr", .)))
另一种选择是dplyr::select_all():
df %>% select_all(~gsub("mean", "mean_expr", .))
并且使用 magritrr 你可以
library(magrittr)
names(df)[df %>% names %>% grep(pattern = "mean")] %<>% paste0("_expr")
df
# A tibble: 6 x 5
gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv foo.treated.mean_expr
* <chr> <dbl> <dbl> <dbl> <dbl>
1 0610005C13Rik 1.16 2.66 0.906 2.41
2 0610007P14Rik 0.121 187. 0.186 181.
3 0610009B22Rik 0.206 142. 0.159 139.
4 0610009L18Rik 0.505 16.7 0.552 11.9
5 0610009O20Rik 0.334 69.9 0.157 76.8
6 0610010B08Rik 0.544 4.78 0.290 2.24