或者，您可以将键列转换为具有反映原始列名顺序的水平的因子：

df %>% 
    gather(key="key", value="value", -sample) %>%
    mutate(key=factor(key, levels=names(df)[-1])) %>% # add this line to convert the key to a factor
    group_by(sample, key) %>%
    summarize(mean = mean(value)) %>%
    filter(sample == 1)

# A tibble: 3 x 3
# Groups:   sample [1]
#  sample    key       mean
#   <dbl> <fctr>      <dbl>
#1      1   y1.1  0.8310786
#2      1   y2.1 -1.2596933
#3      1  y10.1  0.8208812

tidyverse包现在支持优雅的解决方案：

library(tidyverse)
    N <- 4
    df <- data.frame(sample = c(1,1,2,2),
                    y1.1 = rnorm(N), y2.1 = rnorm(N), y10.1 = rnorm(N))
    df %>% 
        gather("key", "value", -sample, factor_key = T) %>% 
        group_by(sample, key) %>%
        summarise(mean = mean(value))

这导致了

# A tibble: 6 x 3
    # Groups:   sample [2]
    sample key      mean
    <dbl> <fct>   <dbl>
    1      1 y1.1   0.0894
    2      1 y2.1   0.551 
    3      1 y10.1  0.254 
    4      2 y1.1  -0.555 
    5      2 y2.1  -1.36  
    6      2 y10.1 -0.794

我通过使用查找表找到了一个可行的解决方案。它似乎对我很有效，因为我可以提取列名并为列名分配一个有序编号，然后与我的data.frame配对。
解决方案如下：

lookup <- tibble(key = c("y1.1", "y2.1", "y10.1"),
                 index = c(1,2,3))

> left_join(dfg, lookup, by="key")
# A tibble: 6 x 4
  sample   key       mean index
   <dbl> <chr>      <dbl> <dbl>
1      1  y1.1  0.2936335     1
2      1 y10.1  0.6170505     3
3      1  y2.1 -0.2250543     2
4      2  y1.1  1.3652070     1
5      2 y10.1  0.9889233     3
6      2  y2.1  0.5216553     2

如果您的列确实是按照它所包含的数字排序的，这应该可以工作：

library(readr)

df %>% 
  gather(key="key", value="value", -sample) %>%
  group_by(sample, key)         %>%
  summarize(mean = mean(value)) %>%
  arrange(parse_number(key))    %>%  # <- sorting by number contained in key
  filter(sample == 1)

# # A tibble: 3 x 3
# # Groups:   sample [1]
#     sample   key       mean
# <dbl> <chr>      <dbl>
#   1      1  y1.1 -0.9236688
#   2      1  y2.1 -0.2168337
#   3      1 y10.1  0.5041981

还有一种方法是使用您想要排序的键列的定制版本来arrange Dataframe ：

library(dplyr)
library(tidyr)

df %>% 
  gather(key="key", value="value", -sample) %>%
  group_by(sample, key) %>%
  summarize(mean = mean(value)) %>%
  arrange(as.numeric(stringr::str_replace(key, "y", "")), .by_group = TRUE)

#> # A tibble: 6 x 3
#> # Groups:   sample [2]
#>   sample   key        mean
#>    <dbl> <chr>       <dbl>
#> 1      1  y1.1  0.07001689
#> 2      1  y2.1  1.15349430
#> 3      1 y10.1  1.18266024
#> 4      2  y1.1  0.42616604
#> 5      2  y2.1  1.05891682
#> 6      2 y10.1 -0.12561209

如果我们合并前面建议答案的思想，并且使用pivot_longer()，因为它不过时，我们可以添加一个步骤来设置键as_factor()而不是类字符。如果我们将其保留为字符，它将被重新排序为字母数字顺序。

library(tidyverse)

N <- 4
df <- data.frame(sample = c(1,1,2,2),
                 y1.1 = rnorm(N), y2.1 = rnorm(N), y10.1 = rnorm(N))

dfg <- df |> 
  pivot_longer(2:4, names_to = "key", values_to = "value") |> 
  mutate(key = as_factor(key)) |> 
  group_by(sample, key) |> 
  summarize(mean = mean(value)) |> 
  ungroup()

dfg

# A tibble: 6 × 3
  sample key     mean
   <dbl> <fct>  <dbl>
1      1 y1.1  -0.789
2      1 y2.1   1.16 
3      1 y10.1 -0.187
4      2 y1.1   0.962
5      2 y2.1   0.673
6      2 y10.1  0.502