如何计算按列名分组的数据帧/矩阵中的平均值和标准差

Question

示例数据：

我想计算每个唯一列名称的平均值和标准偏差（忽略NA）以获得如下输出：

可再现数据:

df <- data.frame(c(1, 3.7, 1.5, 3.7, 2.9, 7),
                 c(2, 1, 0.4, NA, 1.1, 4.9),
                 c(1.1, 1, 2.1, 3.6, 10, 6.9),
                 c(4, 3.3, 3.3, 3.1, 7.1, 5.4),
                 c(9.8, NA, 6, 5.6, 7.7, 4.9))
names(df) <- c("sun", "sun", "sun", "sky", "sky")

我得到的最接近的是

#for mean
sapply(split.default(df, names(df)), rowMeans, na.rm = TRUE) 

#for sd
sapply(split.default(df, names(df)), function(x) apply(x, 1, sd, na.rm=TRUE))

这是我从这篇文章中得到的，但我不知道如何调整它来得到我想要的。我知道我可以取rowmeans的平均值来得到每组的平均值，但这不适用于标准差。

Answer 1

这里是＜code＞dplyr＜code＞中的另一种方法，将类似命名列的值放入一列中，然后对它们计算＜code＞mean＜code＞和＜code＞sd＜code＞。

library(dplyr)
library(tidyr)

df %>%
  pivot_longer(cols = everything(), 
               names_to = '.value') %>%
  summarise(across(.fns = list(mean = ~mean(., na.rm = TRUE), 
                              sd = ~sd(., na.rm = TRUE))))

#  sun_mean sun_sd sky_mean sky_sd
#     <dbl>  <dbl>    <dbl>  <dbl>
#1     3.17   2.68     5.47   2.10

如果您希望均值和标准差值在单独的列中，您可以添加到上述答案中-

 %>% pivot_longer(cols = everything(), names_to = c('col', '.value'), names_sep = '_')

#  col    mean    sd
#  <chr> <dbl> <dbl>
#1 sun    3.17  2.68
#2 sky    5.47  2.10

Answer 2

这里有一个< code>tidyverse解决方案

library(tidyverse)

df <- data.frame(c(1, 3.7, 1.5, 3.7, 2.9, 7),
                 c(2, 1, 0.4, NA, 1.1, 4.9),
                 c(1.1, 1, 2.1, 3.6, 10, 6.9),
                 c(4, 3.3, 3.3, 3.1, 7.1, 5.4),
                 c(9.8, NA, 6, 5.6, 7.7, 4.9))
names(df) <- c("sun", "sun", "sun", "sky", "sky")

df %>%
  #Pivotting data
  pivot_longer(cols = everything()) %>%
  #Grouping by sun/sky
  group_by(name) %>% 
  #Caluclating mean and sg grouped by sun/sky
  summarise(
    mean = mean(value,na.rm = T),
    sd = sd(value,na.rm = T)
  )

  name   mean    sd
  <chr> <dbl> <dbl>
1 sky    5.47  2.10
2 sun    3.17  2.68

Answer 3

我们可以使用

t(sapply(split.default(df, names(df)), function(x)  {
    x1 <- unlist(x)
    data.frame(mean = mean(x1, na.rm = TRUE), sd = sd(x1, na.rm = TRUE))}))

-输出

       mean     sd      
sky 5.472727 2.102423
sun 3.170588 2.677631

或者使用< code>data.table

library(data.table)
melt(setDT(df), measure = patterns("^sun", "^sky"), 
  value.name = c("sun", "sky"))[, c(list(categ = c("mean", "sd")), 
    lapply(.SD, function(x) c(mean = mean(x, na.rm = TRUE), 
     sd = sd(x, na.rm = TRUE)))), .SDcols = sun:sky]
   categ      sun      sky
1:  mean 3.170588 5.472727
2:    sd 2.677631 2.102423