我可以在没有“apply()"函数的情况下跨行应用 R 标准差吗? [英] Can I apply R standard deviation across rows without `apply()` function?

问题描述

library(tidyverse)
df <- tibble(col1 = c(5, 2), col2 = c(6, 4), col3 = c(9, 9))
# # A tibble: 2 x 3
#    col1  col2  col3
#   <dbl> <dbl> <dbl>
# 1     5     6     9
# 2     2     4     9

df %>% mutate(col4 = apply(.[, c(1, 3)], 1, sum))
df %>% mutate(col4 = rowSums(.[c(1, 3)], na.rm = TRUE))

最近 R 的 apply() 函数给我带来了麻烦.目前我将尽量减少它的使用并使用替代品.@akrun 告诉我我可以使用 rowSums() 而不是 apply() 作为示例，如上所示.

Lately R's apply() function has been trouble for me. For the time being I'm going to minimize it's use and use alternatives. @akrun educated me that I could use rowSums() instead of apply() as shown above, as an example.

但是有没有一种方法可以应用跨列的标准偏差，就像我在下面所做的那样.显然我的 imaginary::rowSd 函数将无法工作.编的.

But is there a way to apply, say, standard deviation across columns, like I do below. Obviously my imaginary::rowSd function is not going to work. It's made up.

df %>% mutate(col4 = apply(.[, c(1, 3)], 1, sd))
df %>% mutate(col4 = imaginary::rowSd(.[c(1, 3)], na.rm = TRUE))

不使用 apply() 的方法是什么?虽然我对这个包和 map() 函数知之甚少，但我在想 purrr.也许有一个更简单/更优雅的解决方案.

What is an approach that would work, without using apply()? I'm thinking purrr although I've little knowledge on this package and the map() functions. Maybe there's an even easier/elegant solution.

我应该提到我不能使用列名，因为我从中提取信息的数据库中的名称经常发生变化.我只能使用列号，因为相对列位置在我从中提取数据的数据库中不会改变.

I should've mentioned I can't use column names because the names often change in the database I pull my info from. I can only use column numbers, because relative column position doesn't change in the database I pull data from.

推荐答案

一个更简单的选择是 matrixStats 中的 rowSds，但它只适用于 matrix，因此将数据集的子集转换为 matrix 并应用 rowSds

An easier option is rowSds from matrixStats, but it works only on a matrix, so convert the subset of dataset to matrix and apply rowSds

library(matrixStats)
library(dplyr)
df %>%
    mutate(col4 = rowSds(as.matrix(.[c(1, 3)]))) 
# A tibble: 2 x 4
#   col1  col2  col3  col4
#  <dbl> <dbl> <dbl> <dbl>
#1     5     6     9  2.83
#2     2     4     9  4.95

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