使用 for 循环或应用函数来分离和求和变量答案

【问题标题】：Use for-loop or apply function to separate and sum variables使用 for 循环或应用函数来分离和求和变量
【发布时间】：2019-09-01 11:55:15
【问题描述】：

我有一个数据框，其中有一组变量，我需要在其中提取每个组并创建一个新变量，该变量是每个组的行总和。每个组都可以被识别，因为它们包含一个从 0 到 5 的数字。

目前我正在通过以下方式分别提取和汇总每个组：

sum_cols <- function(x, col1, col2, col3){
  x[[col1]] + x[[col2]] + x[[col3]]
}

test %>%
  select(contains('0')) %>%
  mutate(
    p0 = sum_cols(x = ., col1 = 1, col2 = 2, col3 = 3)
  ) %>% pull(p0)

然后对每个组重复此步骤，替换数字 1、2、3、4、5，然后将每个新创建的向量组合到数据帧中。

我假设有更快/更好的方法来执行此步骤，我玩过 for 循环但没有设法让它工作。或者，使用split() 或apply 系列中的函数拆分数据的某种方式，但不确定什么是最好的。有什么建议将不胜感激？

谢谢

以下是数据样本：

test <- data.frame(Time = c(0, 0.001, 0.002, 0.003, 0.004, 0.005, 0.006, 0.007, 0.008, 0.009),
                   X.P0..X = c(2.26, 1.722, 2.798, 1.722, 1.722, 2.798, 1.722, 1.722, 2.26, 2.26), 
                   X.P0..Y = c(1.434, 1.434, 0.641, 0.641, 0.107, -0.411, -0.683, -0.423, -0.151, 0.369), 
                   X.P0..Z = c(3.188, 3.188, 3.188, 3.726, 3.188, 2.65, 3.188, 3.188, 3.188, 3.188), 
                   X.P1..X = c(15.656, 15.656, 15.656, 15.118, 15.118, 15.656, 15.656, 15.118, 15.656, 15.656),
                   X.P1..Y = c(268.808, 268.033, 269.322, 269.063, 269.318, 269.065, 269.065, 269.322, 268.813, 269.055), 
                   X.P1..Z = c(14.096, 14.096, 14.096, 14.634, 14.634, 14.634, 14.096, 14.634, 14.634, 14.634), 
                   X.P2..X = c(1.063, 1.063, 1.063, 1.063, 1.063, 1.063,  1.063, 1.063, 1.063, 1.063), 
                   X.P2..Y = c(-0.292, -1.068, -1.07, -0.813, -1.329, -0.551, -1.07, -0.813, -1.074, -1.074), 
                   X.P2..Z = c(-1.143, -1.681, -1.681, -1.143, -1.681, -1.143, -1.681, -1.681, -1.681, -1.143), 
                   X.P3..X = c(-0.417, -0.955, -1.493, -1.493, -0.955, -0.955, -0.955, -0.955, -1.493, -0.955), 
                   X.P3..Y = c(0.041, 0.297, 0.288, 0.554, 0.565, 0.291, 0.295, 0.298, 0.554, 0.554),
                   X.P3..Z = c(0.726, 1.802, 1.802, 1.264, 1.264, 1.264, 1.264, 0.726, 1.264, 0.726),
                   X.P4..X = c(-7.411,  -6.873, -7.411, -7.411, -6.873, -7.411, -7.411, -7.411, -6.873, -7.411),
                   X.P4..Y = c(278.284, 278.28, 279.06, 278.8, 278.803, 279.323, 278.801, 278.284, 279.062, 278.547),
                   X.P4..Z = c(3.511,  4.587, 3.511, 4.587, 3.511, 4.587, 4.587, 4.587, 5.125, 4.587),
                   X.P5..X = c(-1.843, -1.843, -1.843, -1.843, -1.843, -1.843, -1.843, -1.843, -1.843, -1.843),
                   X.P5..Y = c(0.964, 0.964, 0.447, 0.701, 0.439, 0.708, 0.183, 0.701, 0.959, 0.701), 
                   X.P5..Z = c(-3.631, -3.631, -3.631, -3.631, -4.169, -3.093,  -3.093, -4.169, -4.169, -3.631)
                   )

【问题讨论】：

标签： r

【解决方案1】：

使用map_dfc循环遍历paste0('P',0:5)、select和transmute当前P然后将Ps组合在一起

library(dplyr)
library(purrr)
#use map if you need the output as a list
map_dfc(paste0('P',0:5), 
         ~select(test, contains(.x)) %>% transmute(!!.x := rowSums(., na.rm = TRUE)))

      P0      P1     P2    P3      P4     P5
1  6.882 298.560 -0.372 0.350 274.384 -4.510
2  6.344 297.785 -1.686 1.144 275.994 -4.510
3  6.627 299.074 -1.688 0.597 275.160 -5.027
4  6.089 298.815 -0.893 0.325 275.976 -4.773
5  5.017 299.070 -1.947 0.874 275.441 -5.573
6  5.037 299.355 -0.631 0.600 276.499 -4.228
7  4.227 298.817 -1.688 0.604 275.977 -4.753
8  4.487 299.074 -1.431 0.069 275.460 -5.311
9  5.297 299.103 -1.692 0.325 277.314 -5.053
10 5.817 299.345 -1.154 0.325 275.723 -4.773

【讨论】：

谢谢！我从未想过使用purrr

【解决方案2】：

我们还可以使用 base R split.default 根据名称中的公共子字符串（“P”后跟数字）拆分列，然后使用 sapply 获取每个组的 rowSums。

df <- test[-1]
data.frame(sapply(split.default(df, sub(".*(P\\d+).*", "\\1", names(df))), rowSums))

#      P0      P1     P2    P3      P4     P5
#1  6.882 298.560 -0.372 0.350 274.384 -4.510
#2  6.344 297.785 -1.686 1.144 275.994 -4.510
#3  6.627 299.074 -1.688 0.597 275.160 -5.027
#4  6.089 298.815 -0.893 0.325 275.976 -4.773
#5  5.017 299.070 -1.947 0.874 275.441 -5.573
#6  5.037 299.355 -0.631 0.600 276.499 -4.228
#7  4.227 298.817 -1.688 0.604 275.977 -4.753
#8  4.487 299.074 -1.431 0.069 275.460 -5.311
#9  5.297 299.103 -1.692 0.325 277.314 -5.053
#10 5.817 299.345 -1.154 0.325 275.723 -4.773

sub 部分获取列名中的公共部分

sub(".*(P\\d+).*", "\\1", names(df))
# [1] "P0" "P0" "P0" "P1" "P1" "P1" "P2" "P2" "P2" "P3" "P3" "P3" "P4" "P4" "P4" 
#     "P5" "P5" "P5"

【讨论】：