如何计算多个数字列的每行平均值答案

【问题标题】：How do I compute the average per row of multiple numeric columns如何计算多个数字列的每行平均值
【发布时间】：2019-06-15 22:05:08
【问题描述】：

我有 8 个年龄类别作为 8 个单独的列。每列都有一个介于 1 和 3 之间的值。我想计算一个新列来保存每行的平均年龄。

这是我的数据：

structure(list(`2.5` = c(0, 0, 0, 1, 1, 2, 1, 2, 0, 0, 1, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 
0, 2, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 2, 0, 0, 0, 
0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 2, 2, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 2, 0, 2, 0, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 
0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 
0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 3, 
0, 0, 1), `9` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 
1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 1, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 3, 0, 1, 
0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 2, 0, 
2, 0, 3, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 
1, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 1
), `15.5` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 
0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 
0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 2, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0), 
    `21.5` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), `29.5` = c(0, 
    1, 2, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 
    0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 
    0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 
    0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 
    0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 1, 2, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 
    0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 
    0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0), `42` = c(0, 0, 0, 
    2, 1, 2, 2, 2, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 1, 1, 0, 2, 
    0, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 
    1, 0, 0, 0, 0, 2, 1, 2, 1, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 
    2, 0, 1, 0, 0, 0, 2, 2, 2, 1, 0, 2, 0, 0, 1, 0, 0, 2, 0, 
    2, 1, 1, 0, 0, 2, 0, 0, 0, 2, 1, 1, 1, 1, 0, 1, 2, 2, 0, 
    0, 0, 0, 2, 0, 2, 0, 0, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 1, 0, 2, 0, 2, 1, 0, 1, 1, 2, 0, 0, 2, 1, 2, 2, 2, 0, 
    1, 0, 1, 0, 2, 2, 2, 1, 0, 0, 2, 0, 0, 0, 0, 2, 0, 2, 2, 
    2, 2, 1, 2, 0, 2, 0, 2, 0, 2, 2, 1, 0, 0, 0, 2, 2, 0, 2, 
    0, 0, 2, 2, 0, 0, 0, 0, 2, 1, 2, 0, 0, 1, 2, 0, 0, 0, 1, 
    1, 2, 2, 1, 0, 0, 0, 2, 1, 1, 2), `57` = c(0, 1, 0, 0, 0, 
    0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 2, 1, 
    0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 2, 
    0, 0, 2, 0, 0, 0, 0, 1, 0, 2, 0, 1, 0, 2, 2, 0, 0, 0, 0, 
    0, 0, 2, 0, 0, 0, 0, 1, 2, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 
    0, 0, 2, 0, 0, 2, 0, 0, 0, 1, 1, 0, 2, 0, 0, 0, 2, 0, 1, 
    2, 0, 2, 0, 1, 1, 0, 0, 0, 2, 0, 0, 1, 2, 2, 2, 0, 2, 0, 
    0, 0, 0, 0, 1, 2, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 2, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 2, 2, 2, 0, 0, 0, 0, 0, 0, 
    0, 0, 2, 0, 1, 0, 2, 0, 0, 0, 2, 2, 0, 0, 0, 1, 0, 0, 1, 
    0, 0, 1, 1, 2, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 
    0, 1, 2, 0, 1, 0, 1, 0, 0), `72` = c(2, 0, 0, 0, 1, 0, 0, 
    0, 0, 0, 0, 2, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 2, 
    0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 0, 1, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 2, 0, 0, 0, 2, 
    0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 
    0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 2, 1, 0, 0, 
    0, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 1, 0, 
    0, 2, 1, 0, 0, 0, 0)), row.names = c(NA, -204L), class = c("data.table", 
"data.frame"))

【问题讨论】：

标签： r

【解决方案1】：

首先，创建一个临时数据框，您可以在其中根据列名计算年龄。然后，用rowSums 计算平均年龄。（假设您的数据框名为d。）

tmp <- do.call(cbind, lapply(seq(d), function(x) d[x] * as.numeric(colnames(d)[x])))
d$mu <- rowSums(tmp) / rowSums(d)

head(d)
#   2.5 9 15.5 21.5 29.5 42 57 72       mu
# 1   0 0    0    0    0  0  0  2 72.00000
# 2   0 0    0    0    1  0  1  0 43.25000
# 3   0 0    0    0    2  0  0  0 29.50000
# 4   1 0    0    0    0  2  0  0 28.83333
# 5   1 0    0    0    1  1  0  1 36.50000
# 6   2 0    0    0    0  2  0  0 22.25000

【讨论】：

啊抱歉，可能和data.table有关，先试试d <- as.data.frame(d)
是的，我刚刚想通了 :) 当我将其更改为数据框时，一切正常！

【解决方案2】：

apply 是一个有用的选项，1 告诉它按行计算。它似乎也与data.tables 配合得很好：

df$means <- apply(df, 1, function(r) sum(r * as.double(names(df))) / sum(r))

#### OUTPUT ####

     2.5 9 15.5 21.5 29.5 42 57 72    means
  1:   0 0    0    0    0  0  0  2 72.00000
  2:   0 0    0    0    1  0  1  0 43.25000
  3:   0 0    0    0    2  0  0  0 29.50000
  4:   1 0    0    0    0  2  0  0 28.83333
  5:   1 0    0    0    1  1  0  1 36.50000
 ---                                       
200:   0 0    0    0    0  0  1  1 64.50000
201:   3 0    0    0    0  2  0  0 18.30000
202:   0 0    1    0    0  1  1  0 38.16667
203:   0 0    0    0    1  1  0  0 35.75000
204:   1 1    0    0    0  2  0  0 23.87500

【讨论】：

【解决方案3】：

这是一个基本的 R 单线，我们将数据框列中的值乘以其名称，计算列值的总和，然后除以它的 rowSums。

df$result <- colSums(t(df) * as.numeric(names(df)))/rowSums(df)

head(df)
#  2.5 9 15.5 21.5 29.5 42 57 72   result
#1   0 0    0    0    0  0  0  2 72.00000
#2   0 0    0    0    1  0  1  0 43.25000
#3   0 0    0    0    2  0  0  0 29.50000
#4   1 0    0    0    0  2  0  0 28.83333
#5   1 0    0    0    1  1  0  1 36.50000
#6   2 0    0    0    0  2  0  0 22.25000

【讨论】：