查找R中数据框中每3列的平均值答案

【问题标题】：Finding average for every 3 columns in dataframe in R查找R中数据框中每3列的平均值
【发布时间】：2019-08-03 06:24:22
【问题描述】：

我想在包含 60 列的数据框中找到每 3 列的平均值，这样一个新的数据框就会有 20 列左右。我将示例数据如下：此外，如果我可以将它们作为字符串数组传递，新变量也会有所帮助。

structure(list(`1961` = c(0, 0, 0, 0, 0, 0, 0, 0, 4.633, 54.247, 
0, 0, 0, 0, 0, 0, 0, 31.036, 3.18, 19.862), `1962` = c(0, 0, 
0, 0, 0, 0, 0, 0, 4.168, 63.587, 0, 0, 0, 0, 0, 0, 0, 28.169, 
2.913, 17.273), `1963` = c(0, 0, 0, 0, 0, 0, 0, 0, 3.284, 56.888, 
0, 0, 0, 0, 0, 0, 0, 26.667, 2.653, 16.586), `1964` = c(0, 0, 
0, 0, 0, 0, 0, 0, 2.689, 48.722, 0, 0, 0, 0, 0, 0, 0, 25.483, 
3.873, 15.708), `1965` = c(0, 0, 0, 0, 0, 0, 0, 0, 3.304, 33.838, 
0, 0, 0, 0, 0, 0, 0, 28.164, 3.927, 11.147), `1966` = c(0, 0, 
0, 0, 0, 0, 0, 0, 2.871, 26.695, 0, 0, 0, 0, 0, 0, 0, 28.962, 
4.434, 14.056), `1967` = c(0, 0, 0, 0, 0, 0, 0, 0, 2.752, 36.246, 
0, 0, 0, 0, 0, 0, 0, 30.877, 4.739, 14.765), `1968` = c(0, 0, 
0, 0, 0, 0, 0, 0, 3.537, 33.368, 0, 0, 0, 0, 0, 0, 0, 25.628, 
5.445, 14.372), `1969` = c(0, 0, 0, 0, 0, 0, 0, 0, 2.484, 35.711, 
0, 0, 0, 0, 0, 0, 0, 27.123, 5.286, 15.527)), row.names = c("Almonds, with shell", 
"Anise, badian, fennel, coriander", "Apples", "Apricots", "Areca nuts", 
"Asparagus", "Avocados", "Bananas", "Barley", "Bastfibres, other", 
"Beans, dry", "Beans, green", "Berries nes", "Broad beans, horse beans, dry", 
"Buckwheat", "Cabbages and other brassicas", "Carrots and turnips", 
"Cashew nuts, with shell", "Cassava", "Castor oil seed"), class = "data.frame")

【问题讨论】：

标签： r multiple-columns mean

【解决方案1】：

或者，这可以使用基础 R 来解决：

sapply(seq(2, ncol(mydf), 3), function(j) rowMeans(mydf[, j+(-1:1)]))

或

sapply(seq(1, ncol(mydf), 3), function(j) rowMeans(mydf[, j+(0:2)]))

                                      [,1]      [,2]      [,3]
Almonds, with shell               0.000000  0.000000  0.000000
Anise, badian, fennel, coriander  0.000000  0.000000  0.000000
Apples                            0.000000  0.000000  0.000000
Apricots                          0.000000  0.000000  0.000000
Areca nuts                        0.000000  0.000000  0.000000
Asparagus                         0.000000  0.000000  0.000000
Avocados                          0.000000  0.000000  0.000000
Bananas                           0.000000  0.000000  0.000000
Barley                            4.028333  2.954667  2.924333
Bastfibres, other                58.240667 36.418333 35.108333
Beans, dry                        0.000000  0.000000  0.000000
Beans, green                      0.000000  0.000000  0.000000
Berries nes                       0.000000  0.000000  0.000000
Broad beans, horse beans, dry     0.000000  0.000000  0.000000
Buckwheat                         0.000000  0.000000  0.000000
Cabbages and other brassicas      0.000000  0.000000  0.000000
Carrots and turnips               0.000000  0.000000  0.000000
Cashew nuts, with shell          28.624000 27.536333 27.876000
Cassava                           2.915333  4.078000  5.156667
Castor oil seed                  17.907000 13.637000 14.888000

好处是聚合完全基于位置，而不是列名。这与 A.S.K.'s approach 相反，后者要求列名可以强制转换为连续的整数序列，即年份序列。

但是，上面的代码返回一个矩阵，而不是一个 data.frame。此外，OP 已请求将新变量作为字符串数组传递。

library(magrittr)   # piping used to improve readability
new_cols <- c("Mean_A", "Mean_B", "Mean_C")
sapply(seq(1, ncol(mydf), 3), function(j) rowMeans(mydf[, j+(0:2)])) %>% 
  as.data.frame() %>% 
  set_names(new_cols)

                                    Mean_A    Mean_B    Mean_C
Almonds, with shell               0.000000  0.000000  0.000000
Anise, badian, fennel, coriander  0.000000  0.000000  0.000000
Apples                            0.000000  0.000000  0.000000
Apricots                          0.000000  0.000000  0.000000
Areca nuts                        0.000000  0.000000  0.000000
Asparagus                         0.000000  0.000000  0.000000
Avocados                          0.000000  0.000000  0.000000
Bananas                           0.000000  0.000000  0.000000
Barley                            4.028333  2.954667  2.924333
Bastfibres, other                58.240667 36.418333 35.108333
Beans, dry                        0.000000  0.000000  0.000000
Beans, green                      0.000000  0.000000  0.000000
Berries nes                       0.000000  0.000000  0.000000
Broad beans, horse beans, dry     0.000000  0.000000  0.000000
Buckwheat                         0.000000  0.000000  0.000000
Cabbages and other brassicas      0.000000  0.000000  0.000000
Carrots and turnips               0.000000  0.000000  0.000000
Cashew nuts, with shell          28.624000 27.536333 27.876000
Cassava                           2.915333  4.078000  5.156667
Castor oil seed                  17.907000 13.637000 14.888000

顺便说一句：重新考虑数据结构

数据集看起来更像是一个矩阵而不是 data.frame，即所有列都是相同的数据类型。否则，将无法通过列进行聚合。也许，数据应该被视为矩阵，我们可以从矩阵运算中受益，例如rowMeans()。

矩阵的行和列也可以命名：

library(magrittr)
new_cols <- c("Mean_A", "Mean_B", "Mean_C")
sapply(seq(1, ncol(mydf), 3), function(j) rowMeans(mydf[, j+(0:2)])) %>% 
  set_colnames(new_cols)

                                    Mean_A    Mean_B    Mean_C
Almonds, with shell               0.000000  0.000000  0.000000
Anise, badian, fennel, coriander  0.000000  0.000000  0.000000
Apples                            0.000000  0.000000  0.000000
Apricots                          0.000000  0.000000  0.000000
Areca nuts                        0.000000  0.000000  0.000000
Asparagus                         0.000000  0.000000  0.000000
Avocados                          0.000000  0.000000  0.000000
Bananas                           0.000000  0.000000  0.000000
Barley                            4.028333  2.954667  2.924333
Bastfibres, other                58.240667 36.418333 35.108333
Beans, dry                        0.000000  0.000000  0.000000
Beans, green                      0.000000  0.000000  0.000000
Berries nes                       0.000000  0.000000  0.000000
Broad beans, horse beans, dry     0.000000  0.000000  0.000000
Buckwheat                         0.000000  0.000000  0.000000
Cabbages and other brassicas      0.000000  0.000000  0.000000
Carrots and turnips               0.000000  0.000000  0.000000
Cashew nuts, with shell          28.624000 27.536333 27.876000
Cassava                           2.915333  4.078000  5.156667
Castor oil seed                  17.907000 13.637000 14.888000

打印输出看起来类似于 data.frame 解决方案，但底层数据结构现在是一个矩阵。

或者，数据可以在整形后以长格式存储（这是对gather() 的调用在A.S.K.'s approach 中所做的）。然后，列名成为数据对象，可以这样操作。

【讨论】：

【解决方案2】：

我们可以使用tidyverse将数据从宽旋转到长，然后按3年序列分组。

library(dplyr)
library(tidyr)
library(tibble)

df.averaged = df %>%
  # Extract rownames as their own column
  rownames_to_column("product") %>%
  # Convert from wide to long: one row per product per year
  gather(year, value, -product) %>%
  # Add a column with "year group" (every 3 years go into one group)
  mutate(year = as.numeric(year),
         year.group = (floor((year + 1) / 3) * 3) - 1) %>%
  # Group by product and year group
  group_by(product, year.group) %>%
  # Get averages
  summarize(value = mean(value)) %>%
  # Convert back from long to wide, if desired
  spread(year.group, value)

编辑：对于移动平均线，我们可以使用RcppRoll 包（另请参阅this answer）。

library(RcppRoll)
df.moving.window = df %>%
  # Extract rownames as their own column
  rownames_to_column("product") %>%
  # Convert from wide to long: one row per product per year
  gather(year, value, -product) %>%
  # Order by product, then year
  arrange(product, year) %>%
  # Compute the rolling average
  group_by(product) %>%
  mutate(value = roll_mean(value, n = 3, align = "right", fill = NA)) %>%
  # Convert back from long to wide, if desired
  spread(year, value)

【讨论】：

谢谢！你能解释一下这个(floor((year + 1) / 3) * 3) - 1)。我跟不上。
将每年映射到其 3 组中的第一年。floor(year / 3) * 3 会将 0、1 和 2 转换为 0； 3、4、5至3；等等。 + 1 和 - 1 只是稍微移动窗口，以便 1961-3 进入同一组（否则将是 1959-61、1962-4 等）。
谢谢，太好了，有没有像每列 3 年的移动平均值这样的东西。

【解决方案3】：

这是另一种比较稳健的方式：

n <- 3
i <- seq(1, length(DF), n)

DF2 <- data.frame(nut = rownames(DF))
DF2[, paste0('NewCol', seq_along(i))] <- lapply(i, function (j) rowMeans(DF[, j:min(j+2, length(DF))]))

DF2

                                nut   NewCol1   NewCol2   NewCol3
1               Almonds, with shell  0.000000  0.000000  0.000000
2  Anise, badian, fennel, coriander  0.000000  0.000000  0.000000
3                            Apples  0.000000  0.000000  0.000000
4                          Apricots  0.000000  0.000000  0.000000
5                        Areca nuts  0.000000  0.000000  0.000000
6                         Asparagus  0.000000  0.000000  0.000000
7                          Avocados  0.000000  0.000000  0.000000
8                           Bananas  0.000000  0.000000  0.000000
9                            Barley  4.028333  2.954667  2.924333
10                Bastfibres, other 58.240667 36.418333 35.108333
11                       Beans, dry  0.000000  0.000000  0.000000
12                     Beans, green  0.000000  0.000000  0.000000
13                      Berries nes  0.000000  0.000000  0.000000
14    Broad beans, horse beans, dry  0.000000  0.000000  0.000000
15                        Buckwheat  0.000000  0.000000  0.000000
16     Cabbages and other brassicas  0.000000  0.000000  0.000000
17              Carrots and turnips  0.000000  0.000000  0.000000
18          Cashew nuts, with shell 28.624000 27.536333 27.876000
19                          Cassava  2.915333  4.078000  5.156667
20                  Castor oil seed 17.907000 13.637000 14.888000

需要指出的是，输出是data.frame。 lapply() 函数返回一个列表。然后将这些列表分配回DF2 中的新列。

最重要的是j:min(j+2, length(DF))。这部分将允许代码是否有 2 列或 3 列。

【讨论】：

【解决方案4】：

有趣的问题！使用purrr 和hablar 再次拍摄。为每 3 列创建一个整数列表。在每 3 列上应用行均值并组合成新的 df。从原始 df 复制行名。

代码

library(tidyverse)
library(hablar)
library(magrittr)

l <- map(seq(1, ncol(df), 3), ~seq(.x, .x + 2))

map_dfc(l, ~df %>% transmute(mean = row_mean_(.x))) %>% 
  set_rownames(rownames(df))

结果

                                      mean     mean1     mean2
Almonds, with shell               0.000000  0.000000  0.000000
Anise, badian, fennel, coriander  0.000000  0.000000  0.000000
Apples                            0.000000  0.000000  0.000000
Apricots                          0.000000  0.000000  0.000000
Areca nuts                        0.000000  0.000000  0.000000
Asparagus                         0.000000  0.000000  0.000000
Avocados                          0.000000  0.000000  0.000000
Bananas                           0.000000  0.000000  0.000000
Barley                            4.028333  2.954667  2.924333
Bastfibres, other                58.240667 36.418333 35.108333
Beans, dry                        0.000000  0.000000  0.000000
Beans, green                      0.000000  0.000000  0.000000
Berries nes                       0.000000  0.000000  0.000000
Broad beans, horse beans, dry     0.000000  0.000000  0.000000
Buckwheat                         0.000000  0.000000  0.000000
Cabbages and other brassicas      0.000000  0.000000  0.000000
Carrots and turnips               0.000000  0.000000  0.000000
Cashew nuts, with shell          28.624000 27.536333 27.876000
Cassava                           2.915333  4.078000  5.156667
Castor oil seed                  17.907000 13.637000 14.888000

【讨论】：