查找R中数据框中每3列的平均值
Finding average for every 3 columns in dataframe in R
我想在包含 60 列的数据框中找到每 3 列的平均值，这样新的数据框就会有 20 列左右。我将样本数据如下：
此外，新变量，如果我可以将它们作为字符串数组传递，将会有所帮助。
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")
我们可以使用 tidyverse 将数据从宽到长旋转，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 ）。
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)
或者，这可以使用 base R 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
好处是聚合仅基于位置，而不是基于列名。这与 相反，后者要求可以将列名强制转换为连续的整数序列，即年份序列。
不过，上面的代码returns是一个矩阵，不是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
数据集看起来更像矩阵而不是 data.frame，即所有列都是相同的数据类型。否则，就无法通过跨列的方式进行聚合。也许，数据应该被视为矩阵，我们可以从矩阵运算中受益，比如 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 中所做的）。然后，列名成为数据对象，可以这样操作。
这是另一种比较稳健的方法：
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returns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。
有趣的问题！使用 purrr 和 hablar 进行了另一次拍摄。为每 3 列创建一个整数列表。对每 3 列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
我想在包含 60 列的数据框中找到每 3 列的平均值，这样新的数据框就会有 20 列左右。我将样本数据如下： 此外，新变量，如果我可以将它们作为字符串数组传递，将会有所帮助。
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")
我们可以使用 tidyverse 将数据从宽到长旋转，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
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)
或者，这可以使用 base R 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
好处是聚合仅基于位置，而不是基于列名。这与
不过，上面的代码returns是一个矩阵，不是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
数据集看起来更像矩阵而不是 data.frame，即所有列都是相同的数据类型。否则，就无法通过跨列的方式进行聚合。也许，数据应该被视为矩阵，我们可以从矩阵运算中受益，比如 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
这是另一种比较稳健的方法：
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returns一个列表。然后将这些列表分配回 DF2.
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。
有趣的问题！使用 purrr 和 hablar 进行了另一次拍摄。为每 3 列创建一个整数列表。对每 3 列应用行均值并组合成新的 df。从原始 df.
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