以下是在基本 R 中执行此操作的多种方法，包括替代 aggregate 方法。下面的示例返回意味着每月，我认为这是您所要求的。虽然，可以使用相同的方法来返回每个人的平均值：
使用ave：
my.data <- read.table(text = '
Name Month Rate1 Rate2
Aira 1 12 23
Aira 2 18 73
Aira 3 19 45
Ben 1 53 19
Ben 2 22 87
Ben 3 19 45
Cat 1 22 87
Cat 2 67 43
Cat 3 45 32
', header = TRUE, stringsAsFactors = FALSE, na.strings = 'NA')
Rate1.mean <- with(my.data, ave(Rate1, Month, FUN = function(x) mean(x, na.rm = TRUE)))
Rate2.mean <- with(my.data, ave(Rate2, Month, FUN = function(x) mean(x, na.rm = TRUE)))
my.data <- data.frame(my.data, Rate1.mean, Rate2.mean)
my.data
使用by：
my.data <- read.table(text = '
Name Month Rate1 Rate2
Aira 1 12 23
Aira 2 18 73
Aira 3 19 45
Ben 1 53 19
Ben 2 22 87
Ben 3 19 45
Cat 1 22 87
Cat 2 67 43
Cat 3 45 32
', header = TRUE, stringsAsFactors = FALSE, na.strings = 'NA')
by.month <- as.data.frame(do.call("rbind", by(my.data, my.data$Month, FUN = function(x) colMeans(x[,3:4]))))
colnames(by.month) <- c('Rate1.mean', 'Rate2.mean')
by.month <- cbind(Month = rownames(by.month), by.month)
my.data <- merge(my.data, by.month, by = 'Month')
my.data
使用lapply 和split：
my.data <- read.table(text = '
Name Month Rate1 Rate2
Aira 1 12 23
Aira 2 18 73
Aira 3 19 45
Ben 1 53 19
Ben 2 22 87
Ben 3 19 45
Cat 1 22 87
Cat 2 67 43
Cat 3 45 32
', header = TRUE, stringsAsFactors = FALSE, na.strings = 'NA')
ly.mean <- lapply(split(my.data, my.data$Month), function(x) c(Mean = colMeans(x[,3:4])))
ly.mean <- as.data.frame(do.call("rbind", ly.mean))
ly.mean <- cbind(Month = rownames(ly.mean), ly.mean)
my.data <- merge(my.data, ly.mean, by = 'Month')
my.data
使用sapply 和split：
my.data <- read.table(text = '
Name Month Rate1 Rate2
Aira 1 12 23
Aira 2 18 73
Aira 3 19 45
Ben 1 53 19
Ben 2 22 87
Ben 3 19 45
Cat 1 22 87
Cat 2 67 43
Cat 3 45 32
', header = TRUE, stringsAsFactors = FALSE, na.strings = 'NA')
my.data
sy.mean <- t(sapply(split(my.data, my.data$Month), function(x) colMeans(x[,3:4])))
colnames(sy.mean) <- c('Rate1.mean', 'Rate2.mean')
sy.mean <- data.frame(Month = rownames(sy.mean), sy.mean, stringsAsFactors = FALSE)
my.data <- merge(my.data, sy.mean, by = 'Month')
my.data
使用aggregate：
my.data <- read.table(text = '
Name Month Rate1 Rate2
Aira 1 12 23
Aira 2 18 73
Aira 3 19 45
Ben 1 53 19
Ben 2 22 87
Ben 3 19 45
Cat 1 22 87
Cat 2 67 43
Cat 3 45 32
', header = TRUE, stringsAsFactors = FALSE, na.strings = 'NA')
my.summary <- with(my.data, aggregate(list(Rate1, Rate2), by = list(Month),
FUN = function(x) { mon.mean = mean(x, na.rm = TRUE) } ))
my.summary <- do.call(data.frame, my.summary)
colnames(my.summary) <- c('Month', 'Rate1.mean', 'Rate2.mean')
my.summary
my.data <- merge(my.data, my.summary, by = 'Month')
my.data
编辑：2020 年 6 月 28 日
这里我使用aggregate 按组获取整个matrix 的列means，其中组在外部vector 中定义：
my.group <- c(1,2,1,2,2,3,1,2,3,3)
my.data <- matrix(c( 1, 2, 3, 4, 5,
10, 20, 30, 40, 50,
2, 4, 6, 8, 10,
20, 30, 40, 50, 60,
20, 18, 16, 14, 12,
1000, 1100, 1200, 1300, 1400,
2, 3, 4, 3, 2,
50, 40, 30, 20, 10,
1001, 2001, 3001, 4001, 5001,
1000, 2000, 3000, 4000, 5000), nrow = 10, ncol = 5, byrow = TRUE)
my.data
my.summary <- aggregate(list(my.data), by = list(my.group), FUN = function(x) { my.mean = mean(x, na.rm = TRUE) } )
my.summary
# Group.1 X1 X2 X3 X4 X5
#1 1 1.666667 3.000 4.333333 5.000 5.666667
#2 2 25.000000 27.000 29.000000 31.000 33.000000
#3 3 1000.333333 1700.333 2400.333333 3100.333 3800.333333