Mean per group in a data.frame [duplicate]
I have a data.frame and I need to calculate the mean per group (i.e. per Month, below).
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
My desired output is like below, where the values for Rate1 and Rate2 are the group means. Please disregard the value, I have made it up for the example.
Name Rate1 Rate2
Aira 23.21 12.2
Ben 45.23 43.9
Cat 33.22 32.2
This type of operation is exactly what aggregate was designed for:
d <- 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)
aggregate(d[, 3:4], list(d$Name), mean)
Group.1 Rate1 Rate2
1 Aira 16.33333 47.00000
2 Ben 31.33333 50.33333
3 Cat 44.66667 54.00000
Here we aggregate columns 3 and 4 of data.frame d, grouping by d$Name, and applying the mean function.
Or, using a formula interface:
aggregate(. ~ Name, d[-2], mean)
Or use group_by & summarise_at from the dplyr package:
library(dplyr)
d %>%
group_by(Name) %>%
summarise_at(vars(-Month), funs(mean(., na.rm=TRUE)))
# A tibble: 3 x 3
Name Rate1 Rate2
<fct> <dbl> <dbl>
1 Aira 16.3 47.0
2 Ben 31.3 50.3
3 Cat 44.7 54.0
See ?summarise_at for the many ways to specify the variables to act on. Here, vars(-Month) says all variables except Month.
In more recent versions of tidyverse/dplyr, using summarise(across(...)) is preferred to summarise_at:
d %>%
group_by(Name) %>%
summarise(across(-Month, mean, na.rm = TRUE))
You can also use package plyr, which is somehow more versatile:
library(plyr)
ddply(d, .(Name), summarize, Rate1=mean(Rate1), Rate2=mean(Rate2))
Name Rate1 Rate2
1 Aira 16.33333 47.00000
2 Ben 31.33333 50.33333
3 Cat 44.66667 54.00000
A third great alternative is using the package data.table, which also has the class data.frame, but operations like you are looking for are computed much faster.
library(data.table)
mydt <- structure(list(Name = c("Aira", "Aira", "Aira", "Ben", "Ben", "Ben", "Cat", "Cat", "Cat"), Month = c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), Rate1 = c(15.6396600443877, 2.15649279424609, 6.24692918928743, 2.37658797276116, 34.7500663272292, 3.28750138697048, 29.3265553981065, 17.9821839334431, 10.8639802575958), Rate2 = c(17.1680489538369, 5.84231656330206, 8.54330866437461, 5.88415184986176, 3.02064294862551, 17.2053351400752, 16.9552950199166, 2.56058000170089, 15.7496228048122)), .Names = c("Name", "Month", "Rate1", "Rate2"), row.names = c(NA, -9L), class = c("data.table", "data.frame"))
Now to take the mean of Rate1 and Rate2 for all 3 months, for each person (Name): First, decide which columns you want to take the mean of
colstoavg <- names(mydt)[3:4]
Now we use lapply to take the mean over the columns we want to avg (colstoavg)
mydt.mean <- mydt[,lapply(.SD,mean,na.rm=TRUE),by=Name,.SDcols=colstoavg]
mydt.mean
Name Rate1 Rate2
1: Aira 8.014361 10.517891
2: Ben 13.471385 8.703377
3: Cat 19.390907 11.755166
Here are a variety of ways to do this in base R including an alternative aggregate approach. The examples below return means per month, which I think is what you requested. Although, the same approach could be used to return means per person:
Using 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
Using 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$Month, FUN = function(x) colMeans(x[,3:4]))))
colnames(by.month) <- c('Rate1.mean', 'Rate2.mean')
by.month <- cbind(Month = rownames(by.month), by.month)
my.data <- merge(my.data, by.month, by = 'Month')
my.data
Using lapply and split:
my.data <- read.table(text = '
Name Month Rate1 Rate2
Aira 1 12 23
Aira 2 18 73
Aira 3 19 45
Ben 1 53 19
Ben 2 22 87
Ben 3 19 45
Cat 1 22 87
Cat 2 67 43
Cat 3 45 32
', header = TRUE, stringsAsFactors = FALSE, na.strings = 'NA')
ly.mean <- lapply(split(my.data, my.data$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
Using sapply and 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
Using 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
EDIT: June 28, 2020
Here I use aggregate to obtain the column means of an entire matrix by group where group is defined in an external vector:
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