dplyr-summarise.R

# What's the most natural way to express this code in base R?
library(dplyr, warn.conflicts = FALSE)
mtcars %>%
  group_by(cyl) %>%
  summarise(mean = mean(disp), n = n())
#> # A tibble: 3 x 3
#>     cyl  mean     n
#>   <dbl> <dbl> <int>
#> 1     4  105.    11
#> 2     6  183.     7
#> 3     8  353.    14

# tapply() ----------------------------------------------------------------
data.frame(
  cyl = sort(unique(mtcars$cyl)),
  mean = tapply(mtcars$disp, mtcars$cyl, mean),
  n = tapply(mtcars$disp, mtcars$cyl, length)
)
#>   cyl     mean  n
#> 4   4 105.1364 11
#> 6   6 183.3143  7
#> 8   8 353.1000 14
# - hard to generalise to more than one group because tapply() will
#   return an array
# - is `sort(unique(mtcars$cyl))` guaranteed to be in the same order as
#   the tapply() output?

# aggregate() -------------------------------------------------------------
df_mean <- aggregate(mtcars["disp"], mtcars["cyl"], mean)
df_length <- aggregate(mtcars["disp"], mtcars["cyl"], length)
names(df_mean)[2] <- "mean"
names(df_length)[2] <- "n"
merge(df_mean, df_length, by = "cyl")
#>   cyl     mean  n
#> 1   4 105.1364 11
#> 2   6 183.3143  7
#> 3   8 353.1000 14

# + generalises in stratightforward to multiple grouping variables and
#   multiple summary variables
# - need to manually rename summary variables

# Could also use formula interface
# https://twitter.com/tjmahr/status/1231255000766005248
df_mean <- aggregate(disp ~ cyl, mtcars, mean)
df_length <- aggregate(disp ~ cyl, mtcars, length)

# by() --------------------------------------------------------------------
mtcars_by <- by(mtcars, mtcars$cyl, function(df) {
  data.frame(cyl = df$cyl[[1]], mean = mean(df$disp), n = nrow(df))
})
do.call(rbind, mtcars_by)
#>   cyl     mean  n
#> 4   4 105.1364 11
#> 6   6 183.3143  7
#> 8   8 353.1000 14

# + generalises easily to more/different summaries
# - need to know about anonymous functions + do.call + rbind

# by() = split() + lapply()
mtcars_by <- lapply(split(mtcars, mtcars$cyl), function(df) {
  data.frame(cyl = df$cyl[[1]], mean = mean(df$disp), n = nrow(df))
})
do.call(rbind, mtcars_by)
#>   cyl     mean  n
#> 4   4 105.1364 11
#> 6   6 183.3143  7
#> 8   8 353.1000 14

# Manual indexing approahes -------------------------------------------------
# from https://twitter.com/fartmiasma/status/1231258479865647105
cyl_counts <- sort(unique(mtcars$cyl))
tabl <- sapply(cyl_counts, function(ct) {
  with(mtcars, c(cyl = ct, mean = mean(disp[cyl == ct]), n = sum(cyl == ct)))
})
as.data.frame(t(tabl))
#>   cyl     mean  n
#> 1   4 105.1364 11
#> 2   6 183.3143  7
#> 3   8 353.1000 14

# - coerces all results (and grouping var) to common type

# Similar approach from
# https://gist.github.com/hadley/c430501804349d382ce90754936ab8ec#gistcomment-3185680
s <- lapply(cyl_counts, function(cyl) {
  indx <- mtcars$cyl == cyl
  data.frame(cyl = cyl, mean = mean(mtcars$disp[indx]), n = sum(indx))
})
do.call(rbind, s)
#>   cyl     mean  n
#> 1   4 105.1364 11
#> 2   6 183.3143  7
#> 3   8 353.1000 14

# - harder to generalise to multiple grouping vars (need to use Map())

Adding the link of the tweet where this was also discussed, so that we can remove it from the vignette, because CRAN url checks.

https://twitter.com/hadleywickham/status/1231252596712771585

dfs <- split(mtcars, mtcars$cyl)                                      # ~ group_by
res <- t(sapply(dfs, function(df) c(mean(df$disp), length(df$disp)))) # ~ summarise
res <- as.data.frame(cbind(as.numeric(names(dfs)), res),              # attach cyl values
                     row.names=1:nrow(res))                           # name rows 
colnames(res) <- c("cyl", "mean", "n")                                # name cols

The result:

#>res
#  cyl     mean  n
#1   4 105.1364 11
#2   6 183.3143  7
#3   8 353.1000 14

@hadley In your examples, instead of do.call one could also use Reduce instead.

@hadley To your question in your tapply() example - whether the order of tapply's output is guaranteed to be the same like sort(unique(mtcars$cyl)):

Initially, I thought "yes!", because tapply uses split from which it overtakes the output order. split in turn uses factor to sort the names of the groups, which in turn uses order() to define the order of the levels. sort.default uses in its code also order for sorting. Thus, the output order of the groups must be identical, since tapply() as well as sort(unique()) both use the same function order() to order their output.
In case the names are integers, sort.int() should always lead to the same output order ...

> split.default
function (x, f, drop = FALSE, sep = ".", lex.order = FALSE, ...) 
{
    if (!missing(...)) 
        .NotYetUsed(deparse(...), error = FALSE)
    if (is.list(f)) 
        f <- interaction(f, drop = drop, sep = sep, lex.order = lex.order)
    else if (!is.factor(f)) 
        f <- as.factor(f)
    else if (drop) 
        f <- factor(f)
    storage.mode(f) <- "integer"
    if (is.null(attr(x, "class"))) 
        return(.Internal(split(x, f)))
    ind <- .Internal(split(seq_along(x), f))
    lapply(ind, function(i) x[i])
}

> factor
function (x = character(), levels, labels = levels, exclude = NA, 
    ordered = is.ordered(x), nmax = NA) 
{
    if (is.null(x)) 
        x <- character()
    nx <- names(x)
    if (missing(levels)) {
        y <- unique(x, nmax = nmax)
        ind <- order(y)
        levels <- unique(as.character(y)[ind])
    }
    force(ordered)
    if (!is.character(x)) 
        x <- as.character(x)
    levels <- levels[is.na(match(levels, exclude))]
    f <- match(x, levels)
    if (!is.null(nx)) 
        names(f) <- nx
    if (missing(labels)) {
        levels(f) <- as.character(levels)
    }
    else {
        nlab <- length(labels)
        if (nlab == length(levels)) {
            nlevs <- unique(xlevs <- as.character(labels))
            at <- attributes(f)
            at$levels <- nlevs
            f <- match(xlevs, nlevs)[f]
            attributes(f) <- at
        }
        else if (nlab == 1L) 
            levels(f) <- paste0(labels, seq_along(levels))
        else stop(gettextf("invalid 'labels'; length %d should be 1 or %d", 
            nlab, length(levels)), domain = NA)
    }
    class(f) <- c(if (ordered) "ordered", "factor")
    f
}

> sort.default
function (x, decreasing = FALSE, na.last = NA, ...) 
{
    if (is.object(x)) 
        x[order(x, na.last = na.last, decreasing = decreasing)]
    else sort.int(x, na.last = na.last, decreasing = decreasing, 
        ...)
}

But then I realized after all this - that if the cyl column in mtcars would not be numeric but a factor with customized levels, the ordering methods might differ, and indeed it does:

# if the `cyl` column would be a factor - with a custom ordering of its levels
mtcars_ <- mtcars
mtcars_$cyl <- as.factor(mtcars_$cyl)
levels(mtcars_$cyl) <- c("8", "6", "4")

# then the ordering by `tapply()` differs from the ordering by `sort()`
data.frame(
  cyl = sort(unique(mtcars_$cyl)),
  mean = tapply(mtcars_$disp, mtcars_$cyl, mean),
  n = tapply(mtcars_$disp, mtcars_$cyl, length)
)
 
#   cyl     mean  n
# 8   8 105.1364 11
# 6   6 183.3143  7
# 4   4 353.1000 14

@gwangjinkim Reduce works pair wise, which is potential slower than providing all data frames to rbind at once (although I don't know if rbind actually takes advantage of that). And nice spotting with the order problem, thanks!

@hadley Thank you, too! I often used Reduce (because I was coming from Lisp languages, where you have reduce). do.call is actually Lisp's apply - while the apply functions in R are more like a mapcar in LIsp ... - so to use do.call in such situations is then better! Good to know, thank you!

Thank you for the interesting question.
To me the most natural way to express the counts in a group variable in base R is to use table(), but it coerces the grouped variable into a factor. Since I didn't see its use in the previous answers, here is my trial:

dat            <- setNames(as.data.frame(table(mtcars$cyl)), c("cyl", "n"))
dat$cyl      <- as.numeric(as.character(dat$cyl))
dat$mean     <- sapply(dat$cyl, function(x) mean(with(mtcars, disp[cyl == x])))
dat            <- dat[, c(1,3,2)]
dat
#   cyl     mean  n
# 1   4 105.1364 11
# 2   6 183.3143  7
# 3   8 353.1000 14

Note a downside of the formula based aggregate syntax — it drops missing values like a modelling function:

aggregate(cbind(Ozone, Temp) ~ Month, data = airquality, length)
#>   Month Ozone Temp
#> 1     5    26   26
#> 2     6     9    9
#> 3     7    26   26
#> 4     8    26   26
#> 5     9    29   29
aggregate(airquality[c("Ozone", "Temp")], airquality["Month"], length)
#>   Month Ozone Temp
#> 1     5    31   31
#> 2     6    30   30
#> 3     7    31   31
#> 4     8    31   31
#> 5     9    30   30

^{Created on 2022-10-24 with reprex v2.0.2}