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# 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())
@llrs

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@llrs llrs commented Feb 22, 2020

The second example doesn't return the same result as the other solutions, you used mpg instead of disp column for the mean.

I would use this or make a for loop to avoid the final call to rbind and to create a new data.frame for each case.

s <- lapply(unique(mtcars$cyl), function(x){
    k <- mtcars$cyl == x
    n <- sum(k)
    m <- mean(mtcars$disp[k]) # if several columns it could be used inside an apply call.
    data.frame(cyl = x, mean = m, n = n)}
)
do.call(rbind, s)
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@hadley hadley commented Feb 22, 2020

@llrs added your approach — thanks! How would you use a for loop here?

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@llrs llrs commented Feb 22, 2020

Like this if you want to be memory efficient:

keys <- unique(mtcars$cyl)
n <- vector("numeric", length(keys))
m <- vector("numeric", length(keys))
for (x in seq_along(keys)) {
    k <- mtcars$cyl == keys[x]
    n[x] <- sum(k)
    m[x] <- mean(mtcars$disp[k]) # if several columns it could be used inside an apply call.
}
data.frame(cyl = keys, mean = m, n = n)
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@TimTeaFan TimTeaFan commented Feb 22, 2020

The aggregate approach could be optimized the following way:

aggregate(disp ~ cyl,
          mtcars,
          function(x) c(mean = mean(x), n = length(x)))

#>  cyl disp.mean   disp.n
#> 1   4  105.1364  11.0000
#> 2   6  183.3143   7.0000
#> 3   8  353.1000  14.0000

++ It's much less verbose than the original aggregate approach from above and easier to generalize than the twitter approach with separate calls with df <-
++ no need to adjust the naming of the variables
-- it will return all variables in the same format, that means <dbl> as soon as there is one variable included that can't be coerced to integer.
-- The result is a data.frame with two columns (cyl and disp) the latter is a matrix.

To remedy the last point, we could wrap the aggregate in a with call, but that would be again more verbose:

with(aggregate(disp ~ cyl,
               mtcars,
               function(x) c(mean = mean(x), n = length(x))),
     as.data.frame(cbind(cyl, disp)))
#>  cyl     mean  n
#>1   4 105.1364 11
#>2   6 183.3143  7
#>3   8 353.1000 14
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@AlbanOtt AlbanOtt commented Feb 23, 2020

Just after I left the university, I would probably have written something like that :

cyl_u = unique(mtcars$cyl)
res=c()
for(cyl in cyl_u){
  keepit=mtcars$cyl==cyl
  mean=mean(mtcars[keepit,"disp"])
  n=sum(keepit)
  res=rbind(res,
            data.frame(cyl,mean,n))
}
res

Yes it might be shameful but you said : "How would you use a for loop here?" so here I am...

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@Myfanwy Myfanwy commented Feb 24, 2020

I learned Base R mostly after the fact, but here's how I was taught to do it, FWIW: write a function as if there were only one group, and make sure it returns the answer in the format you want. Then apply it to all the groups. Can replace some of the below with by, or combine different parts, but just wanted to convey the thought process most importantly (do for one group first, make sure it works, then apply to all groups):

onecar = function(x) {
  data.frame(mean_per_cyl = mean(x$disp),
             n = nrow(x))
}

mtsplit = split(mtcars, mtcars$cyl) # could obviously move this step into the function
summ_cyls = do.call(rbind, lapply(mtsplit, onecar)) 

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@dwoll dwoll commented Feb 25, 2020

merge() has an argument suffixes which eliminates the need to manually rename the aggregated variables:

df_mean <- aggregate(mtcars["disp"], mtcars["cyl"], mean)
df_length <- aggregate(mtcars["disp"], mtcars["cyl"], length)
merge(df_mean, df_length, by = "cyl", suffixes = c("_mean", "_n"))
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@alexpavlakis alexpavlakis commented Feb 27, 2020

imo this is the clearest base R approach to the problem (probably not the fastest though).

summariseByGroup <- function(groupData, numericData) {
  group <- sort(unique(groupData))
  nGroups <- length(group)
  n <- vector('integer', nGroups)
  mean <- vector('numeric', nGroups)
  for(g in seq_along(group)) {
    for(i in seq_along(numericData)) {
      if(groupData[i] == group[g]) {
        n[g] <- n[g] + 1
        mean[g] <- (mean[g]*(n[g] - 1) + numericData[i])/n[g]
      }
    }
  }
  data.frame(group, mean, n)
}

summariseByGroup(mtcars$cyl, mtcars$disp)
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