Calculate the mean of every 13 rows in data frame

Question

I have a data frame with 2 columns and 3659 row df

I am trying to reduce the data set by averaging every 10 or 13 rows in this data frame, so I tried the following :

# number of rows per group
n=13
# number of groups
n_grp=nrow(df)/n
round(n_grp,0)
# row indices (one vector per group)
idx_grp <- split(seq(df), rep(seq(n_grp), each = n))

# calculate the col means for all groups
res <- lapply(idx_grp, function(i) {
  # subset of the data frame
  tmp <- dat[i]
  # calculate row means
  colMeans(tmp, na.rm = TRUE)
})
# transform list into a data frame
dat2 <- as.data.frame(res)

However, I can't divide my number of rows by 10 or 13 because data length is not a multiple of split variable. So I am not sure what should do then (I just want may be to calculate the mean of the last group -even with less than 10 elements)

I also tried this one, but the results are the same:

df1=split(df, sample(rep(1:301, 10)))

Answer 1

Here's a solution using aggregate() and rep().

df <- data.frame(a=1:12, b=13:24 );
df;
##     a  b
## 1   1 13
## 2   2 14
## 3   3 15
## 4   4 16
## 5   5 17
## 6   6 18
## 7   7 19
## 8   8 20
## 9   9 21
## 10 10 22
## 11 11 23
## 12 12 24
n <- 5;
aggregate(df, list(rep(1:(nrow(df) %/% n + 1), each = n, len = nrow(df))), mean)[-1];
##      a    b
## 1  3.0 15.0
## 2  8.0 20.0
## 3 11.5 23.5

The important part of this solution that handles the issue of non-divisibility of nrow(df) by n is specifying the len parameter (actually the full parameter name is length.out) of rep(), which automatically caps the group vector to the appropriate length.

Answer 2

If df is a data.table, you can use %/% to group as in

library(data.table)
setDT(df)
n <- 13 # every 13 rows

---

df[, mean(z), by= (seq(nrow(df)) - 1) %/% n]

if instead you want every nTH row, use %% instead of %/%

df[, mean(z), by= (seq(nrow(df)) - 1) %% n]