在列组上应用函数

Question

How can I use apply or a related function to create a new data frame that contains the results of the row averages of each pair of columns in a very large data frame?

I have an instrument that outputs n replicate measurements on a large number of samples, where each single measurement is a vector (all measurements are the same length vectors). I'd like to calculate the average (and other stats) on all replicate measurements of each sample. This means I need to group n consecutive columns together and do row-wise calculations.

For a simple example, with three replicate measurements on two samples, how can I end up with a data frame that has two columns (one per sample), one that is the average each row of the replicates in dat$a , dat$b and dat$c and one that is the average of each row for dat$d , dat$e and dat$f .

Here's some example data

dat <- data.frame( a = rnorm(16), b = rnorm(16), c = rnorm(16), d = rnorm(16), e = rnorm(16), f = rnorm(16))

            a          b            c          d           e          f
1  -0.9089594 -0.8144765  0.872691548  0.4051094 -0.09705234 -1.5100709
2   0.7993102  0.3243804  0.394560355  0.6646588  0.91033497  2.2504104
3   0.2963102 -0.2911078 -0.243723116  1.0661698 -0.89747522 -0.8455833
4  -0.4311512 -0.5997466 -0.545381175  0.3495578  0.38359390  0.4999425
5  -0.4955802  1.8949285 -0.266580411  1.2773987 -0.79373386 -1.8664651
6   1.0957793 -0.3326867 -1.116623982 -0.8584253  0.83704172  1.8368212
7  -0.2529444  0.5792413 -0.001950741  0.2661068  1.17515099  0.4875377
8   1.2560402  0.1354533  1.440160168 -2.1295397  2.05025701  1.0377283
9   0.8123061  0.4453768  1.598246016  0.7146553 -1.09476532  0.0600665
10  0.1084029 -0.4934862 -0.584671816 -0.8096653  1.54466019 -1.8117459
11 -0.8152812  0.9494620  0.100909570  1.5944528  1.56724269  0.6839954
12  0.3130357  2.6245864  1.750448404 -0.7494403  1.06055267  1.0358267
13  1.1976817 -1.2110708  0.719397607 -0.2690107  0.83364274 -0.6895936
14 -2.1860098 -0.8488031 -0.302743475 -0.7348443  0.34302096 -0.8024803
15  0.2361756  0.6773727  1.279737692  0.8742478 -0.03064782 -0.4874172
16 -1.5634527 -0.8276335  0.753090683  2.0394865  0.79006103  0.5704210

I'm after something like this

            X1          X2
1  -0.28358147 -0.40067128
2   0.50608365  1.27513471
3  -0.07950691 -0.22562957
4  -0.52542633  0.41103139
5   0.37758930 -0.46093340
6  -0.11784382  0.60514586
7   0.10811540  0.64293184
8   0.94388455  0.31948189
9   0.95197629 -0.10668118
10 -0.32325169 -0.35891702
11  0.07836345  1.28189698
12  1.56269017  0.44897971
13  0.23533617 -0.04165384
14 -1.11251880 -0.39810121
15  0.73109533  0.11872758
16 -0.54599850  1.13332286

which I did with this, but is obviously no good for my much larger data frame...

data.frame(cbind(
apply(cbind(dat$a, dat$b, dat$c), 1, mean),
apply(cbind(dat$d, dat$e, dat$f), 1, mean)
))

I've tried apply and loops and can't quite get it together. My actual data has some hundreds of columns.

Answer 1

This may be more generalizable to your situation in that you pass a list of indices. If speed is an issue (large data frame) I'd opt for lapply with do.call rather than sapply :

x <- list(1:3, 4:6)
do.call(cbind, lapply(x, function(i) rowMeans(dat[, i])))

Works if you just have col names too:

x <- list(c('a','b','c'), c('d', 'e', 'f'))
do.call(cbind, lapply(x, function(i) rowMeans(dat[, i])))

EDIT

Just happened to think maybe you want to automate this to do every three columns. I know there's a better way but here it is on a 100 column data set:

dat <- data.frame(matrix(rnorm(16*100), ncol=100))

n <- 1:ncol(dat)
ind <- matrix(c(n, rep(NA, 3 - ncol(dat)%%3)), byrow=TRUE, ncol=3)
ind <- data.frame(t(na.omit(ind)))
do.call(cbind, lapply(ind, function(i) rowMeans(dat[, i])))

EDIT 2 Still not happy with the indexing. I think there's a better/faster way to pass the indexes. here's a second though not satisfying method:

n <- 1:ncol(dat)
ind <- data.frame(matrix(c(n, rep(NA, 3 - ncol(dat)%%3)), byrow=F, nrow=3))
nonna <- sapply(ind, function(x) all(!is.na(x)))
ind <- ind[, nonna]

do.call(cbind, lapply(ind, function(i)rowMeans(dat[, i])))

Answer 2

mean for rows from vectors a,b,c

 rowMeans(dat[1:3])

means for rows from vectors d,e,f

 rowMeans(dat[4:6])

all in one call you get

results<-cbind(rowMeans(dat[1:3]),rowMeans(dat[4:6]))

if you only know the names of the columns and not the order then you can use:

rowMeans(cbind(dat["a"],dat["b"],dat["c"]))
rowMeans(cbind(dat["d"],dat["e"],dat["f"]))

#I dont know how much damage this does to speed but should still be quick

Answer 3

A similar question was asked here by @david: averaging every 16 columns in r (now closed), which I answered by adapting @TylerRinker's answer above, following a suggestion by @joran and @Ben. Because the resulting function might be of help to OP or future readers, I am copying that function here, along with an example for OP's data.

# Function to apply 'fun' to object 'x' over every 'by' columns
# Alternatively, 'by' may be a vector of groups
byapply <- function(x, by, fun, ...)
{
    # Create index list
    if (length(by) == 1)
    {
        nc <- ncol(x)
        split.index <- rep(1:ceiling(nc / by), each = by, length.out = nc)
    } else # 'by' is a vector of groups
    {
        nc <- length(by)
        split.index <- by
    }
    index.list <- split(seq(from = 1, to = nc), split.index)

    # Pass index list to fun using sapply() and return object
    sapply(index.list, function(i)
            {
                do.call(fun, list(x[, i], ...))
            })
}

Then, to find the mean of the replicates:

byapply(dat, 3, rowMeans)

Or, perhaps the standard deviation of the replicates:

byapply(dat, 3, apply, 1, sd)

Update

by can also be specified as a vector of groups:

byapply(dat, c(1,1,1,2,2,2), rowMeans)

Answer 4

该rowMeans解决方案会更快，但为了完整性这里是你如何可能做到这一点与apply ：

t(apply(dat,1,function(x){ c(mean(x[1:3]),mean(x[4:6])) }))

Answer 5

Inspired by @joran's suggestion I came up with this (actually a bit different from what he suggested, though the transposing suggestion was especially useful):

Make a data frame of example data with p cols to simulate a realistic data set (following @TylerRinker's answer above and unlike my poor example in the question)

p <- 99 # how many columns?
dat <- data.frame(matrix(rnorm(4*p), ncol = p))

Rename the columns in this data frame to create groups of n consecutive columns, so that if I'm interested in the groups of three columns I get column names like 1,1,1,2,2,2,3,3,3, etc or if I wanted groups of four columns it would be 1,1,1,1,2,2,2,2,3,3,3,3, etc. I'm going with three for now (I guess this is a kind of indexing for people like me who don't know much about indexing)

n <- 3 # how many consecutive columns in the groups of interest?
names(dat) <- rep(seq(1:(ncol(dat)/n)), each = n, len = (ncol(dat)))

Now use apply and tapply to get row means for each of the groups

dat.avs <- data.frame(t(apply(dat, 1, tapply, names(dat), mean)))

The main downsides are that the column names in the original data are replaced (though this could be overcome by putting the grouping numbers in a new row rather than the colnames) and that the column names are returned by the apply-tapply function in an unhelpful order.

Further to @joran's suggestion, here's a data.table solution:

p <- 99 # how many columns?
dat <- data.frame(matrix(rnorm(4*p), ncol = p))
dat.t <-  data.frame(t(dat))

n <- 3 # how many consecutive columns in the groups of interest?
dat.t$groups <- as.character(rep(seq(1:(ncol(dat)/n)), each = n, len = (ncol(dat))))

library(data.table)
DT <- data.table(dat.t)
setkey(DT, groups)
dat.av <- DT[, lapply(.SD,mean), by=groups]

Thanks everyone for your quick and patient efforts!

Answer 6

There is a beautifully simple solution if you are interested in applying a function to each unique combination of columns, in what known as combinatorics.

combinations <- combn(colnames(df),2,function(x) rowMeans(df[x]))

To calculate statistics for every unique combination of three columns, etc., just change the 2 to a 3. The operation is vectorized and thus faster than loops, such as the apply family functions used above. If the order of the columns matters, then you instead need a permutation algorithm designed to reproduce ordered sets: combinat::permn