Stats on every n rows for each column

问题:

I would like to calculate the mean and standard deviation for every nth (in my case every 6) rows (or samples). The following function gives me the means for every 6 rows (96 rows gives me 16 mean values)

colMeans(matrix(data.trim$X0, nrow=6)) 

I would like to do this for ALL columns (a total of 1280 mean values). I tried running this function:

colMeans(matrix(data.trim, nrow=6)) 

but this does not work at all and I get the following error message:

Error in colMeans(matrix(data.trim, nrow = 6)) : 'x' must be numeric

In addition: Warning message:

In matrix(data.trim, nrow = 6) : data length [80] is not a sub-multiple or multiple of the number of rows [6]

回答1:

You can apply the function to each column with sapply:

sapply(iris[1:4], function(x) colMeans(matrix(x, nrow=6)))       Sepal.Length Sepal.Width Petal.Length Petal.Width  [1,]     4.950000    3.383333     1.450000   0.2333333  [2,]     4.850000    3.316667     1.483333   0.2000000  [3,]     5.183333    3.633333     1.316667   0.2500000 

...

[23,]     6.533333    2.950000     5.583333   1.9333333 [24,]     6.516667    3.033333     5.316667   2.1333333 [25,]     6.383333    3.033333     5.266667   2.1333333 

Compare with creating the means of the first six rows manually:

colMeans(iris[1:6, 1:4]) Sepal.Length  Sepal.Width Petal.Length  Petal.Width     4.9500000    3.3833333    1.4500000    0.2333333  

You can also do this with aggregate given the proper by argument:

aggregate(iris[1:4], by=list((seq(nrow(iris))-1) %/% 6), FUN=mean)    Group.1 Sepal.Length Sepal.Width Petal.Length Petal.Width 1        0     4.950000    3.383333     1.450000   0.2333333 2        1     4.850000    3.316667     1.483333   0.2000000 3        2     5.183333    3.633333     1.316667   0.2500000 

...

This works by creating a vector which identifies the groups to be averaged:

(seq(nrow(iris))-1) %/% 6   [1]  0  0  0  0  0  0  1  1  1  1  1  1  2  2  2  2  2  2  3  3  3  3  3  3  4  4  4  4  4  4  5  5  5  5  5  5  6  6  6  6  6  6  7  7  7  7  7  7  8  8  8  8  [53]  8  8  9  9  9  9  9  9 10 10 10 10 10 10 11 11 11 11 11 11 12 12 12 12 12 12 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 16 16 16 17 17 [105] 17 17 17 17 18 18 18 18 18 18 19 19 19 19 19 19 20 20 20 20 20 20 21 21 21 21 21 21 22 22 22 22 22 22 23 23 23 23 23 23 24 24 24 24 24 24 

The sapply solution returns a matrix, whereas the aggregate solution returns a data frame, in case one is more desirable.

回答2:

I think a possible reason that you got Error, warning message is because you applied it directly on the data.frame. For example

set.seed(48) d1 <- as.data.frame(matrix(sample(1:40, 80*96, replace=T), ncol=80)) rowMeans(matrix(d1, ncol=6, byrow=T)) #Error in rowMeans(matrix(d1, ncol = 6, byrow = T)) : 'x' must be numeric #In addition: Warning message: #In matrix(d1, ncol = 6, byrow = T) : #  data length [80] is not a sub-multiple or multiple of the number of rows [14] 

You could unlist the data.frame

 res <- rowMeans(matrix(unlist(d1), ncol=6, byrow=T))  dim(res) <- c(96/6, 80) length(res) #[1] 1280 

Crosschecking the results from @Matthew Lundberg's method

res1 <- sapply(d1, function(x) colMeans(matrix(x, nrow=6)))  all.equal(res,res1, check.attributes=F) [1] TRUE 

文章来源: Stats on every n rows for each column