如何为每列分配因子并根据因子水平计算行均值

Question

I have a data frame which I want to calculate row-means according to certain conditions,
example:

df= 
rownames    A1  A2  A3  B1  B2  B3
r1  1   2   3   4   5   6
r2  1   2   3   4   5   6
r3  1   2   3   4   5   6

treatment= factor(rep(c("A","B"),each=3))

I want to get the means of each row based on the factor level of treatment, data as below:

rownames    A   B
r1  2   5
r2  2   5
r3  2   5

Any thoughts on this?

Answer 1

We can do this with melt from data.table . The measure argument can take multiple patterns to convert to 'long' format, grouped by 'rownames' and specifying the columns in .SDcols , loop through the Subset of Data.table ( .SD ) and get the mean

library(data.table)
melt(setDT(df), measure = patterns("^A", "^B"), value.name = c('A', 'B'))[, 
              lapply(.SD, mean), rownames, .SDcols = A:B]
#   rownames A B
#1:       r1 2 5
#2:       r2 2 5
#3:       r3 2 5

NOTE: The output is a data.table which can be converted to data.frame ( setDF ) as showed in the OP's output

---

Another option is split from base R

sapply(split.default(df[-1], sub("\\d+", "", names(df)[-1])), rowMeans)
#     A B
#[1,] 2 5
#[2,] 2 5
#[3,] 2 5

data

df <- structure(list(rownames = c("r1", "r2", "r3"), A1 = c(1L, 1L, 
1L), A2 = c(2L, 2L, 2L), A3 = c(3L, 3L, 3L), B1 = c(4L, 4L, 4L
), B2 = c(5L, 5L, 5L), B3 = c(6L, 6L, 6L)), .Names = c("rownames", 
"A1", "A2", "A3", "B1", "B2", "B3"), class = "data.frame", row.names = c(NA, 
-3L))

Answer 2

In base R using grep :

sapply(levels(treatment), function(a) rowMeans(df[,grep(a, names(df))]))

#     A B
#[1,] 2 5
#[2,] 2 5
#[3,] 2 5