Question

我遇到了问题，但我确信这对于非常熟悉R的人来说非常容易。我有一个3008 x 3008的矩阵。我想要的是每行中每8列的总和。所以基本上你最终会得到一个新的矩阵，现在是367 x 367.

这是一个小例子：

           C.1 C.2 C.3 C.4 C.5 C.6
    row1    1   2   1   2   5   6
    row1    1   2   3   4   5   6
    row1    2   6   3   4   5   6
    row1    1   2   3   4   10   6

所以说我想在每一行中每3列加上这些，我想最终得到：

           C.1 C.2
    row1    4   13
    row1    6   15
    row1   11   15
    row1    6   20

Answer 1

# m is your matrix
n <- 8
grp <- seq(1, ncol(m), by=n)
sapply(grp, function(x) rowSums(m[, x:(x+n-1)]))

如果您是R. grp的新手，则会有一些解释是一系列数字，它们为每组列提供起点：1,9,17等，如果您想要每8列求和一次。

sapply电话可以理解如下。对于grp中的每个数字，它调用rowSums函数，并将与该组编号对应的矩阵列传递给它。因此，当grp为1时，它获得第1-8列的行总和;当grp为9时，它获取第9-16列的行总和，依此类推。这些是向量，sapply然后将它们组合成一个矩阵。

Answer 2

将矩阵转换为数组，然后使用apply和rowSums。

mat <- structure(c(1L, 1L, 2L, 1L, 2L, 2L, 6L, 2L, 1L, 3L, 3L, 3L, 2L, 4L, 4L, 4L, 5L, 5L, 5L, 10L, 6L, 6L, 6L, 6L), 
                 .Dim = c(4L, 6L), 
                 .Dimnames = list(c("row1", "row2", "row3", "row4"), c("C.1", "C.2", "C.3", "C.4", "C.5", "C.6")))

n <- 3 #this needs to be a factor of the number of columns
a <- array(mat,dim=c(nrow(mat),n,ncol(mat)/n))
apply(a,3,rowSums)
#      [,1] [,2]
# [1,]    4   13
# [2,]    6   15
# [3,]   11   15
# [4,]    6   20

Answer 3

#Create sample data:
df <- matrix(rexp(200, rate=.1), ncol=20)

#Choose the number of columns you'd like to sum up (e.g., 3 or 8)
number_of_columns_to_sum <- 3

df2 <- NULL #Set to null so that you can use cbind on the first value below
for (i in seq(1,ncol(df), by = number_of_columns_to_sum)) {
  df2 <- cbind(df2, rowSums(df[,i:(i+number_of_columns_to_sum-1)]))
}

Answer 4

另一种选择：虽然它可能不那么优雅

mat <- structure(c(1L, 1L, 2L, 1L, 2L, 2L, 6L, 2L, 1L, 3L, 3L, 3L, 2L, 4L, 4L, 4L, 5L, 5L, 5L, 10L, 6L, 6L, 6L, 6L), 
                 .Dim = c(4L, 6L), 
                 .Dimnames = list(c("row1", "row1", "row1", "row1"), c("C.1", "C.2", "C.3", "C.4", "C.5", "C.6")))

new<- data.frame((mat[,1]+mat[,2]+mat[,3]),(mat[,4]+mat[,5]+mat[,6]))
names(new)<- c("C.1","C.2")
new

R对每行中的某些列求和

4 个答案: