Question

我有一个超过1000行和100列的大矩阵。在每行中只有6-10列具有值，其余为零。我想创建一个只有5列的矩阵，它取每行中5个连续列的值。例如：

A = structure(c(0, 1L, 6L, 0, 2L, 0, 2L, 0, 1L, 4L, 1L, 3L, 7L, 2L, 6L, 2L, 4L, 0, 3L, 0, 3L, 5L, 1L, 4L, 0, 4L, 6L, 2L, 0, 0, 5L, 0, 3L, 5L, 0, 0, 0, 4L, 6L, 7L, 0, 7L, 5L, 7L, 8L, 6L, 0, 0, 8L, 9L, 0, 0, 0, 9L, 1L, 0 , 0, 0, 0, 2L, 7L, 0, 2L, 0, 0, 1L, 8L, 4, 0, 0), .Dim = c(5L, 14L))

#A =
#     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
#[1,]    0    0    1    2    3    4    5    0    0     6     0     0     7     1
#[2,]    1    2    3    4    5    6    0    0    7     0     0     0     0     8
#[3,]    6    0    7    0    1    2    3    4    5     0     0     0     2     4
#[4,]    0    1    2    3    4    0    5    6    7     8     9     0     0     0
#[5,]    2    4    6    0    0    0    0    7    8     9     1     2     0     0

我想要这个矩阵：

B = structure(c(1L, 1L, 1L, 5L, 7L, 2L, 2L, 2L, 6L, 8L, 3L, 3L, 3L, 7L, 9L, 4L, 4L, 4L, 8L, 1L, 5L, 5L, 5L, 9L, 2L), .Dim = c(5L, 5L))


#B = 
#     [,1] [,2] [,3] [,4] [,5]
#[1,]    1    2    3    4    5
#[2,]    1    2    3    4    5
#[3,]    1    2    3    4    5
#[4,]    5    6    7    8    9
#[5,]    7    8    9    1    2

我的代码：

df = data.frame(A)
B = do.call(rbind, lapply(1:NROW(df), function(i) df[i,][(df[i,])!=0][1:5]))
# or
B = t(apply(X = df, MARGIN = 1, function(x) x[x!=0][1:5]))

我的代码适用于A的前两行，但其余行无效。我还考虑过获取无零的列索引，然后查看是否有5个连续的列（它们之间没有任何间隙）并检索它们的值。任何帮助非常感谢！

Answer 1

您可以使用rollapply：

library(zoo)
t(apply(A,1,function(x) {x[match(T,rollapply(!!x,5,all)) + (0:4)]}))

#      [,1] [,2] [,3] [,4] [,5]
# [1,]    1    2    3    4    5
# [2,]    1    2    3    4    5
# [3,]    1    2    3    4    5
# [4,]    5    6    7    8    9
# [5,]    7    8    9    1    2

如果您的行没有任何5的序列，它会崩溃，如果您希望处理它，请更新您的帖子。

或者相同但更漂亮：

library(purrr)
Adf       <- as.data.frame(t(A)) # data.frame fits more this data conceptually, you have different series, and it's better to put them in columns
res_df  <- map_df(Adf,~ {.x[match(T,rollapply(.x!=0,5,all))+(0:4)]})
res_mat <- as.matrix(t(unname(res_df))) # if you want to go back to ugly :)

Answer 2

以下是使用rle

的选项

t(apply(A, 1, function(x) {
      rl <- rle(x !=0)
    head(x[inverse.rle(within.list(rl, values[!(values & lengths >= 5)] <- FALSE))], 5)}))
#      [,1] [,2] [,3] [,4] [,5]
#[1,]    1    2    3    4    5
#[2,]    1    2    3    4    5
#[3,]    1    2    3    4    5
#[4,]    5    6    7    8    9
#[5,]    7    8    9    1    2

Answer 3

编辑：错过了一些细节，这是我尝试使用apply和基础库的新尝试：

cumfun <- function(x){           
  y<-ifelse(x>0,1,0)
  tmp<-cumsum(y)
  pos<-which(tmp-cummax((!y)*tmp)==5)
  x[(pos-4) : pos]
}

B<-t(apply(A,1,cumfun))

> B
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    2    3    4    5
[2,]    1    2    3    4    5
[3,]    1    2    3    4    5
[4,]    5    6    7    8    9
[5,]    7    8    9    1    2

Answer 4

library(zoo)
t(apply(A, MAR = 1, function(x, n = 5) x[which(rollsum(!!x, n)==n)[1]+seq_len(n)-1]))

     [,1] [,2] [,3] [,4] [,5]
[1,]    1    2    3    4    5
[2,]    1    2    3    4    5
[3,]    1    2    3    4    5
[4,]    5    6    7    8    9
[5,]    7    8    9    1    2

使用R中的条件减少列数

4 个答案: