Question

假设我有一个矩阵A

    [,1] [,2] [,3] [,4] [,5] [,6]
A    0   1    2    3    4    5
     6   7    8    9    8    9

如何将其转换为R中的n x 4矩阵？第一列是常量，第二列是行id，第三列是列id，第四列是单元格值。例如

Z      [,1] [,2] [,3] [,4]
 [1,]   A    1    1    0
 [2,]   A    1    2    1
 [3,]   A    1    3    2
 [4,]   A    1    4    3
 [5,]   A    1    5    4
 [6,]   A    1    6    5
 [7,]   A    2    1    6
 [8,]   A    2    2    7
 [9,]   A    2    3    8
 [10,]  A    2    4    9
 [11,]  A    2    5    8
 [12,]  A    2    6    9

我尝试了很少的功能，申请等等，但结果并没有接近我的预期。

Answer 1

在R中，这样的事情会起作用。

A <- matrix(c(0:9, 8, 9), 2, byrow = TRUE)  ## your matrix
tA <- t(A)
cbind("A", c(col(tA)), c(row(tA)), c(tA))
#       [,1] [,2] [,3] [,4]
#  [1,] "A"  "1"  "1"  "0" 
#  [2,] "A"  "1"  "2"  "1" 
#  [3,] "A"  "1"  "3"  "2" 
#  [4,] "A"  "1"  "4"  "3" 
#  [5,] "A"  "1"  "5"  "4" 
#  [6,] "A"  "1"  "6"  "5" 
#  [7,] "A"  "2"  "1"  "6" 
#  [8,] "A"  "2"  "2"  "7" 
#  [9,] "A"  "2"  "3"  "8" 
# [10,] "A"  "2"  "4"  "9" 
# [11,] "A"  "2"  "5"  "8" 
# [12,] "A"  "2"  "6"  "9"

Answer 2

虽然理查德的回答非常简洁，但你可能没有这些功能让我们的伪代码：

# looks like python
init_A = yourmatrixA
init_R = Matrix.init(4, init_A.nrow*init_A.ncol) # ncol = get number of columns
C = constant   

count = 1
rown = 1
for current_row in init_A.rows:
    coln = 1
    for current_col in current_row: # Assumes current_row is a list
        init_R.set(count, 1, C) # set constant in first row
        init_R.set(count, 2, rown)
        init_R.set(count, 3, coln)
        init_R.set(count, 4, current_col) # current_col is a value
        coln += 1
        count += 1
    rown += 1

init_R.print()

未经测试

Answer 3

此外：

dimnames(A) <- list(row=1:nrow(A),col=1:ncol(A))
cbind(id="A",as.data.frame.table(A,responseName="value"))

#   id row col value
#1   A   1   1     0
#2   A   2   1     6
#3   A   1   2     1
#4   A   2   2     7
#5   A   1   3     2
#6   A   2   3     8
#7   A   1   4     3
#8   A   2   4     9
#9   A   1   5     4
#10  A   2   5     8
#11  A   1   6     5
#12  A   2   6     9

Answer 4

使用melt

library(reshape2)
cbind(id='A', melt(A)) 
#it is better to have the result as a `data.frame`
#as there are `character` columns too
#If you want it as `matrix`, use `as.matrix`

 as.matrix(cbind(id='A', melt(A)))
 #Note that now all the `columns are `character` class

使用常量和单元格地址转换矩阵

4 个答案: