[英]R: Matrix row operations
I have a matrix A that has a large number of rows and columns (below one example of such a matrix) that occasionally has a full row of 0 values (as in row 4 at this particular example). 我有一个矩阵A,它具有大量的行和列(在这样的矩阵的一个示例下面),偶尔具有一整行的0值(如在此特定示例中的第4行)。
I want to have a function that checks all rows of A and allows me to perform an operation on each element of these rows. 我想拥有一个检查A的所有行并允许我对这些行的每个元素执行操作的函数。 Is there an easy way to do that?
有没有简单的方法可以做到这一点?
I also wonder if matrix is the right data structure for this. 我也想知道矩阵是否是正确的数据结构。 It feels not quite right, perhaps data frames are better for that?
感觉不太正确,也许数据帧对此更好?
A = matrix(
c(0, 0, 1, 0, 0, 0, 0,
1, 0, 1, 1, 0, 0, 0,
0, 0, 0, 1, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1,
0, 0, 0, 1, 1, 0, 1), nrow=7,ncol=7,byrow = TRUE)
For every row of that matrix I want to determine if there are only 0's in it. 对于该矩阵的每一行,我都想确定其中是否只有0。 If so, I want to set (for each element) the value 1/N (where N is the ncol(A)).
如果是这样,我想为每个元素设置值1 / N(其中N是ncol(A))。
Sudo code: 须藤代码:
If (sum(row of A) == 0) then row_of_A = 1/ncol(A) 如果(sum(A行)== 0),则row_of_A = 1 / ncol(A)
Apparently you want this: 显然,您需要这样做:
A[rowSums(A != 0) == 0,] <- 1/ncol(A)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#[1,] 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#[2,] 1.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000 0.0000000
#[3,] 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000
#[4,] 0.1428571 0.1428571 0.1428571 0.1428571 0.1428571 0.1428571 0.1428571
#[5,] 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000
#[6,] 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 1.0000000
#[7,] 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 0.0000000
Explanation: 说明:
A != 0
checks all matrix elements and returns a logical matrix with TRUE
for non-zero elements. A != 0
检查所有矩阵元素,并为非零元素返回逻辑为TRUE
的逻辑矩阵。 FALSE
/ TRUE
is coerced to 0/1. FALSE
/ TRUE
强制为0/1。 Benchmarks to show that apply
is slower: 基准表明
apply
较慢:
set.seed(42); A = matrix(sample(0:1, 5e4, TRUE), nrow=1e4)
library(microbenchmark)
microbenchmark(A[rowSums(A != 0) == 0,],
A[!apply(A != 0, 1, any),],
A[apply(A == 0, 1, all),])
#Unit: microseconds
# expr min lq mean median uq max neval cld
# A[rowSums(A != 0) == 0, ] 572.202 593.298 620.7931 624.248 629.638 780.387 100 a
# A[!apply(A != 0, 1, any), ] 14978.248 16124.652 17261.9530 17441.054 18129.975 22469.219 100 b
# A[apply(A == 0, 1, all), ] 15182.122 16149.751 17616.8010 16561.657 17997.703 75148.079 100 b
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