R - 在不使用循环的情况下循环遍历不同的矩阵！帮助简单的代码

所以我有两个单独的矩阵（mat1和mat2），我需要通过它们才能进行检查。我需要将结果存储到第三个矩阵中。R - 在不使用循环的情况下循环遍历不同的矩阵！帮助简单的代码

我觉得我的代码很长的目的。

我想提供一些避免循环的建议。

所以我的第一个矩阵看起来像这样（到底dput）

wit5.001 wit5.002 wit5.003 wit5.004 wit5.005 wit5.006 wit5.007 wit5.008 wit5.009 wit5.010 
[1,]  1  1  1  1  1  1  1   1  1  1 
[2,]  1  1  1  1  1  1  1  1  1  1 
[3,]  1  1  1  1  1  1  1  1  1  1 
[4,]  1  1  1  1  1  1  1  1  1  1 
[5,]  1  1  1  1  1  1  1  0  1  1 
[6,]  1  1  1  1  1  1  1  0  0  0 
[7,]  0  1  1  1  1  1  1  1  1  1 
[8,]  1  1  1  1  1  1  1  1  1  1 
[9,]  1  1  1  1  1  1  1  1  1  1 
[10,]  1  1  1  1  1  1  1  1  1  1

我的第二个矩阵具有类似的结构。

这里我创建了我的第三个矩阵 - 为了存储检查的结果。

matCheck <- matrix('ok', ncol = ncol(mat1), nrow = nrow(mat1))

这里是我的循环 - 这是我想避免

for(j in 1:ncol(mat1)){ 
    for(i in 1:nrow(mat1)){ 

    if(mat1[i,j] == 1 & mat2[i,j] == 1) 
    {matCheck[i,j] <- 'ok'} 

    if(mat1[i,j] != 1 & mat2[i,j] == 1) 
    {matCheck[i,j] <- '!'} 

     } 
    }

检查

 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] 
[1,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" 
[2,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" 
[3,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" 
[4,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" 
[5,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" 
[6,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "!" "!" "ok" 
[7,] "!" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" 
[8,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" 
[9,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" 
[10,] "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok" "ok"

任何建议的结果？

这里是矩阵1

mat1 = structure(c(1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1), .Dim = c(10L, 
10L), .Dimnames = list(NULL, c("wit5.001", "wit5.002", "wit5.003", 
"wit5.004", "wit5.005", "wit5.006", "wit5.007", "wit5.008", "wit5.009", 
"wit5.010")))

这里是矩阵2

mat2 = structure(c(1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 
1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 
1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 
0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 
0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0), .Dim = c(10L, 
10L), .Dimnames = list(NULL, c("wit5.020", "wit5.021", "wit5.022", 
"wit5.023", "wit5.024", "wit5.025", "wit5.026", "wit5.027", "wit5.028", 
"wit5.029")))

来源

2015-06-19 giacomo

@Frank - 对不起，我的错 - 你是对的，他们不是不同的维度（我在我的子集中犯了一个错误）。我更正了代码。 – giacomo

我会试试你们的主张，谢谢 – giacomo

是不是很简单！哦，不......在所有的努力后把我的循环;）谢谢@弗兰克 - 把它作为一个答案 – giacomo

对于示例给出的，可以构造该结果作为

matCheck <- (mat1 | !mat2)

这等同于初始化matCheck为真，然后填写虚假!mat1 & mat2（如在OP的循环）。括号是可选的，但更容易阅读（我认为）。

来源

2015-06-19 16:34:10 Frank

R - 在不使用循环的情况下循环遍历不同的矩阵！帮助简单的代码

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