Question

我有一个矩阵，

set.seed(100)
x <- matrix(sample(11, 110, replace = T), nrow = 11)

和另一个矩阵，

y <- cbind(1:11, sample(110, 11))

我想将x的每个元素与y的第一列相匹配。如果匹配应返回y其他0的第二列。最终出局应与x具有相同的维度。

我尝试使用带有3个嵌套循环的for-loop执行此操作，但有没有有效的方法呢？

预期输出

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]   26   85    4   40   96    9    4   21   26    21
 [2,]    9   26   21   75    4   88  103   85   21    26
 [3,]   96  103  103   85   21    9    9   88   85   103
 [4,]   93   21   88   88   85   96   75    4    9    93
 [5,]    4   75   21   96    9    9  103   96   26    26
 [6,]    4    9   85   40   26    4   26   40  103    96
 [7,]   21   26   96   88   26   75   96   40   85    75
 [8,]  103   26   26   26    9   40   40   93  103    40
 [9,]   96   75    4   85    9   75   75   96    4    75
[10,]   88    4   40   21   26  103   96   21   88    93
[11,]   96   75   26   85   96   26   85    9   93     4

Answer 1

这是一种可能性：

x[!x%in%y[,1]] <- 0 # set all elements to zero which are not in first column of y
sapply(1:nrow(y),function(i) x[x==y[i,1]] <<- y[i,2]) #replace all matches with second column of y
#> x
#      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,]   26   85    4   40   96   21    4   21   26    21
# [2,]   21   26   21   75    4   88  103   85   21    26
# [3,]   96  103  103   85   21   21   21   88   85   103
# [4,]   93   21   88   88   85   96   75    4   21    93
# [5,]    4   75   21   96   21   21  103   96   26    26
# [6,]    4   21   85   40   26    4   26   40  103    96
# [7,]   21   26   96   88   26   75   96   40   85    75
# [8,]  103   26   26   26   21   40   40   93  103    40
# [9,]   96   75    4   85   21   75   75   96    4    75
#[10,]   88    4   40   21   26  103   96   21   88    93
#[11,]   96   75   26   85   96   26   85   21   93     4

Answer 2

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的命题

replace

在R中的Matrix下匹配（查找）

2 个答案: