如何在R中将二进制矩阵转换为逻辑矩阵？
二进制矩阵包含诸如“是”或“否”，“ 1”或“ 0”之类的值，或代表其他大多数相反值的任何其他两个值，并且全局接受的逻辑值为FALSE和TRUE。因此，要将二进制矩阵转换为逻辑矩阵，我们可以使用ifelse函数并将一类二进制变量转换为适当的逻辑值，而其余的则返回遗漏的值。在R中，这是一项非常简单的任务，请查看以下示例以了解如何完成。
例1
> M1<-matrix(sample(c("No","Yes"),40,replace=TRUE),nrow=20)> M1
输出结果[,1] [,2][1,] "No" "Yes"
[2,] "No" "No"
[3,] "No" "Yes"
[4,] "Yes" "Yes"
[5,] "Yes" "Yes"
[6,] "No" "No"
[7,] "Yes" "No"
[8,] "Yes" "Yes"
[9,] "No" "No"
[10,] "No" "Yes"
[11,] "No" "No"
[12,] "Yes" "Yes"
[13,] "No" "No"
[14,] "Yes" "Yes"
[15,] "No" "Yes"
[16,] "No" "No"
[17,] "Yes" "No"
[18,] "Yes" "No"
[19,] "Yes" "No"
[20,] "Yes" "Yes"
将M1转换为逻辑矩阵-
> M1[,]<-ifelse(M1 %in% c("No"),FALSE,TRUE)> M1
输出结果[,1] [,2][1,] "FALSE" "TRUE"
[2,] "FALSE" "FALSE"
[3,] "FALSE" "TRUE"
[4,] "TRUE" "TRUE"
[5,] "TRUE" "TRUE"
[6,] "FALSE" "FALSE"
[7,] "TRUE" "FALSE"
[8,] "TRUE" "TRUE"
[9,] "FALSE" "FALSE"
[10,] "FALSE" "TRUE"
[11,] "FALSE" "FALSE"
[12,] "TRUE" "TRUE"
[13,] "FALSE" "FALSE"
[14,] "TRUE" "TRUE"
[15,] "FALSE" "TRUE"
[16,] "FALSE" "FALSE"
[17,] "TRUE" "FALSE"
[18,] "TRUE" "FALSE"
[19,] "TRUE" "FALSE"
[20,] "TRUE" "TRUE"
例2
> M2<-matrix(sample(c("0","1"),40,replace=TRUE),nrow=20)> M2
输出结果[,1] [,2][1,] "1" "1"
[2,] "0" "1"
[3,] "1" "0"
[4,] "0" "0"
[5,] "1" "0"
[6,] "1" "1"
[7,] "1" "0"
[8,] "0" "1"
[9,] "0" "0"
[10,] "0" "0"
[11,] "0" "1"
[12,] "0" "0"
[13,] "0" "1"
[14,] "0" "0"
[15,] "1" "1"
[16,] "0" "1"
[17,] "0" "1"
[18,] "1" "0"
[19,] "1" "0"
[20,] "1" "0"
将M2转换为逻辑矩阵-
> M2[,]<-ifelse(M2 %in% c("0"),FALSE,TRUE)> M2
输出结果[,1] [,2][1,] "TRUE" "TRUE"
[2,] "FALSE" "TRUE"
[3,] "TRUE" "FALSE"
[4,] "FALSE" "FALSE"
[5,] "TRUE" "FALSE"
[6,] "TRUE" "TRUE"
[7,] "TRUE" "FALSE"
[8,] "FALSE" "TRUE"
[9,] "FALSE" "FALSE"
[10,] "FALSE" "FALSE"
[11,] "FALSE" "TRUE"
[12,] "FALSE" "FALSE"
[13,] "FALSE" "TRUE"
[14,] "FALSE" "FALSE"
[15,] "TRUE" "TRUE"
[16,] "FALSE" "TRUE"
[17,] "FALSE" "TRUE"
[18,] "TRUE" "FALSE"
[19,] "TRUE" "FALSE"
[20,] "TRUE" "FALSE"
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