如何在 R 中將二進位制矩陣轉換成邏輯矩陣？

二進位制矩陣包含是或否、1 或 0 之類的值，或其他表示相反情況的值，而公認的邏輯值為假或真。因此，要將二進位制矩陣轉換成邏輯矩陣，我們可使用 ifelse 函式，並將二進位制變數的一類轉換成適當的邏輯值，其他值返回未列出的值。這是 R 中一個非常簡單的任務，檢視以下示例以瞭解如何執行此操作。

示例 1

即時演示

> M1<-matrix(sample(c("No","Yes"),40,replace=TRUE),nrow=20)
> M1

輸出

[,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"

將 M1 轉換成邏輯矩陣 −

> M1[,]<-ifelse(M1 %in% c("No"),FALSE,TRUE)
> M1

輸出

[,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"

示例 2

即時演示

> M2<-matrix(sample(c("0","1"),40,replace=TRUE),nrow=20)
> M2

輸出

[,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"

將 M2 轉換成邏輯矩陣 −

> M2[,]<-ifelse(M2 %in% c("0"),FALSE,TRUE)
> M2

輸出

[,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"

Nizamuddin Siddiqui

更新日期： 04-Mar-2021

837 次瀏覽

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