R 编程:如何按行替换值?
R programming: How to do replace values by row?
如何使用 R 对值进行逐行替换?
我有一个矩阵,我想用索引向量替换它的一些值。问题是 R 会自动按列提取值,而不是按行提取值。
您将在下面找到我的代码和结果:
Matrix=matrix(rep(0,42),nrow=6,ncol=7,byrow=TRUE)
v=c(1,7,11,16,18)
Matrix[v]=1
Matrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1 1 0 0 0 0 0
[2,] 0 0 0 0 0 0 0
[3,] 0 0 0 0 0 0 0
[4,] 0 0 1 0 0 0 0
[5,] 0 1 0 0 0 0 0
[6,] 0 0 1 0 0 0 0
我真正想要得到的是这个意思的逐行版本:
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1 0 0 0 0 0 1
[2,] 0 0 0 1 0 0 0
[3,] 0 1 0 1 0 0 0
[4,] 0 0 0 0 0 0 0
[5,] 0 0 0 0 0 0 0
[6,] 0 0 0 0 0 0 0
>
显然 R 默认情况下会按列替换值。
获取值的逐行替换的最佳方法是什么?
谢谢!
您可以重做 1 以使它们按行排列,或者您可以执行以下操作:
Matrix=matrix(rep(0,42),nrow=6,ncol=7,byrow=TRUE)
v=c(1,7,11,16,18)
Matrix<-t(Matrix)
Matrix[v]=1
Matrix<-t(Matrix)
Matrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1 0 0 0 0 0 1
[2,] 0 0 0 1 0 0 0
[3,] 0 1 0 1 0 0 0
[4,] 0 0 0 0 0 0 0
[5,] 0 0 0 0 0 0 0
[6,] 0 0 0 0 0 0 0
您可以将一维索引重新计算为行索引和列索引。假设您已经计算了矩阵 Ind
第一列中的行索引和 Ind
第二列中的列索引,您可以执行 Matrix[Ind] <- 1
Matrix <- matrix(rep(0,42),nrow=6,ncol=7,byrow=TRUE)
v <- c(1,7,11,16,18)
Row <- (v-1) %/% ncol(Matrix) +1
Col <- (v-1) %% ncol(Matrix) +1
Matrix[cbind(Row,Col)] <- 1
Matrix
# [,1] [,2] [,3] [,4] [,5] [,6] [,7]
# [1,] 1 0 0 0 0 0 1
# [2,] 0 0 0 1 0 0 0
# [3,] 0 1 0 1 0 0 0
# [4,] 0 0 0 0 0 0 0
# [5,] 0 0 0 0 0 0 0
# [6,] 0 0 0 0 0 0 0
我们可以做到
+(matrix(seq_along(Matrix) %in% v, ncol=ncol(Matrix), nrow=nrow(Matrix), byrow=TRUE))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#[1,] 1 0 0 0 0 0 1
#[2,] 0 0 0 1 0 0 0
#[3,] 0 1 0 1 0 0 0
#[4,] 0 0 0 0 0 0 0
#[5,] 0 0 0 0 0 0 0
#[6,] 0 0 0 0 0 0 0
如何使用 R 对值进行逐行替换? 我有一个矩阵,我想用索引向量替换它的一些值。问题是 R 会自动按列提取值,而不是按行提取值。 您将在下面找到我的代码和结果:
Matrix=matrix(rep(0,42),nrow=6,ncol=7,byrow=TRUE)
v=c(1,7,11,16,18)
Matrix[v]=1
Matrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1 1 0 0 0 0 0
[2,] 0 0 0 0 0 0 0
[3,] 0 0 0 0 0 0 0
[4,] 0 0 1 0 0 0 0
[5,] 0 1 0 0 0 0 0
[6,] 0 0 1 0 0 0 0
我真正想要得到的是这个意思的逐行版本:
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1 0 0 0 0 0 1
[2,] 0 0 0 1 0 0 0
[3,] 0 1 0 1 0 0 0
[4,] 0 0 0 0 0 0 0
[5,] 0 0 0 0 0 0 0
[6,] 0 0 0 0 0 0 0
>
显然 R 默认情况下会按列替换值。 获取值的逐行替换的最佳方法是什么?
谢谢!
您可以重做 1 以使它们按行排列,或者您可以执行以下操作:
Matrix=matrix(rep(0,42),nrow=6,ncol=7,byrow=TRUE)
v=c(1,7,11,16,18)
Matrix<-t(Matrix)
Matrix[v]=1
Matrix<-t(Matrix)
Matrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1 0 0 0 0 0 1
[2,] 0 0 0 1 0 0 0
[3,] 0 1 0 1 0 0 0
[4,] 0 0 0 0 0 0 0
[5,] 0 0 0 0 0 0 0
[6,] 0 0 0 0 0 0 0
您可以将一维索引重新计算为行索引和列索引。假设您已经计算了矩阵 Ind
第一列中的行索引和 Ind
第二列中的列索引,您可以执行 Matrix[Ind] <- 1
Matrix <- matrix(rep(0,42),nrow=6,ncol=7,byrow=TRUE)
v <- c(1,7,11,16,18)
Row <- (v-1) %/% ncol(Matrix) +1
Col <- (v-1) %% ncol(Matrix) +1
Matrix[cbind(Row,Col)] <- 1
Matrix
# [,1] [,2] [,3] [,4] [,5] [,6] [,7]
# [1,] 1 0 0 0 0 0 1
# [2,] 0 0 0 1 0 0 0
# [3,] 0 1 0 1 0 0 0
# [4,] 0 0 0 0 0 0 0
# [5,] 0 0 0 0 0 0 0
# [6,] 0 0 0 0 0 0 0
我们可以做到
+(matrix(seq_along(Matrix) %in% v, ncol=ncol(Matrix), nrow=nrow(Matrix), byrow=TRUE))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#[1,] 1 0 0 0 0 0 1
#[2,] 0 0 0 1 0 0 0
#[3,] 0 1 0 1 0 0 0
#[4,] 0 0 0 0 0 0 0
#[5,] 0 0 0 0 0 0 0
#[6,] 0 0 0 0 0 0 0