在不包括主对角线值的矩阵上采样最大值
Sample the maximum value on a matrix excluding value on main diagonal
我有一个矩阵 X,它的最大值沿主对角线。
首先,要对第 i 行进行采样,并沿第 i 行选择最大值,不包括主对角线值,即 max != X[i,i].
下面的代码通常会产生结果,但经常会出错:
Error in if (MAX < l[k]) { : missing value where TRUE/FALSE needed
# initial values
n = 10
pop = runif(n,min =0,max =1)
D = matrix(rnorm(n*n,0,0.2),nrow=n)
str_mat = abs(D)
for (l in 1:n) {
str_mat[l,l] = 1
}
int_mat = matrix(rbinom(n*n,1,z),n,n) ##z takes the values 0.1 - 0.9
for (j in 1:n) {
int_mat[j,j] = 1
}
X = (int_mat*str_mat)*pop
b = c(1:n) #creating a vector with the length being the dimensions of the matrix
a = sample(b,1)## sampling one value from the vector
if (sum(int_mat[a,])< n)
{
### int_mat is a binary matrix
break
}}
l = X[a,]
## Ensuring the maximum value picked is not on the main diagonal
MAX = 0
j = 1
for (k in 1:length(l)) {
if(k!=a) {
if (MAX<l[k]) {
MAX = l[k]
j = k
}
}
}
感谢大家的贡献。我想出了如何解决这个问题。如下;
X=(int_mat*str_mat)*pop ## creating a matrix of interaction, competition strength and population densities
repeat{
IntRowsums=rowSums(int_mat)
introwsums_greater=which(IntRowsums>2,arr.ind = T)
if (length(introwsums_greater)>1){
a= sample(introwsums_greater,1)
}else{
a=introwsums_greater
}
if (sum(int_mat[a,])< n){
break
}}
q= ABX[a,]
j_k=which(q!=q[a] & q!=0,arr.ind = T) ## from the sampled row in str_mat check the position of all zeros
k=sample(j_k,1)
我有一个矩阵 X,它的最大值沿主对角线。 首先,要对第 i 行进行采样,并沿第 i 行选择最大值,不包括主对角线值,即 max != X[i,i].
下面的代码通常会产生结果,但经常会出错:
Error in if (MAX < l[k]) { : missing value where TRUE/FALSE needed
# initial values
n = 10
pop = runif(n,min =0,max =1)
D = matrix(rnorm(n*n,0,0.2),nrow=n)
str_mat = abs(D)
for (l in 1:n) {
str_mat[l,l] = 1
}
int_mat = matrix(rbinom(n*n,1,z),n,n) ##z takes the values 0.1 - 0.9
for (j in 1:n) {
int_mat[j,j] = 1
}
X = (int_mat*str_mat)*pop
b = c(1:n) #creating a vector with the length being the dimensions of the matrix
a = sample(b,1)## sampling one value from the vector
if (sum(int_mat[a,])< n)
{
### int_mat is a binary matrix
break
}}
l = X[a,]
## Ensuring the maximum value picked is not on the main diagonal
MAX = 0
j = 1
for (k in 1:length(l)) {
if(k!=a) {
if (MAX<l[k]) {
MAX = l[k]
j = k
}
}
}
感谢大家的贡献。我想出了如何解决这个问题。如下;
X=(int_mat*str_mat)*pop ## creating a matrix of interaction, competition strength and population densities
repeat{
IntRowsums=rowSums(int_mat)
introwsums_greater=which(IntRowsums>2,arr.ind = T)
if (length(introwsums_greater)>1){
a= sample(introwsums_greater,1)
}else{
a=introwsums_greater
}
if (sum(int_mat[a,])< n){
break
}}
q= ABX[a,]
j_k=which(q!=q[a] & q!=0,arr.ind = T) ## from the sampled row in str_mat check the position of all zeros
k=sample(j_k,1)