我正在尝试生成一个名为StepstoShop的函数,该函数将生成表示x和y坐标的n * 2矩阵。允许的方向只有北,南,东和西(不允许对角线移动)
这是我到目前为止开发的代码
#include <RcppArmadilloExtensions/sample.h>
#include <cstdlib>
#include <ctime>
using namespace Rcpp;
// [[Rcpp::depends(RcppArmadillo)]]
// [[Rcpp::export]]
NumericMatrix StepstoShop(double Steps){
NumericMatrix SampleGen(Steps,2);
int n=SampleGen.size();
int colnum= rand()%1;
for( int i=0; i<n; i++)
{ SampleGen(i,colnum)=(rand()%3)-1;
}
return SampleGen;
}
我已经尝试在通过行的for循环中的列中索引0到1之间的随机数分配,以获取4个方向的预期结果 (0,1)(1,0)(-1,0)(0,-1),但是我在8个方向上混合了。
任何指导,将不胜感激
谢谢
我追溯性地包含了R代码,以说明我正在尝试重新创建的内容
# compute path
n <- 1000
rw <- matrix(0, ncol = 2, nrow = n)
# generate the indices to set the deltas
indx <- cbind(seq(n), sample(c(1, 2), n, TRUE))
# now set the values
rw[indx] <- sample(c(-1, 1), n, TRUE)
# cumsum the columns
rw[,1] <- cumsum(rw[, 1])
rw[, 2] <- cumsum(rw[, 2])
plot(0,type="n",xlab="x",ylab="y",main="Random Walk Simulation In Two
Dimensions",col=1:10,xlim=range(-10,15),ylim=range(-40,40))
# use 'segments' to color each path
segments(head(rw[, 1], -1)
, head(rw[, 2], -1)
, tail(rw[, 1], -1)
, tail(rw[, 2], -1)
, col = rainbow(nrow(rw) -1) # a range of colors
)
end<-cbind(-10,30)
start<-cbind(0,0)
points(start,pch=16,col="green", cex = 3)
points(end,pch=16,col="red", cex = 3)
想法是将该代码重复100,000次并计算到达端点的概率
谢谢
答案 0 :(得分:1)
我会用
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
IntegerMatrix StepstoShop(double Steps){
IntegerVector possible_x = IntegerVector::create(0, 1, -1, 0);
IntegerVector possible_y = IntegerVector::create(1, 0, 0, -1);
IntegerMatrix SampleGen(Steps, 2);
int ind;
for (int i = 0; i < Steps; i++) {
ind = R::runif(0, 4);
SampleGen(i, 0) = possible_x[ind];
SampleGen(i, 1) = possible_y[ind];
}
return SampleGen;
}