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首先 - 很抱歉这么长的帖子。我想具体一点！感谢阅读 :) 我有两个向量，如下所示：

popd = vector(mode='numeric', 100) popr = vector(mode='numeric', 100)

这些代表了 2 个种群内的初始分布。我想定义这些，例如

popd[]=0.01
popr[]=0.01

我还定义了 e=0.05，然后种群将根据以下 for 循环进行转换：

loop <- for(i in 1:100)
{π <- function(S)
    {x <- sum(popd[1:S])
    return((100-S)*x)}
µ <- function(A)
    {share <- vector(mode='numeric', (101-A))
    share[] = A:100
    return(share%*%popr[A:100])}
for(S in 1:100){vectorπ <- vector(mode='numeric', 100)
    vectorπ[S]=π(S)}
for(A in 1:100){vectorµ <- vector(mode='numeric', 100)
    vectorµ[A]=µ(A)}
av.payoffd <- sum(popd * vectorπ)
av.payoffr <- sum(popr * vectorµ)
newd <- vector(mode='numeric', 100)
    for(S in 1:100){newd[S]=(popd[S]/(av.payoffd))}
newr <- vector(mode='numeric', 100)
    for(A in 1:100){newr[A]=(popr[A]/(av.payoffr))}
newpopd <- vectorπ * newd
    test1[i,2]=(sum(newpopd * (1:100)))
newpopr <- vectorµ * newr
    test1[i,4]=(sum(newpopr * (1:100)))
mutationmatrix <- function(e)
    {mut <- c(e, 1-(2*e), e)
    return(matrix(c(1-e, e, (rep(c(rep098, mut), times=98)), rep098, e, 1-e), nrow=100))}
mutpopd <- function(e)
    {mutationmatrix(e) %*% newpopd}
    test1[i,3]=(sum(mutpopd(e) * (1:100)))
mutpopr <- function(e)
    {mutationmatrix(e) %*% newpopr}
    test1[i,5]=(sum(mutpopr(e) * (1:100)))
if(i<100){
    popd=mutpopd(e)
    popr=mutpopr(e)
}
if(i==100){
    for(i in 1:100){
    popdistcomp1[i,3]=mutpopd(e)[i,1]
    popdistcomp1[i,6]=mutpopr(e)[i,1]}
}
}

test1 和 popdistcomp1 都是我已经定义的矩阵。但是，当我询问结果时，我得到：

test1
         i meanoffer meanmutoffer meanminaccept meanmutminaccept
  [1,]   1       NaN          NaN           100            99.95
  [2,]   2       NaN          NaN           100            99.95
  [3,]   3       NaN          NaN           100            99.95
  [4,]   4       NaN          NaN           100            99.95
  [5,]   5       NaN          NaN           100            99.95
  [6,]   6       NaN          NaN           100            99.95
  [7,]   7       NaN          NaN           100            99.95
  [8,]   8       NaN          NaN           100            99.95
  [9,]   9       NaN          NaN           100            99.95
 [10,]  10       NaN          NaN           100            99.95

和

popdistcomp1
       p(i)   initialpopd mutatedpopd q(a)   initialpopr mutatedpopr
  [1,] "p1"   "0.01"      "NaN"       "q1"   "0.01"      "0"        
  [2,] "p2"   "0.01"      "NaN"       "q2"   "0.01"      "0"        
  [3,] "p3"   "0.01"      "NaN"       "q3"   "0.01"      "0"        
  [4,] "p4"   "0.01"      "NaN"       "q4"   "0.01"      "0"        
  [5,] "p5"   "0.01"      "NaN"       "q5"   "0.01"      "0"        
  [6,] "p6"   "0.01"      "NaN"       "q6"   "0.01"      "0"        
  [7,] "p7"   "0.01"      "NaN"       "q7"   "0.01"      "0"        
  [8,] "p8"   "0.01"      "NaN"       "q8"   "0.01"      "0"        
  [9,] "p9"   "0.01"      "NaN"       "q9"   "0.01"      "0"        
 [10,] "p10"  "0.01"      "NaN"       "q10"  "0.01"      "0"

出于篇幅考虑，我只展示了前 10 个条目，但它们以大致相同的方式继续，尽管我可能应该说 popdistcomp1 在 mutatedpopr 下的最后两个条目是 0.05 和 0.95。

我不确定的是“NaN”条目的来源。我想某处的代码有问题，或者可能是我的循环结构有问题。我是一个完整的初学者，已经使用 R 大约 5 天了，如果我很愚蠢，请道歉。

谢谢你的帮助，

露西

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1 回答 1

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一些开始的建议：

R是高度矢量化的，所以避免编写大量的“微函数”。一方面，您很幸运，对象popd碰巧对函数可见（因为您没有将它们传递给函数）。试试my.pi <- sum(popd[1:S]*(100-S)吧。

for不返回值，所以不要输入foo<-for(.....

不要使用希腊符号。它没有帮助，很可能无法转移到其他系统。

最后，做一些调试工作。看看在计算meanofferaka之前的每个计算test1[,4]，检查每个步骤的输出，并查看NaN第一次出现的位置。您的步骤非常简单，因此错误应该很容易弹出。

于 2013-04-26T17:04:01.870 回答