Question

我有这个功能

   gillespieLO = function (N, n, ...) 
{
  tt = 0
  x = N$M
  S = t(N$Post - N$Pre)
  u = nrow(S)
  v = ncol(S)
  tcrit = 2
  tvec = vector("numeric", n)
  xmat = matrix(ncol = u, nrow = n + 1)
  xmat[1, ] = x
  for (i in 1:n) {
    h = N$h(x, tt, ...)
    tt = tt + rexp(1, sum(h))
    j = sample(v, 1, prob = h)
    x = x + S[, j]
    tvec[i] = tt
    xmat[i + 1, ] = x
    poo = min(which(tvec>tcrit))
    if (i == poo){
      x[4] = x[4]+(10000/10)
    }
  }
  print(min(which(tvec>tcrit)))
  return(list(t = tvec, x = xmat))
}



LO = list()
LO$M = c(i = 1, ri = 0, I = 50, Lac = 20, ILac = 0, o = 1, Io = 0, RNAP = 100, RNAPo = 0, r = 0, z = 0)
LO$Pre = matrix(c(1,0,0,0,0,0,0,0,0,0,0,
                  0,1,0,0,0,0,0,0,0,0,0,
                  0,0,1,1,0,0,0,0,0,0,0,
                  0,0,0,0,1,0,0,0,0,0,0,
                  0,0,1,0,0,1,0,0,0,0,0,
                  0,0,0,0,0,0,1,0,0,0,0,
                  0,0,0,0,0,1,0,1,0,0,0,
                  0,0,0,0,0,0,0,0,1,0,0,
                  0,0,0,0,0,0,0,0,1,0,0,
                  0,0,0,0,0,0,0,0,0,1,0,
                  0,0,0,1,0,0,0,0,0,0,1,
                  0,1,0,0,0,0,0,0,0,0,0,
                  0,0,1,0,0,0,0,0,0,0,0,
                  0,0,0,0,1,0,0,0,0,0,0,
                  0,0,0,0,0,0,0,0,0,1,0,
                  0,0,0,0,0,0,0,0,0,0,1), ncol=11, byrow=TRUE)
LO$Post = matrix(c(1,1,0,0,0,0,0,0,0,0,0,
                   0,1,1,0,0,0,0,0,0,0,0,
                   0,0,0,0,1,0,0,0,0,0,0,
                   0,0,1,1,0,0,0,0,0,0,0,
                   0,0,0,0,0,0,1,0,0,0,0,
                   0,0,1,0,0,1,0,0,0,0,0,
                   0,0,0,0,0,0,0,0,1,0,0,
                   0,0,0,0,0,1,0,1,0,0,0,
                   0,0,0,0,0,1,0,1,0,1,0,
                   0,0,0,0,0,0,0,0,0,1,1,
                   0,0,0,0,0,0,0,0,0,0,1,
                   0,0,0,0,0,0,0,0,0,0,0,
                   0,0,0,0,0,0,0,0,0,0,0,
                   0,0,0,1,0,0,0,0,0,0,0,
                   0,0,0,0,0,0,0,0,0,0,0,
                   0,0,0,0,0,0,0,0,0,0,0), ncol=11, byrow=TRUE)
LO$h = function(x,t,th=c(0.02,0.1,0.005,0.1,1,0.01,0.1,0.01,0.03,0.1,1e-05,0.01,0.002,0.01,0.001))
{
  with(as.list(c(x, th)), {
    return(c(th[1]*i, th[2]*ri, th[3]*I*Lac, th[4]*ILac, th[5]*I*o, th[6]*Io, th[7]*o*RNAP,
             th[8]*RNAPo, th[9]*RNAPo, th[10]*r, th[11]*Lac*z, th[12]*ri, th[13]*I, 
             th[13]*ILac, th[14]*r, th[15]*z))
  })
}

把

gillespieLO(LO,15)

给出

gillespieLO(LO,15)
[1] 11
$t
 [1] 0.001399092 0.053674441 0.132884069 0.156973856 0.365723847 0.700767071 1.178798507
 [8] 1.529528682 1.663988976 1.719413303 2.352115017 2.353169914 2.355999647 2.364615018
[15] 2.367255303

$x
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
 [1,]    1    0   50   20    0    1    0  100    0     0     0
 [2,]    1    0   49   20    0    0    1  100    0     0     0
 [3,]    1    0   48   19    1    0    1  100    0     0     0
 [4,]    1    0   47   18    2    0    1  100    0     0     0
 [5,]    1    0   46   17    3    0    1  100    0     0     0
 [6,]    1    0   47   18    2    0    1  100    0     0     0
 [7,]    1    0   46   17    3    0    1  100    0     0     0
 [8,]    1    0   45   16    4    0    1  100    0     0     0
 [9,]    1    0   44   15    5    0    1  100    0     0     0
[10,]    1    0   43   14    6    0    1  100    0     0     0
[11,]    1    0   42   13    7    0    1  100    0     0     0
[12,]    1    0   41   12    8    0    1  100    0     0     0
[13,]    1    0   40 1011    9    0    1  100    0     0     0
[14,]    1    0   39 1010   10    0    1  100    0     0     0
[15,]    1    0   38 1009   11    0    1  100    0     0     0
[16,]    1    0   37 1008   12    0    1  100    0     0     0

但我得到错误，

1: In min(which(tvec > tcrit)) :
  no non-missing arguments to min; returning Inf
And I want to adapt it so that I can find the minimum value of `i` in my loop such that `tt>tcrit`

所以我想做像min这样的事情（我这样（tt＆gt; tcrit））但我不知道如何在R中写这个。

很抱歉没有多大帮助，因为我一直在努力寻找这个，但我不知道如何正确解释。

Answer 1

循环之前：

firstIalreadyFound <- FALSE

然后，将此行作为循环中的第一行：

if (tt > tcrit & !firstIalreadyFound){
    firstIalreadyFound <- TRUE
    # here, add the code you want to execute for this first i
}

只要满足条件，它就会打印出来。

Answer 2

如果没有传递有效（非NA）值，min函数将返回Inf。

tcrit的值为2。值将在行tvec的循环中分配给tvec[i] = tt。假设在第一次迭代中，分配给tvec[i]的值小于1（小于tcrit）。在该迭代中，which(tvec>tcrit)不会返回任何有效值，因为没有值大于tcrit。 min返回Inf，因为没有有效值传递给它。您可以通过简单的测试来尝试这种情况：

a <- c(1,2,2,3,4)
> min(which(a>5))
[1] Inf

因此，为了避免警告，最好检查any中是否存在tvec匹配条件，然后仅申请min，否则min无效。修改后的代码：

if(any(which(tvec>tcrit))){
  poo = min(which(tvec>tcrit))
  if (i == poo){
    x[4] = x[4]+(10000/10)
  }
  print(min(which(tvec>tcrit)))
}

R哪个循环重复产生的值大于临界值

2 个答案: