我正在尝试返回一个 vector 中的值,该 vector 按加权随机分布返回(即,如果 vector 有 1,2,3,它将给出 1/6 的机会和 2/6 的机会2 和 3/6 的机会)。我收到
的编译时错误Protein.cpp:809:54: error: cannot convert ‘__gnu_cxx::__normal_iterator<double*, std::vector<double> >’ to ‘double’ in assignment
Protein.cpp:814:24: error: could not convert ‘((Protein*)this)->Protein::fitness1.std::vector<_Tp, _Alloc>::operator[]<double, std::allocator<double> >((std::vector<double, std::allocator<double> >::size_type)index)’ from ‘double’ to ‘std::vector<double>’
当我编译代码并试图修复它时运气很差。如果有人可以看一下该功能并帮助我,我将不胜感激。再次感谢您抽出宝贵时间!
函数
vector<double> Protein::Init_Seq_Recursive(State& s, int d)
{
vector<float> sequence;
double index;
float initial;
int i =0;
for (i; i < k; i++)
{
s[d] = i;
if (d == (L - 1))
{
int ind = Index(s);
Ef[ind] = GetEf(s);
Eb1[ind] = GetEb1(s);
Eb2[ind] = GetEb2(s);
double zf = exp(beta_f*Ef[ind]);
double zb1 = exp(beta_b*Eb1[ind]);
double zb2 = exp(beta_b*Eb2[ind]);
fitness1[ind] = (1. + f_ub*zb1 + f_ub*f_uf*zf*zb1)/( 1. + zb1 + zf*zb1 );
fitness2[ind] = (1. + f_ub*zb2 + f_ub*f_uf*zf*zb2)/( 1. + zb2 + zf*zb2 );
}
else
Init_Fitness_Recursive(s, d + 1);
}
if(i==k-1){
vector<float> weights;
float running_total = 0;
for(int y = 0; y<pow(k,L); y++){
running_total = running_total+fitness1[y];
weights.push_back(running_total);
}
double rnd = (static_cast <double> (rand()) / static_cast <double> (RAND_MAX))*running_total;
for(int y = 0; y<fitness1.size(); y++){
if(rnd<weights[y]){
index = find(fitness1.begin(), fitness1.end(), rnd);
}
}
}
return(fitness1[index]);
}
最佳答案
std::find
不返回索引。它返回一个指向找到的元素的迭代器。获取 index
,您需要计算 begin
之间的距离迭代器和你找到的那个。像这样:
vector<double>::iterator found = find(fitness1.begin(), fitness1.end(), rnd);
index = distance(fitness1.begin(), found);
很可能您会使用 double
用于容器中的索引。
此外,您的 return(fitness1[index]);
正在尝试返回单个 double
,但您的返回类型是 vector<double>
.
关于c++ - vector <double> 将 c++ 代码中的错误加倍,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24684020/