2

在做一个大学项目时，我遇到了以下问题：我有两个地图（Kmer1 和 Kmer2），它们由一个字符串（键）和一个整数（值）组成。我必须计算遵循这个公式的距离

[1-(I/U)]*100

Where...
     ...U = the sum of all int values inside Kmer1 U Kmer2
     ...I = the sum of all int values inside Kmer1 ∩ Kmer2

Consider that...
             ... The U and ∩ are made evaluating the keys (strings)
             ... When an element is in both maps:
                 - At the Union we add the one with higher int value
                 - At the Intersection we add the one with lower int value

例子：

Kmer1 = AAB¹ AAC¹ AAG³
Kmer2 = AAG¹ AAT² ABB¹

Union = AAB¹ AAC¹ AAG³ AAT² ABB¹   U= 8
Intersection = AAG¹                I= 1
Distance = 87.5

代码时间！我一直在尝试解决它，但所有解决方案都像.. 部分正确，并非所有情况都包括在内。所以当我试图覆盖它们时，我以无限循环结束，异常上升，if-else 的长长的嵌套（这很糟糕......）无论如何，这是最不糟糕且不起作用的尝试：

设置：

Species::Kmer Kmer1, Kmer2;        //The two following lines get the Kmer from another
Kmer1 = esp1->second.query_kmer(); //object.
Kmer2 = esp2->second.query_kmer(); 

Species::Kmer::const_iterator it1, it2, last1, last2;
it1 = Kmer1.cbegin();           //Both Kmer are maps, therefore they are ordered and
it2 = Kmer2.cbegin();           //whitout duplicates.
last1 = --Kmer1.cend();
last2 = --Kmer2.cend();

double U, I;
U = I = 0;

应用公式的循环：

while (it1 != Kmer1.cend() and it2 != Kmer2.cend()){
    if (it1->first == it2->first) {         
        if (it1->second > it2->second) {
            U += it1->second;
            I += it2->second;
        } else {
            U += it2->second;
            I += it1->second;
        }
        ++it1;
        ++it2;

    } else if (it1->first < it2->first) {
        U += it1->second;
        ++it1;
    } else {
        U += it2->second;
        ++it2;
    }
}

请注意，我不是先创建联合和交集，然后再计算每个的总和，而是直接跳到值的总和。我知道也许这并不难，但我一直在努力解决它，但我几乎被困住了......

I've uploaded the whole code at Github: (Maybe it helps)
    - There is a makefile to build the code
    - There is a file called input.txt with a sample for this specific problem
    - Also inside the input.txt, after line13 (fin) I've added the expected output
    - Executing ./program.exe < input.txt should be enough to test it.

https://github.com/PauGalopa/Cpp-Micro-Projects/tree/master/Release

重要的 是！我知道几乎所有的 STL 功能都可以在几行中做到这一点，但是......由于这是一个大学项目，我受制于教学大纲的限制，所以请考虑我只被允许使用“地图" "string" "vector" 等等。不，我不能使用“算法”（我真的希望我可以）我会在评论中澄清关于我可以做什么或使用哪些事情的任何疑问。

4

4 回答 4

2

在主 while 循环之后添加这两个循环。

while (it1 != Kmer1.cend()){
    U += it1->second;
    it1++;
}
while (it2 != Kmer2.cend()){
    U += it2->second;
    it2++;
}
于 2020-05-06T12:16:53.807 回答
2

这是一个相当简单的解决方案，仅使用 的一些属性std::map，没有迭代器。我希望你被允许使用这种解决方案。

#include <iostream>
#include <map>
#include <string>

int main () {
    std::map <std::string, int> A = {{"AAB", 1}, {"AAC", 1}, {"AAG", 3}};
    std::map <std::string, int> B = {{"AAG", 1}, {"AAT", 2}, {"ABB", 1}};

    std::map <std::string, int> Union;
    int sum_A = 0, sum_B = 0, sum_Union = 0, sum_Inter = 0;;

    for (auto &x: A) {
        Union[x.first] = std::max (Union[x.first], x.second);
        sum_A += x.second;
    }
    for (auto &x: B) {
        Union[x.first] = std::max (Union[x.first], x.second);
        sum_B += x.second;
    }   
    for (auto &x: Union) {
        sum_Union += x.second;
    }
    sum_Inter = sum_A + sum_B - sum_Union;
    double distance = 100.0 * (1.0 - double(sum_Inter)/sum_Union);

    std::cout << "sum_Union = " << sum_Union << " sum_Inter = " << sum_Inter << "\n";
    std::cout << "Distance = " << distance << "\n";
}
于 2020-05-06T12:46:33.123 回答
1

一种稍微干净一点unordered_mapping但仍然适用的方法是将tomapping的所有元素和共享元素添加到. 然后添加to的所有非共享元素：Kmer1UIKmer2U

for(it1 = Kmer1.cbegin(); it1 != Kmer1.cend(); it1++) {
    auto other = Kmer2.find(it1->first);
    if(other == Kmer2.cend()) {
        U += it1->second;
    } else {
        U += max(it1->second, other->second);
        I += min(it1->second, other->second);
    }
}
for(it2 = Kmer2.cbegin(); it2 != Kmer2.cend(); it2++) {
    if(Kmer1.count(it2->first) == 0) {
        U += it2->second
    }
}

对于正确实现的unordered_mapping（哈希表），find操作将是O(1)，而不是O(log(n)，使其更快一点。

于 2020-05-06T12:33:49.077 回答
1

这个循环应该工作：

while ( true ){
    bool end1 = it1 == Kmer1.cend();
    bool end2 = it2 == Kmer2.cend();
    if( end1 and end2 )
        break;

    if( end2 or it1->first < it2->first ) {
        U += (it1++)->second;
        continue;
    }
    if( end1 or it2->first < it1->first ) {
        U += (it2++)->second;
        continue;
    }
    auto p = std::minmax( (it1++)->second, (it2++)->second );
    I += p.first;
    U += p.second;
}
于 2020-05-06T12:41:10.417 回答