C++ 中两个映射之间的同时联合和交集

Question

While doing a college project I came upon the following problem: I have two maps (Kmer1 and Kmer2) which are composed by a string(key) and a int(value). I have to calculate the distance which follows this formula

[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

Example:

Kmer1 = AAB¹ AAC¹ AAG³
Kmer2 = AAG¹ AAT² ABB¹

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

Code time. I've been trying to solve it but all solutions are like., partially right. not all cases are covered, So when I tried to cover them, I ended in infinite loops, exceptions rising. long long nests of if-else (which were awfull.,) anyway: here is the the least worse and non working try:

The setup:

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;

The loop where the formula is applied:

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;
    }
}

Note that instead of first creating the Union and the intersection and then doing the total sum of each, I jumped directly to the sum of the values. I know maybe it's not that hard but I've been trying to solve it but I'm pretty much stuck...

---

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

---

IMPORTANT Yes. I'm aware of almost all the STL functionality that could do this in a few lines BUT... Since this is a college project i'm bound to the limitations of the sillabus so consider that I'm only allowed to use "map" "string" "vector" and a few more, No. I can't use "algorithm" (I really wish I could) I'll clarify any doubt about which things I can do or use in the coments.

Answer 1

Add these two loops just after your main while loop.

while (it1 != Kmer1.cend()){
    U += it1->second;
    it1++;
}
while (it2 != Kmer2.cend()){
    U += it2->second;
    it2++;
}

Answer 2

Here is a rather simple solution, using only some propoerties of std::map , with no iterator. I hope you are allowed to use this kind of solution.

#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";
}

Answer 3

A slightly cleaner approach for unordered_mapping , but which would still work with mapping would be to add all the elements of Kmer1 to U , and shared elements to I . Then add all the unshared elements of Kmer2 to U :

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
    }
}

For a properly implemented unordered_mapping (hash table), the find operation will be O(1) , not O(log(n) , making it a bit faster.

Answer 4

This loop should work:

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;
}