我有几个看起来像这样的数据:
I have several data that looks like this:
Vector1_elements = T,C,A
Vector2_elements = C,G,A
Vector3_elements = C,G,T
..... up to ...
VectorK_elements = ...
#Note also that the member of each vector is always 3.
我想要做的是在 Vector1 到 VectorK 中创建所有元素组合.因此最终我们希望得到这个输出(使用 Vector1,2,3):
What I want to do is to create all combination of elements in Vector1 through out VectorK. Hence in the end we hope to get this output (using Vector1,2,3):
TCC
TCG
TCT
TGC
TGG
TGT
TAC
TAG
TAT
CCC
CCG
CCT
CGC
CGG
CGT
CAC
CAG
CAT
ACC
ACG
ACT
AGC
AGG
AGT
AAC
AAG
AAT
我现在遇到的问题是我的以下代码通过对循环进行硬编码来实现.由于向量的数量可以变化,我们需要一种灵活的方法来获得相同的结果.有吗?
The problem I am having now is that the following code of mine does that by hardcoding the loops. Since number of Vectors can be varied, we need a flexible way to get the same result. Is there any?
我的这段代码最多只能处理 3 个向量(硬编码):
This code of mine can only handle up to 3 Vectors (hardcoded):
#include <iostream>
#include <vector>
#include <fstream>
#include <sstream>
using namespace std;
int main ( int arg_count, char *arg_vec[] ) {
vector <string> Vec1;
Vec1.push_back("T");
Vec1.push_back("C");
Vec1.push_back("A");
vector <string> Vec2;
Vec2.push_back("C");
Vec2.push_back("G");
Vec2.push_back("A");
vector <string> Vec3;
Vec3.push_back("C");
Vec3.push_back("G");
Vec3.push_back("T");
for (int i=0; i<Vec1.size(); i++) {
for (int j=0; j<Vec2.size(); j++) {
for (int k=0; k<Vec1.size(); k++) {
cout << Vec1[i] << Vec2[i] << Vec3[k] << endl;
}
}
}
return 0;
}
这样做可以解决问题:
void printAll(const vector<vector<string> > &allVecs, size_t vecIndex, string strSoFar)
{
if (vecIndex >= allVecs.size())
{
cout << strSoFar << endl;
return;
}
for (size_t i=0; i<allVecs[vecIndex].size(); i++)
printAll(allVecs, vecIndex+1, strSoFar+allVecs[vecIndex][i]);
}
致电:
printAll(allVecs, 0, "");
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