如何解决c + +:有没有更快的方法来获取map / unordered_map的交集?
在c ++中是否有更快的方法来执行以下操作,以便我能胜过python的实现?
- 获取两个map / unordered_map键的交集
- 对于这些相交的键,计算它们各自set / unordered_set中元素之间的成对差异 一些可能有用的信息:
- hash_DICT1具有大约O(10000)个密钥,每个集合中大约有O(10)个元素。
- hash_DICT2具有大约O(1000)个密钥,每个集合中大约有O(1)个元素。
例如:
map <int,set<int>> hash_DICT1;
hash_DICT1[1] = {1,2,3};
hash_DICT1[2] = {4,5,6};
map <int,set<int>> hash_DICT2;
hash_DICT2[1] = {11,12,13};
hash_DICT2[3] = {4,6};
vector<int> output_vector
= GetPairDiff(hash_DICT1,hash_DICT2)
= [11-1,12-1,13-1,11-2,12-2,13-2,11-3,12-3,13-3] // only hashkey=1 is intersect,so only compute pairwise difference of the respective set elements.
= [10,11,9,10,8,10] // Note that i do want to keep duplicates,if any. Order does not matter.
GetPairDiff
功能。
vector<int> GetPairDiff(
unordered_map <int,set<int>> &hash_DICT1,unordered_map <int,set<int>> &hash_DICT2) {
// Init
vector<int> output_vector;
int curr_key;
set<int> curr_set1,curr_set2;
// Get intersection
for (const auto &KEY_SET:hash_DICT2) {
curr_key = KEY_SET.first;
// Find pairwise difference
if (hash_DICT1.count(curr_key) > 0){
curr_set1 = hash_DICT1[curr_key];
curr_set2 = hash_DICT2[curr_key];
for (auto it1=curr_set1.begin(); it1 != curr_set1.end(); ++it1) {
for (auto it2=curr_set2.begin(); it2 != curr_set2.end(); ++it2) {
output_vector.push_back(*it2 - *it1);
}
}
}
}
}
主跑
int main (int argc,char ** argv) {
// Using unordered_map
unordered_map <int,set<int>> hash_DICT_1;
hash_DICT_1[1] = {1,3};
hash_DICT_1[2] = {4,6};
unordered <int,set<int>> hash_DICT_2;
hash_DICT_2[1] = {11,13};
hash_DICT_2[3] = {4,6};
GetPairDiff(hash_DICT_1,hash_DICT_1);
}
像这样编译
g++ -o ./CompareRunTime.out -Ofast -Wall -Wextra -std=c++11
欢迎使用其他数据结构,例如map
或unordered_set
。
但是我确实尝试了所有4种置换,并发现GetPairDiff
给出的置换运行速度最快,但远不及python的实现快
hash_DICT1 = { 1 : {1,3},2 : {4,6} }
hash_DICT2 = { 1 : {11,13},3 : {4,6} }
def GetPairDiff(hash_DICT1,hash_DICT2):
vector = []
for element in hash_DICT1.keys() & hash_DICT2.keys():
vector.extend(
[db_t-qry_t
for qry_t in hash_DICT2[element]
for db_t in hash_DICT1[element] ])
return vector
output_vector = GetPairDiff(hash_DICT1,hash_DICT2)
性能比较:
python : 0.00824 s
c++ : 0.04286 s
c ++的实现大约需要花费5倍的时间!!
解决方法
- 您在应该使用
const&
的地方进行了大量复制。 - 您不保存搜索结果。您应该使用
find
而不是count
,然后使用结果。
如果您事先知道要存储的元素数量,可以
push_back
将{li> vector
转换为reserve()
,从而更快。
解决这些问题可能会导致以下情况(需要C ++ 17):
#include <iostream>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using container = std::unordered_map<int,std::unordered_set<int>>;
std::vector<int> GetPairDiff(const container& hash_DICT1,const container& hash_DICT2) {
// Init
std::vector<int> output_vector;
// Get intersection
for(auto& [curr_key2,curr_set2] : hash_DICT2) {
// use find() instead of count()
if(auto it1 = hash_DICT1.find(curr_key2); it1 != hash_DICT1.end()) {
auto& curr_set1 = it1->second;
// Reserve the space you know you'll need for this iteration. Note:
// This might be a pessimizing optimization so try with and without it.
output_vector.reserve(curr_set1.size() * curr_set2.size() +
output_vector.size());
// Calculate pairwise difference
for(auto& s1v : curr_set1) {
for(auto& s2v : curr_set2) {
output_vector.emplace_back(s2v - s1v);
}
}
}
}
return output_vector;
}
int main() {
container hash_DICT1{{1,{1,2,3}},{2,{4,5,6}}};
container hash_DICT2{{1,{11,12,13}},{3,6}}};
auto result = GetPairDiff(hash_DICT1,hash_DICT2);
for(int v : result) {
std::cout << v << '\n';
}
}
对于使用g++ -std=c++17 -O3
编译的我的计算机上的这些容器,这是python版本的8倍以上。
这是同一程序的C ++ 11版本:
#include <iostream>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using container = std::unordered_map<int,const container& hash_DICT2) {
// Init
std::vector<int> output_vector;
// Get intersection
for(auto& curr_pair2 : hash_DICT2) {
auto& curr_key2 = curr_pair2.first;
auto& curr_set2 = curr_pair2.second;
// use find() instead of count()
auto it1 = hash_DICT1.find(curr_key2);
if(it1 != hash_DICT1.end()) {
auto& curr_set1 = it1->second;
// Reserve the space you know you'll need for this iteration. Note:
// This might be a pessimizing optimization so try with and without it.
output_vector.reserve(curr_set1.size() * curr_set2.size() +
output_vector.size());
// Calculate pairwise difference
for(auto& s1v : curr_set1) {
for(auto& s2v : curr_set2) {
output_vector.emplace_back(s2v - s1v);
}
}
}
}
return output_vector;
}
int main() {
container hash_DICT1{{1,hash_DICT2);
for(int v : result) {
std::cout << v << '\n';
}
}
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