python collections.Counter
对象跟踪对象的数量。
>> from collections import Counter
>> myC = Counter()
>> myC.update("cat")
>> myC.update("cat")
>> myC["dogs"] = 8
>> myC["lizards"] = 0
>> print(myC)
{"cat": 2, "dogs": 8, "lizards": 0}
是否有一个类似的C ++对象,可以在其中轻松跟踪类型的出现次数?也许是map
到string
?请记住,以上只是一个示例,在C ++中,它将泛化为其他类型以供计数。
答案 0 :(得分:6)
您可以使用std::map
,例如:
#include <iostream>
#include <map>
int main()
{
std::map<std::string,int> counter;
counter["dog"] = 8;
counter["cat"]++;
counter["cat"]++;
counter["1"] = 0;
for (auto pair : counter) {
cout << pair.first << ":" << pair.second << std::endl;
}
}
输出:
1:0
cat:2
dog:8
答案 1 :(得分:1)
Python3代码:
import collections
stringlist = ["Cat","Cat","Cat","Dog","Dog","Lizard"]
counterinstance = collections.Counter(stringlist)
for key,value in counterinstance.items():
print(key,":",value)
C ++代码:
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
int main()
{
unordered_map <string,int> counter;
vector<string> stringVector {"Cat","Cat","Cat","Dog","Dog","Lizard"};
for (auto stringval: stringVector)
{
if (counter.find(stringval) == counter.end()) // if key is NOT present already
{
counter[stringval] = 1; // initialize the key with value 1
}
else
{
counter[stringval]++; // key is already present, increment the value by 1
}
}
for (auto keyvaluepair : counter)
{
// .first to access key, .second to access value
cout << keyvaluepair.first << ":" << keyvaluepair.second << endl;
}
return 0;
}
输出:
Lizard:1
Cat:3
Dog:2
答案 2 :(得分:1)
您可以使用 CppCounter:
#include <iostream>
#include "counter.hpp"
int main() {
collection::Counter<std::string> counter;
++counter["cat"];
++counter["cat"];
counter["dogs"] = 8;
counter["lizards"] = 0;
std::cout << "{ ";
for (const auto& it: counter) {
std::cout << "\"" << it.first << "\":" << it.second.value() << " ";
}
std::cout << "}" << std::endl;
}
CppCounter 是 collections.Counter 的 C++ 实现:https://gitlab.com/miyamoto128/cppcounter。 它写在 unordered_map 之上,易于使用。
我可以补充一下我是作者吗;)
答案 3 :(得分:0)
如果要使平均查找复杂度保持恒定(使用collections.Counter时),可以使用std::unordered_map。 std::map“通常实现为红黑树”,因此查找的复杂度在容器大小上是对数的。而且我们没有内置库中使用Python实现的红黑树实现。
std::unordered_map<std::string,int> counter;
counter["dog"] = 8;
counter["cat"]++;
counter["cat"]++;
counter["1"] = 0;
for (auto pair : counter) {
cout << pair.first << ":" << pair.second << std::endl;
}