简体   繁体   English

检查元素是否在列表中(包含)

[英]Check if element is in the list (contains)

I've got a list of elements, say, integers and I want to check if my variable (another integer) is one of the elements from the list.我有一个元素列表,比如整数,我想检查我的变量(另一个整数)是否是列表中的元素之一。 In python I'd do:在 python 我会这样做:

my_list = [1,2,3,4] # elements
my_var = 3 # my variable
my_var in my_list # returns boolean

How to do that in C++?如何在 C++ 中做到这一点? I thought of using std::list , but I can find no find method in it.我想过使用std::list ,但我找不到其中的find方法。 I can see such method in std::set structure.我可以在std::set结构中看到这样的方法。

More deeply, the problem is that my program is given some unique ids (a list, a set, whatever) and I iterate over a long list of input data (ids) and check if they are included in the list (boolean value returned for each iteration step).更深入地说,问题是我的程序被赋予了一些唯一的 id(一个列表,一个集合,等等)并且我遍历一长串输入数据(id)并检查它们是否包含在列表中(返回的布尔值每个迭代步骤)。 And I'm not sure how should I do that in C++.而且我不确定在 C++ 中我应该怎么做。

You can use std::find您可以使用std::find

bool found = (std::find(my_list.begin(), my_list.end(), my_var) != my_list.end());

You need to include <algorithm> .您需要包含<algorithm> It should work on standard containers, vectors lists, etc...它应该适用于标准容器、向量列表等...

std::list does not provide a search method. std::list不提供搜索方法。 You can iterate over the list and check if the element exists or use std::find .您可以遍历列表并检查元素是否存在或使用std::find But I think for your situation std::set is more preferable.但我认为对于你的情况std::set更可取。 The former will take O(n) time but later will take O(lg(n)) time to search.前者将花费O(n)时间,但后者将花费O(lg(n))时间进行搜索。

You can simply use:您可以简单地使用:

int my_var = 3;
std::set<int> mySet {1, 2, 3, 4};
if(mySet.find(myVar) != mySet.end()){
      //do whatever
}

你必须#include <algorithm> ,然后你可以使用std::find

They really should add a wrapper.他们真的应该添加一个包装器。 Like this:像这样:

namespace std
{
    template<class _container,
        class _Ty> inline
        bool contains(_container _C, const _Ty& _Val)
        {return std::find(_C.begin(), _C.end(), _Val) != _C.end(); }
};
...
    if( std::contains(my_container, what_to_find) )
    {

    }

A one-liner solution, similar to python, would be (std::set<int> {1, 2, 3, 4}).count(my_var) > 0 .类似于 python 的单行解决方案是(std::set<int> {1, 2, 3, 4}).count(my_var) > 0

Minimal working example最小工作示例

int my_var = 3;
bool myVarIn = (std::set<int> {1, 2, 3, 4}).count(my_var) > 0;
std::cout << std::boolalpha << myVarIn << std::endl;

prints true or false dependent of the value of my_var.根据 my_var 的值打印truefalse

Use std::find , something like:使用std::find ,例如:

if (std::find(std::begin(my_list), std::end(my_list), my_var) != std::end(my_list))
    // my_list has my_var

Declare additional helper function like this:像这样声明额外的辅助函数:

template <class T, class I >
bool vectorContains(const vector<T>& v, I& t)
{
    bool found = (std::find(v.begin(), v.end(), t) != v.end());
    return found;
}

And use it like this:并像这样使用它:

void Project::AddPlatform(const char* platform)
{
    if (!vectorContains(platforms, platform))
        platforms.push_back(platform);
}

Snapshot of example can be found here:示例快照可以在这里找到:

https://github.com/tapika/cppscriptcore/blob/b7f3d62747494a52a440482e841ffb016a3fc56e/SolutionProjectModel/Project.cpp#L13 https://github.com/tapika/cppscriptcore/blob/b7f3d62747494a52a440482e841ffb016a3fc56e/SolutionProjectModel/Project.cpp#L13

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM