std :: vector删除满足某些条件的元素

问题描述

正如标题所述，我要删除/合并满足特定条件的向量中的对象。我的意思是我知道如何从向量中删除整数，例如值99。

As the title says I want to remove/merge objects in a vector which fulfill specific conditions. I mean I know how to remove integers from a vector which have the value 99 for instance.

Scott Meyers的删除习惯用法：

The remove idiom by Scott Meyers:

vector<int> v;
v.erase(remove(v.begin(), v.end(), 99), v.end());

但是，假设有一个包含延迟成员变量的对象向量。现在，我要消除所有延迟相差仅小于特定阈值的对象，并希望将它们组合/合并到一个对象。

But suppose if have a vector of objects which contains a delay member variable. And now I want to eliminate all objects which delays differs only less than a specific threshold and want to combine/merge them to one object.

处理的结果应为所有延迟之差应至少为指定阈值的对象向量。

The result of the process should be a vector of objects where the difference of all delays should be at least the specified threshold.

推荐答案

来抢救！

99将替换为 UnaryPredicate 可以过滤您的延迟，我将使用lambda函数。

99 would be replaced by UnaryPredicate that would filter your delays, which I am going to use a lambda function for.

这是示例：

v.erase(std::remove_if(
    v.begin(), v.end(),
    [](const int& x) {
        return x > 10; // put your condition here
    }), v.end());

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