C ++ - 开发自己的std :: count_if版本 [英] C++ - Developing own version of std::count_if

问题描述

对于任务，我做一些简单的数据采样，以确定哪些样本包含音频计数总能量。我一直在研究 std :: count_if 函数，虽然这在某种程度上适合我的需要，例如：

For a task, I'm doing some simple data sampling to determine which samples contains audio counting the total number of energy. I've been looking into the std::count_if function, and, although this suits my needs to a certain extent, for example:

int foo = std::count_if(
               std::begin(samples), 
               std::end(samples),
               containsSound);

这会计算包含声音的样本总数，但不会指示样本包含声音。我想出了这个解决方案：

This counts the total number of samples that contain sound, but does not give an indication to the samples that contain sound. I came up with this solution:

std::vector<std::string> sound_files = myLib::count_sample_if(
                                              std::begin(samples),
                                              std::end(samples),
                                              samples.DirName, 
                                              containsSOund);

然后存储并推送 samples.DirName 到向量，然后我可以使用它来存储我选择的样本集。

This would then store and push the samples.DirName to a vector which I can then use to only store the sample set of my choice.

是否容易实现？

推荐答案

如果你只是需要可读性/开发速度，关心性能，那么你可以很容易地使用 std :: copy_if 和 std :: transform 来获得你需要的： / p>

If you just need readability / speed of development and you don't care about performance then you can easily use std::copy_if and std::transform to obtain what you need:

std::vector<Song> songs;
std::vector<Song> filtered;
std::vector<std::string> transformed;

std::copy_if(songs.begin(), songs.end(), filtered.begin(), [](const Song &song) { return whatever you need; });
std::transform(filtered.begin(), filtered.end(), transformed.begin(), [](const Song &song) { return song.sample; });

或者您可以使用 std :: for_each ：

std::vector<Song> songs;
std::vector<std::string> transformed;

std::for_each(songs.begin(), songs.end(), [&](const Song &song) { if (song.containsSample()) transformed.push_back(song.sample); });

那么包含声音的样本量只是 transformed.size 。

Then the amount of samples that contains sound is just transformed.size().

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