C++ 脚本处理代码记录

发布于:2024-04-24 ⋅ 阅读:(139) ⋅ 点赞:(0)

1、文件处理相关

1.1、 OpenCV 读取某个文件夹下的所有图片

方法一:

#include <vector>
#include <opencv2/imgproc.hpp>
#include <opencv2/highgui.hpp>

std::vector<cv::Mat> ReadImage(cv::String pattern)
{
    std::vector<cv::String> fn;
    cv::glob(pattern, fn, false);
    std::vector<cv::Mat> images;
    int count = fn.size(); //number of png files in images folder
    for (int i = 0; i < count; i++)
    {
        images.emplace_back(cv::imread(fn[i]));
    }
    return images;
}
int main()
{
    cv::String pattern="E:/YOLOV5/images/*.jpg";
    std::vector<cv::Mat> img_list = ReadImage(pattern);
    return 0;
}

方法二:

#include <iostream>
#include "opencv2/core.hpp"
#include "opencv2/imgproc.hpp"
#include "opencv2/highgui.hpp"
#include <opencv2/core/utils/filesystem.hpp>


int main() {
    std::string img_dir = "E:/YOLOV5/images";
    if(cv::utils::fs::exists(img_dir))
    {
        std::cout << "该文件夹存在" << std::endl;
    }

    // 获取当前文件夹下指定格式的文件
    std::vector<cv::String> img_lists;
    cv::utils::fs::glob(img_dir, "*.jpg", img_lists);
    for(auto name:img_lists)
    {
        std::cout << name << std::endl;
    }
    return 0;
}

1.2、截取目录文件名

#include <iostream>
#include <filesystem>
 
namespace fs = std::filesystem;
 
std::string get_filename_from_path(const std::string& path) {
    return fs::path(path).filename().string();
}
 
int main() {
    std::string path = "E:/Images/filename.ext";
    std::string filename = get_filename_from_path(path);
    std::cout << "Filename: " << filename << std::endl;
    return 0;
}

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