仅迭代文件夹中的几个文件 [英] Iterate through only a couple of files in a folder

问题描述

目前我使用以下方法遍历文件夹中的所有文件：

Currently I loop through all the files in the folder by using:

for eachFile in os.listdir(savePath):

如何快速查看前50个文件？我正在对我的程序进行故障排除，并查看文件夹中的所有~2000个文件需要花费太多时间来循环。



如果我可以减少我看到的文件数量，我可以更快地调试。以下感觉很笨拙...有什么建议使这个超高效？谢谢！

Whats a quick way to look at the only the first 50 files? I am troubleshooting my program and looking at all ~2000 files in the folder takes too much time to loop through.

if I can reduce the number of files I look at I can debug much faster. The following feels clunky... any suggestions on making this ultra efficient? Thank you!

g = []
for eachFile in os.listdir(savePath):
    j = eachFile.strip(".tx")
    g.append(j)

For x in range(0,50):
    

    #Created a list for each file
    w = list(reversed(list(open(savePath + '/' + g[x]))))

推荐答案

将它指向一个文件较少的目录？



或者有一个指向'真实'的牺牲目录dir或你的'减少'目录（视情况而定）使用 mklink [ ^ ]？

point it at a directory with fewer files in it?

or have a sacrificial dir that points to the 'real' dir or your 'reduced' dir (as appropriate) using mklink[^]?

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