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Python遍历子目录查找文件对

directory path python

0

proximacentauri · 技术社区 · 7 年前

我的子文件夹结构很深,如下所示:

a/b/file1.txt
a/b/file1.doc
a/b/file2.txt
a/b/file2.doc
a/c/file3.txt
a/c/file3.doc
a/c/d/file4.txt
a/c/d/file4.doc

到目前为止,我想到的最好的方法是以下看起来效率不高的方法:

files = []
for root, dirs, files in os.walk(path):
    for filename in files:
        if os.path.isdir(os.path.join(os.path.abspath("."), filename)):
            file_list = os.listdir(filename)
            file_list_copy = file_list.copy()
            #for each in file_list of type .txt
            # find .doc of same name in file_list_copy
            #add the 2 to tuple nd append to list

1 回复 | 直到 7 年前

1

0

proximacentauri 7 年前

可能不是最有效的,但有效:

find /path-to-files-root/ -type f -name '*.txt' -exec mv -i {} /new-path-to-files/txt/ \;

然后我跑了:

def get_all_files(path, pattern):
#see https://stackoverflow.com/questions/17282887/getting-files-with-same-name-irrespective-of-their-extension
    datafiles = []
    for root,dirs,files in os.walk(path):
        for file in fnmatch.filter(files, pattern):
            datafiles.append(file)
    return datafiles

txt_files = [f for f in os.listdir(txt_path) if isfile(join(txt_path, f))]
doc_files = [f for f in os.listdir(doc_path) if isfile(join(doc_path, f))]
for i, txt_file in enumerate(txt_files):
    filename = (os.path.splitext(txt_file)[0])
    doc_files = get_all_files(doc_path, '{0}.doc'.format(filename))
    if len(doc_files)== 1:
        doc_file = doc_files[0]
        #do something with txt_file and doc_file