我有一个具有以下结构的目录:
Main directory:
|--2001
|--200101
|--feed_013_01.zip
|--feed_restr_013_01.zip
|--feed_013_04.zip
|--feed_restr_013_04.zip
...
|--feed_013_30.zip
|--feed_restr_013_30.zip
...
|--2021
|--202101
|--feed_013_01.zip
|--feed_restr_013_01.zip
|--feed_013_04.zip
|--feed_restr_013_04.zip
...
|--feed_013_30.zip
|--feed_restr_013_30.zip
我需要按顺序读取和排序zip文件:
feed_restr_013_30.zip, feed_013_30.zip.....feed_restr_013_01.zip, feed_013_01.zip
我目前正在做这样的事情:
def atoi(text):
return int(text) if text.isdigit() else text
def natural_keys(text):
return [atoi(c) for c in re.split(r'(\d+)', text)]
for path, subdirs, files in os.walk(directory):
subdirs.sort(key=natural_keys)
subdirs.reverse()
files.sort(key=natural_keys)
files.reverse()
首先需要所有“rest”文件,我得到的列表如下:
feed_restr_013_30.zip,feed_restr_013_01.zip.....feed_013_30.zip, feed_013_01.zip
更新
我能够用buran和SCKU的答案以及我现有的逻辑来解决这个问题
def atoi(text):
return int(text) if text.isdigit() else text
def parse(fname):
try:
prefix, *middle, n1, n2 = fname.split('_')
except:
prefix, *middle, n1 = fname.split('_')
n2 = ''
return (prefix, n1, [atoi(c) for c in re.split(r'(\d+)',n2)], ''.join(middle))
def get_Files(self, directory, source, keywords):
file_paths = []
for path, subdirs, files in os.walk(directory):
for file in files:
file_name = os.path.join(path, file)
file_paths.append(file_name)
return file_paths
files = get_Files(directory, source, keywords)
files.sort(key=parse, reverse=True)
输出
如果您的目录结构良好且不太大,我建议您立即获取所有文件的路径并对其进行排序:
如果您不想按反向排序年份,可以在键函数中使用
-int(d_y)
等将其反向排序相关问题 更多 >
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