Question

我在目录中有数千个文件，格式为YYYY / MM / DD / HH / MM：

201801010000.txt
201801010001.txt
201801010002.txt

我想保留几个小时，所以我需要每天每小时将60个文件合并为一个。我不知道如何搜索文件名以获取我想要的60个文件。这就是我写的

def concat_files(path):
    file_list = os.listdir(path)
    with open(datetime.datetime.now(), "w") as outfile:
        for filename in sorted(file_list):
            with open(filename, "r") as infile:
                outfile.write(infile.read())

如何命名文件以保留日期？我现在使用datetime但它覆盖了当前的文件名。使用我的代码我将所有文件合并为一个，我应该将每个％60合并到一个不同的文件中。

Answer 1

您可以使用glob来获取所需的文件。它允许您在搜索文件时传入匹配的模式。在下面的最后一行中，它只会找到以'2018010100'开头，有两个字符，以'.txt'结尾的文件

from glob import glob

def concat_files(dir_path, file_pattern):
    file_list = glob(os.path.join(dir_path, file_pattern))
    with open(datetime.datetime.now(), "w") as outfile:
        for filename in sorted(file_list):
            with open(filename, "r") as infile:
                outfile.write(infile.read())

concat_files('C:/path/to/directory', '2018010100??.txt')

Answer 2

你不是那么远，你只需要交换你的逻辑：

file_list = os.listdir(path)
for filename in sorted(file_list):
    out_filename = filename[:-6] + '.txt'
    with open(out_filename, 'a') as outfile:
        with open(path + '/' + filename, 'r') as infile:
            outfile.write(infile.read())

Answer 3

试试这个。

file_list = os.listdir(path)
for f in { f[:-6] for f in file_list }:
    if not f:
        continue
    with open(f + '.txt', 'a') as outfile:
        for file in sorted([ s for s in file_list if s.startswith(f)]):
            with open(path + '/' + file, 'r') as infile:
                outfile.write(infile.read())
            #os.remove(path + '/' + file) # optional

Python合并目录

3 个答案: