我列出了20个文件名,例如['file1.txt', 'file2.txt', ...]。我想编写一个Python脚本将这些文件连接成一个新文件。我可以通过打开每个文件f = open(...),通过调用逐行读取f.readline(),然后将每一行写入该新文件。在我看来,这并不是很“优雅”,尤其是我必须逐行读取/写入的部分。
['file1.txt', 'file2.txt', ...]
f = open(...)
f.readline()
在Python中是否有更“优雅”的方式来做到这一点?
这应该做
对于大文件:
filenames = ['file1.txt', 'file2.txt', ...] with open('path/to/output/file', 'w') as outfile: for fname in filenames: with open(fname) as infile: for line in infile: outfile.write(line)
对于小文件:
filenames = ['file1.txt', 'file2.txt', ...] with open('path/to/output/file', 'w') as outfile: for fname in filenames: with open(fname) as infile: outfile.write(infile.read())
……还有我想到的另一个有趣的东西:
filenames = ['file1.txt', 'file2.txt', ...] with open('path/to/output/file', 'w') as outfile: for line in itertools.chain.from_iterable(itertools.imap(open, filnames)): outfile.write(line)
遗憾的是,最后一种方法留下了一些打开的文件描述符,GC还是应该照顾这些文件描述符。我只是觉得很有趣