Question

嗨，这只是一个简单的问题。我有一个完美无缺的文件串联，但它有点混乱。我想知道是否只有更优雅的方式来写这个：

path = path/to/file/location
with open(path + 'result.txt', 'w') as result, \
        open(path + 'file1.txt') as f1, \
            open(path + 'file2.txt' ) as f2, \
                open(path + 'file3.txt' ) as f3, \
                    open(path + 'file4.txt' ) as f4, \
                        open(path + 'file5.txt' ) as f5, \
                            open(path + 'file6.txt' ) as f6, \
                                open(path + 'file7.txt' ) as f7, \
                                    open(path + 'file8.txt' ) as f8, \
                                        open(path + 'file9.txt' ) as f9, \
                                            open(path + 'file10.txt' ) as f10, \
                                                open(path + 'file11.txt' ) as f11, \
                                                    open(path + 'file12.txt' ) as f12, \
                                                        open(path + 'file13.txt' ) as f13, \
                                                            open(path + 'file14.txt' ) as f14, \
                                                                open(path + 'file15.txt' ) as f15, \
                                                                    open(path + 'file16.txt' ) as f16:
    for line1, line2, line3, line4, line5, line6, line7, line8, \ 
        line9, line10, line11, line12, line13, line14, line15, line16 \
        in zip(f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12,f13,f14,f15,f16):

        result.write('{}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, \
        {}, {}, {}\n'.format(line1.rstrip(), line2.rstrip(), line3.rstrip(), line4.rstrip(), \
        line5.rstrip(), line6.rstrip(), line7.rstrip(), line8.rstrip(), line9.rstrip(), \
        line10.rstrip(), line11.rstrip(), line12.rstrip(), line13.rstrip(), line14.rstrip(), \
        line15.rstrip(), line16.rstrip()))

由于

Answer 1

你总是可以在没有with的情况下完成它，将文件放在一个列表中并在一切完成时手动关闭它们。这也将使format行变得更加简单：

path = "path/to/file/location/"
with open(path + 'result.txt', 'w') as result:
    files = [open(path + 'file%d.txt' % (n+1)) for n in range(16)]
    form = ", ".join('{}' for f in files) + '\n'
    for lines in zip(*files):
        result.write(form.format(*map(str.rstrip, lines)))
    for f in files:
        f.close()

或使用contextlib.ExitStack，如评论中所示。这样，打开的文件将传递给stack，并将在with块之后关闭文件。

with open(path + 'result.txt', 'w') as result, contextlib.ExitStack() as stack:
    files = [stack.enter_context(open(path + 'file%d.txt' % (n+1))) for n in range(16)]
    form = ", ".join('{}' for f in files) + '\n'
    for lines in zip(*files):
        result.write(form.format(*map(str.rstrip, lines)))

Answer 2

您可以按顺序处理文件（这可以解决一次打开太多文件的潜在问题）：

result_names = ['result1','result2']
result_index = 0
old_result_path = path + "file1.txt"
for n in xrange(2,17):
    new_result_path = path + (result_names[result_index] if n<16 else 'result.txt')
    input_path = path + "file%d.txt" % n
    with open( old_result_path, 'r' ) as old_input, \
        open( input_path, 'r' ) as new_input, 
            open( new_result_path, 'w' ) as result:
       for line1, line2 in zip( old_input, new_input ):
           result.write('{}, {}\n'.format(line1.rstrip(), line2.rstrip())
    old_result_path = new_result_path
    result_index = 1 - result_index

这会留下result1.txt和result2.txt，你可能会或可能不关心清理。

优雅的文件组合方式 - Python

2 个答案: