优雅的文件组合方式 – Python

2019-06-23 08:46:39 阅读：154 来源： 互联网

嗨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()))

谢谢

解决方法:

您可以随时执行此操作,将文件放在列表中并在一切完成后手动关闭它们.这也将使格式行更简单：

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()

或者使用contextlib.ExitStack,如评论中所示.这样,打开的文件将传递给堆栈,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)))

标签：python,concatenation
来源： https://codeday.me/bug/20190623/1269202.html

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优雅的文件组合方式 – Python