通过使用Python排序的文本文件 [英] Sorting text file by using Python

问题描述

我有一个文本文件中包含了超过10万线。这样的行：

I have a text file includes over than 10 million lines. Lines like that:

37024469;196672001;255.0000000000
37024469;196665001;396.0000000000
37024469;196664001;396.0000000000
37024469;196399002;85.0000000000
37024469;160507001;264.0000000000
37024469;160506001;264.0000000000

正如你看到的，分隔符是;。我想排序按照第二个元素使用python这个文本文件。我不能使用分割功能。因为它会导致的MemoryError。我该如何管理呢？

As you seen, delimiter is ";". i would like to sort this text file by using python according to the second element. I couldnt use split function. Because it causes MemoryError. how can i manage it ?

推荐答案

不要在内存中的排序10万款。拆分这件事分批来代替：

Don't sort 10 million lines in memory. Split this up in batches instead:

• 运行100 10万行排序（使用文件作为一个迭代器，并结合 islice（）或类似选择一个批次）。写出到别处单独的文件。

• Run 100 100k line sorts (using the file as an iterator, combined with islice() or similar to pick a batch). Write out to separate files elsewhere.

合并排序的文件。这里是你可以通过100打开的文件，它会产生在有序线的合并发生。写一行一个新的文件中的行：

Merge the sorted files. Here is an merge generator that you can pass 100 open files and it'll yield lines in sorted order. Write to a new file line by line:

import operator

def mergeiter(*iterables, **kwargs):
    """Given a set of sorted iterables, yield the next value in merged order

    Takes an optional `key` callable to compare values by.
    """
    iterables = [iter(it) for it in iterables]
    iterables = {i: [next(it), i, it] for i, it in enumerate(iterables)}
    if 'key' not in kwargs:
        key = operator.itemgetter(0)
    else:
        key = lambda item, key=kwargs['key']: key(item[0])

    while True:
        value, i, it = min(iterables.values(), key=key)
        yield value
        try:
            iterables[i][0] = next(it)
        except StopIteration:
            del iterables[i]
            if not iterables:
                raise

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