python - 查找并替换列表宽度 Python 中的空行

Question

我使用的是 Python 2.7.5，我有一个如下所示的列表:

('2047', '1971', '', '1919', '', '1878', '', '', '1760', '1731', '', '1667', '', '')

列表充满了漏洞，我想识别空行并用下一个非空行的数据填充这些空行。如果没有包含数据的下一行，我想将它们设置为 0。像这样:

('2047', '1971', '1919', '1919', '1878', '1878', '1760', '1760', '1760', '1731', '1667', '1667','0','0')

我如何使用 Python 执行此操作？我可以做一个简单的 for 循环来识别空行，但我不知道如何填充空白...任何帮助将非常感激!谢谢马丁

Answer 1

您可以使用 itertools.groupby为此:

>>> from itertools import groupby
>>> t = ('2047', '1971', '', '1919', '', '1878', '', '', '1760', '1731', '', '1667', '', '')
>>> out = []
>>> prev = None
for k, g in groupby(reversed(t)):
    if k == '' and prev is None:
        out.extend('0' for _ in g)
    elif k == '' and prev is not None:
        out.extend(prev for _ in g)
    else:
        for x in g:
            out.append(x)
        prev = x
...             
>>> out.reverse()
>>> out
['2047', '1971', '1919', '1919', '1878', '1878', '1760', '1760', '1760', '1731', '1667', '1667', '0', '0']