python - 从 pandas groupby 操作结果构造一个超集

Question

name_region
bahia                    [10, 11, 12, 1, 2, 3, 4]
distrito_federal      [9, 10, 11, 12, 1, 2, 3, 4]
goias                 [9, 10, 11, 12, 1, 2, 3, 4]
maranhao                 [10, 11, 12, 1, 2, 3, 4]
mato_grosso           [9, 10, 11, 12, 1, 2, 3, 4]
mato_grosso_do_sul    [8, 9, 10, 11, 12, 1, 2, 3]

我上面有一个 pandas 系列，是通过 groupby 操作获得的。第二列代表一年中的月份。我如何构建月份的超集，即 [8, 9, 10, 11, 12, 1, 2, 3, 4] 因为它代表了所有可能出现的月份数据集

--注意:我确实想保持秩序

Answer 1

您可以使用 itertools recipe unique_everseen(保留顺序)如下所示:

>>> [i for i in unique_everseen([z for z in y['months'] for x,y in df.iterrows()])]
[9, 10, 11, 12, 1, 2, 3, 4]

unique_everseen的定义:

import itertools as it
def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in it.ifilterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element