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Accepted
David542
Asked: 2024-04-17 03:24:09 +0800 CST

按列分区，按另一列排序

  • 772

我正在使用以下值模拟 SQL 查询：

rows = [(1, '2021/04', 'Shop 2', 341227.53), (2, '2021/05', 'Shop 2', 315447.24), (3, '2021/06', 'Shop 1', 1845662.35), (4, '2021/04', 'Shop 2', 21487.63), (5, '2021/05', 'Shop 1', 1489774.16), (6, '2021/06', 'Shop 1', 52489.35), (7, '2021/04', 'Shop 1', 154552.82), (8, '2021/05', 'Shop 2', 6548.49), (9, '2021/06', 'Shop 2', 387779.49)]

我想建立一个“窗口”函数的字典。它应该在第三列（前值：“shop1”）上进行分区，并按第二列（前值：“2021/06”）排序。

所以，它应该看起来像这样：

{
    'Shop 1': ['2021/04', '2021/05', ...],
    'Shop 2': [...],
    ...
}

有没有办法做到这一点，这样我就可以定义一个带有两个参数的 lambda 函数，例如：

window_func = lambda partition_func, order_func: ...

上面的partition_func内容是item[2]， order_func 是item[3]。

python

1 个回答

  1. Best Answer

    尝试：

    from itertools import groupby
from operator import itemgetter

rows = [
    (1, "2021/04", "Shop 2", 341227.53),
    (2, "2021/05", "Shop 2", 315447.24),
    (3, "2021/06", "Shop 1", 1845662.35),
    (4, "2021/04", "Shop 2", 21487.63),
    (5, "2021/05", "Shop 1", 1489774.16),
    (6, "2021/06", "Shop 1", 52489.35),
    (7, "2021/04", "Shop 1", 154552.82),
    (8, "2021/05", "Shop 2", 6548.49),
    (9, "2021/06", "Shop 2", 387779.49),
]

window_func = lambda partition_func, order_func: lambda x: {
    k: [order_func(v) for v in g]
    for k, g in groupby(
        sorted(x, key=lambda v: (partition_func(v), order_func(v))), partition_func
    )
}

fn = window_func(itemgetter(2), itemgetter(1))
print(fn(rows))

    印刷：

    {
    "Shop 1": ["2021/04", "2021/05", "2021/06", "2021/06"],
    "Shop 2": ["2021/04", "2021/04", "2021/05", "2021/05", "2021/06"],
}

    编辑：要获取输出中的行列表：

    window_func = lambda partition_func, order_func: lambda x: {
    k: list(g)
    for k, g in groupby(
        sorted(x, key=lambda v: (partition_func(v), order_func(v))), partition_func
    )
}

fn = window_func(itemgetter(2), itemgetter(1))
print(fn(rows))

    印刷：

    {
    "Shop 1": [
        (7, "2021/04", "Shop 1", 154552.82),
        (5, "2021/05", "Shop 1", 1489774.16),
        (3, "2021/06", "Shop 1", 1845662.35),
        (6, "2021/06", "Shop 1", 52489.35),
    ],
    "Shop 2": [
        (1, "2021/04", "Shop 2", 341227.53),
        (4, "2021/04", "Shop 2", 21487.63),
        (2, "2021/05", "Shop 2", 315447.24),
        (8, "2021/05", "Shop 2", 6548.49),
        (9, "2021/06", "Shop 2", 387779.49),
    ],
}
    • 2

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