Python 匹配列表的长度与列不等

Question

你能否将一个长度不等的列表的列表写入一系列列中，同时匹配另一个列表的顺序？

order = ['bx', 'cs', 'lb', 'pc']

totals = [
    [{'unit': 'pc', 'sum': Decimal('3.00')}],
    [{'unit': 'bx', 'sum': Decimal('3.00')}, {'unit': 'pc', 'sum': Decimal('16.00')}],
    [{'unit': 'bx', 'sum': Decimal('6.00')}, {'unit': 'lb', 'sum': Decimal('24.00')}, {'unit': 'pc', 'sum': Decimal('63.00')}],
    [{'unit': 'pc', 'sum': Decimal('36.00')}],
    [{'unit': 'bx', 'sum': Decimal('31.00')}]
]

desired_format = [
    ['', '', '', '3.00 pc'],
    ['3.00 bx', '', '', '16.00 pc'],
    ['6.00 bx', '', '24.00 lb', '63:00 pc'],
    ['', '', '', '36:00 pc'],
    ['31.00 bx', '', '', ''],
]

我已在下面尝试过，但如果 4 个按顺序排列的单元都不存在，它就无法按预期工作。

desired_format = [[]]
for total in totals:
    data = []
    for idx, um in enumerate(total):
        if um['unit'] == order[idx]:
            data.append(str(um['sum'] + ' ' + str(um['unit'])))
        else:
            data.append('')
    desired_format.append(data)

我也试过这个，结果是有太多空列和不均匀/无序的列

desired_format = [[]]
for total in totals:
    data = []
    for um in total:
        for unit in order:
            if um['unit'] == unit:
                data.append(str(um['sum'] + ' ' + str(um['unit'])))
            else:
                data.append('')
    desired_format.append(data)

Answer 1

我建议您根据您的原始帖子使用以下代码:

desired_format = []
for total in totals:
    data = []
    for orderValue in order:
        found = False
        for um in total:
            if um['unit'] == orderValue:
                found = True
                break
        if found:
            data.append(str(um['sum'] + ' ' + str(um['unit'])))
        else:
            data.append('')
    desired_format.append(data)

print(desired_format)