Question

我正在尝试创建一个涉及timedeltas的字典字典，该字典应该转换为天数（四舍五入到1 d.p.）。

输入是OrderedDict和values_list。初始化的OrderedDict和values_list如下所示：

OrderedDict(
    [('Jul 2015', defaultdict(<function __main__.<lambda>>, {})),
     ('Aug 2015', defaultdict(<function __main__.<lambda>>, {})),
     ('Sep 2015', defaultdict(<function __main__.<lambda>>, {})),
     ('Oct 2015', defaultdict(<function __main__.<lambda>>, {})),
     ('Nov 2015', defaultdict(<function __main__.<lambda>>, {})),
     ('Dec 2015', defaultdict(<function __main__.<lambda>>, {}))]
)

values_list = [
    (
        datetime.date(2015, 7, 1),
        datetime.timedelta(1, 11111),
        datetime.timedelta(2, 22222),
        datetime.timedelta(3, 33333)
    ),
    (
        datetime.date(2015, 8, 1),
        datetime.timedelta(4, 44444),
        datetime.timedelta(5, 55555),
        None
    ),
    ...
]

，输出应如下所示：

output_matrix = {
    'Jul 2015': {
        'time_1': 0.1,
        'time_2': 0.3,
        'time_3': 0.4
    }
    'Aug 2015': {
        'time_1': 0.5,
        'time_2': 0.6,
        'time_3': 0
    }
    ...
}

请注意，None timedeltas会转换为0。这看起来很容易，但到目前为止我得到的解决方案真的很难看，而且我确信有更好的方法。

我丑陋的解决方案：

matrix = OrderedDict()
today = timezone.now()
tzinfo = timezone.get_current_timezone()
headers = ['Time 1', 'Time 2', 'Time 3']  # Used elsewhere, could possibly use here?

# Initialise OrderedDict
for i in range(5, -1, -1):
    d = datetime(today.year, today.month, 1, tzinfo=tzinfo) - relativedelta(months=i)
    matrix[d.strftime("%b %Y")] = defaultdict(lambda: 0)

matrix = build_matrix(matrix, values_list)

def build_matrix(matrix, values_list)
    ''' Populates a matrix and converts timedeltas into days. '''
    for month, time1, time2, time3 in values_list:
        month_label = month.strftime("%b %Y")
        if time1 is not None:
            matrix[month_label]['Time 1'] = round(time1.seconds/60/60/24, 1)
        if time2 is not None:
            matrix[month_label]['Time 2'] = round(time2.seconds/60/60/24, 1)
        if time3 is not None:
            matrix[month_label]['Time 3'] = round(time3.seconds/60/60/24, 1)

    return matrix

如何改进此解决方案？

更干净地创建python字典

0 个答案: