如何在timedelta计算中构造输出？

Question

我有2个dicts列表：

hah = [{1:[datetime(2014, 6, 26, 13, 7, 27), datetime(2014, 7, 26, 13, 7,27)]},
                  {2:datetime(2014, 5, 26,13,7,27)}]
dateofstart = [{1:datetime(2013, 6, 26, 13, 7, 27)}, {2:datetime(2013, 6, 
                  26,13,7,27)}]` (just examples).

我需要创建一个包含相同键的 dicts 的新列表，并且值 hah 和 dateofstart <之间的值之间存在差异/强>

它应该是这样的：

[{1:[365, 334]}, {2:[390]}]

当我自己尝试这样做时，我得到了这段代码

dif = list()
diff = list()
for start in dateofstart:
    for time in hah:
        if start.keys() == time.keys():
            starttime = start.values()
            timeofpay = time.values()
            for payments in timeofpay:
                dif.append(starttime.pop(0) - payments)
            diff.append({str(start.keys()):str(dif)})

它运行时出现以下错误：

Traceback (most recent call last):
  File "/Users/mihailbasmanov/Documents/date.py", line 78, in <module>
    dif.append(starttime.pop(0) - payments)
TypeError: unsupported operand type(s) for -: 'datetime.datetime' and 'list'

版：

由我自己管理（差不多）。结果代码如下：

for start in dateofstartbuyer:
    for time in hah:
        if start.keys() == time.keys():
            starttime = start.values()
            starttimetime = starttime.pop(0)
            timeofpay = time.values()
            for payments in timeofpay:
                if type(payments) == list:
                    for wtf in payments:
                        dif.append(wtf - starttimetime)
                else:
                    dif.append(payments - starttimetime)
            key = str(start.keys())
            diff.append({key[1:2]:str(dif)})
            dif = list()
print(diff)

如果您有建议如何提高此代码的效率，欢迎您在评论或答案中发布您的建议。

Answer 1

我假设您提供的示例存在拼写错误，因为：

• hah是list dict，每个字典中只有一个key。相反，它应该是单一字典。 ISN＆＃39;吨？此外，对于键2，值是单个datetime()对象而不是list。它不是[datetime]吗？

  hah = [{1:[datetime(2014, 6, 26, 13, 7, 27), datetime(2014, 7, 26, 13, 7,27)]}, {2:datetime(2014, 5, 26,13,7,27)}]

因此，我假设hah和dateofstart为dict个对象。下面是代码。如果我的任何假设不正确，请告诉我。

>>> hah = {1:[datetime(2014, 6, 26, 13, 7, 27), datetime(2014, 7, 26, 13, 7,27)],
...        2:[datetime(2014, 5, 26,13,7,27)]}
>>> dateofstart = {1:datetime(2013, 6, 26, 13, 7, 27), 2:datetime(2013, 6,26,13,7,27)}
>>> dicts = {}
>>> for key, values in hah.items():
...     dicts[key] = [(value - dateofstart[key]).days for value in values]
...
>>> dicts
{1: [365, 395], 2: [334]}