列表字典的值求和

Question

我有一份字典清单：

[{'Total Incidents': '1', 'CrimeDate': '05/14/2016'}, {'Total Incidents': '1', 'CrimeDate': '05/14/2016'}, {'Total Incidents': '1', 'CrimeDate': '05/14/2016'}, {'Total Incidents': '1', 'CrimeDate': '05/14/2016'}, {'Total Incidents': '1', 'CrimeDate': '05/14/2016'}, {'Total Incidents': '1', 'CrimeDate': '05/14/2016'}, {'Total Incidents': '1', 'CrimeDate': '05/14/2016'}, {'Total Incidents': '1', 'CrimeDate': '05/14/2016'}, {'Total Incidents': '1', 'CrimeDate': '05/14/2016'}]

我需要把'Total Incidents'的de值转换成int，然后把它们加起来。另外，如果日期相等，则函数会将所有日期转换为一个日期。例如，输出应该如下所示：

^{pr2}$

到目前为止，我一直在努力：

dictf = reduce(lambda x, y: dict((k, v + y[k]) for k, v in x.iteritems()), dict1)

dict1是一个包含第一个字典的变量。 但是输出不是我想要的，它给了我这个：

{'Total Incidents': '111111111', 'CrimeDate': '05/14/201605/14/201605/14/201605/14/201605/14/201605/14/201605/14/201605/14/201605/14/2016'}

提前谢谢

Answer 1

使用^{}：

>>> [{'CrimeDate': g[0], 'Total Incidents': sum(map(lambda x: int(x['Total Incidents']), g[1]))} for g in itertools.groupby(d, lambda x: x['CrimeDate'])]
[{'Total Incidents': 9, 'CrimeDate': '05/14/2016'}]