Question

我有这个词典列表：

[{'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'}]

我需要将'Total Incidents'的de值转换为int，并将它们相加。此外，如果日期相同，则该函数将获取所有日期并将其转换为一个日期。例如，输出应该是这样的：

[{'Total Incidents': 9, 'CrimeDate':'05/14/2016'}]

这是我到目前为止所尝试的：

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

dict1是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'}

提前致谢

Answer 1

使用itertools.groupby：

>>> [{'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'}]

求和字典列表的值

1 个答案: