Question

说我有以下列表：

list1 = [{'position':'child','title':'dave','obj':obj,created:'2013-11-20'},
         {'position':'child','title':'james','obj':obj,created:'2013-11-19'},
         {'position':'child','title':'james','obj':obj,created:'2013-11-18'},
         {'position':'child','title':'ian','obj':obj,created:'2013-11-17'},
         {'position':'child','title':'dave','obj':obj,created:'2013-11-16'},
         {'position':'child','title':'dave','obj':obj,created:'2013-11-15'},
         {'position':'child','title':'dave','obj':obj,created:'2013-11-14'}]

正如您所看到的那样，它已经处于日期顺序（我希望保持这种状态）。 stackoverflow上有很多关于如何使用defaultdict按公共条目进行分组的答案，但有没有办法按title分组，同时保留created顺序，如下所示：

list1 = [{'position':'parent','title':'dave root'},
         {'position':'child','title':'dave','obj':obj,created:'2013-11-20'},
         {'position':'parent','title':'james root'},
         {'position':'child','title':'james','obj':obj,created:'2013-11-19'},
         {'position':'child','title':'james','obj':obj,created:'2013-11-18'},
         {'position':'parent','title':'ian root'},
         {'position':'child','title':'ian','obj':obj,created:'2013-11-17'},
         {'position':'parent','title':'dave root'},
         {'position':'child','title':'dave','obj':obj,created:'2013-11-16'},
         {'position':'child','title':'dave','obj':obj,created:'2013-11-15'},
         {'position':'child','title':'dave','obj':obj,created:'2013-11-14'}]

Answer 1

使用itertools.groupby：

list1 = [{'position':'child','title':'dave','obj':'obj','created':'2013-11-20'},
         {'position':'child','title':'james','obj':'obj','created':'2013-11-19'},
         {'position':'child','title':'james','obj':'obj','created':'2013-11-18'},
         {'position':'child','title':'ian','obj':'obj','created':'2013-11-17'},
         {'position':'child','title':'dave','obj':'obj','created':'2013-11-16'},
         {'position':'child','title':'dave','obj':'obj','created':'2013-11-15'},
         {'position':'child','title':'dave','obj':'obj','created':'2013-11-14'}]

import itertools
import operator

list1 = [
    x
    for key, grp in itertools.groupby(list1, key=operator.itemgetter('title'))
    for x in [{'position': 'parent', 'title': key + ' root'}]+ list(grp)
]
for x in list1:
    print(x)

输出：

{'position': 'parent', 'title': 'dave root'}
{'position': 'child', 'created': '2013-11-20', 'obj': 'obj', 'title': 'dave'}
{'position': 'parent', 'title': 'james root'}
{'position': 'child', 'created': '2013-11-19', 'obj': 'obj', 'title': 'james'}
{'position': 'child', 'created': '2013-11-18', 'obj': 'obj', 'title': 'james'}
{'position': 'parent', 'title': 'ian root'}
{'position': 'child', 'created': '2013-11-17', 'obj': 'obj', 'title': 'ian'}
{'position': 'parent', 'title': 'dave root'}
{'position': 'child', 'created': '2013-11-16', 'obj': 'obj', 'title': 'dave'}
{'position': 'child', 'created': '2013-11-15', 'obj': 'obj', 'title': 'dave'}
{'position': 'child', 'created': '2013-11-14', 'obj': 'obj', 'title': 'dave'}

在Python中对列表进行分组但保持顺序

1 个答案: