将字典列表转换成字典列表

Question

I have a list of dictionary.

l = [{'row':1}, {'row':2}, {'row':1}, {'row':3}, {'row':1}]

I want output as :

x = [[{'row':1}, {'row':1}, {'row':1}], [{'row':2}], [{'row':3}]]

I want lists of those dicts whose 'row' value is same.
How can I do that using python?

Answer 1

Using itertools.groupby :

>>> import itertools
>>>
>>> l = [{'row':1}, {'row':2}, {'row':1}, {'row':3}, {'row':1}]
>>> row = lambda d: d['row']
>>> [list(grp) for _, grp in itertools.groupby(sorted(l, key=row), key=row)]
[[{'row': 1}, {'row': 1}, {'row': 1}], [{'row': 2}], [{'row': 3}]]

NOTE You need to pass sorted list to groupby .

Alternative that does not require sorting, using collection.defaultdict :

>>> from collections import defaultdict
>>>
>>> l = [{'row':1}, {'row':2}, {'row':1}, {'row':3}, {'row':1}]
>>> ret = defaultdict(list)
>>> for d in l:
...     ret[d['row']].append(d)
...
>>> ret.values() # list(d.values())   in Python 3.x
[[{'row': 1}, {'row': 1}, {'row': 1}], [{'row': 2}], [{'row': 3}]]

As shx2 commented, the alternative solution result will be arbitrary. If you want sorted result, use following form:

>>> [ret[key] for key in sorted(ret)]
[[{'row': 1}, {'row': 1}, {'row': 1}], [{'row': 2}], [{'row': 3}]]

Answer 2

x_dict = {}
for d in l:
    x_dict[d['row']] = x_dict.get(d['row'], []) + [d]
x = [x_dict[key] for key in sorted(x_dict.keys())]