在python中合并字典列表

问题描述 投票:-2回答:1

我有一个python词典列表。现在,我如何将这些字典合并到python中的单个实体中。字典示例是

input_dictionary = [{"name":"kishore", "playing":["cricket","basket ball"]},
                    {"name":"kishore", "playing":["volley ball","cricket"]},
                    {"name":"kishore", "playing":["cricket","hockey"]},
                    {"name":"kishore", "playing":["volley ball"]},
                    {"name":"xyz","playing":["cricket"]}]

输出应为:

[{"name":"kishore", "playing":["cricket","basket ball","volley ball","hockey"]},{"name":"xyz","playing":["cricket"]}]
python list dictionary python-3.7
1个回答
14
投票

使用itertools.groupby

itertools.groupby

输出:

input_dictionary = [{"name":"kishore", "playing":["cricket","basket ball"]},
                    {"name":"kishore", "playing":["volley ball","cricket"]},
                    {"name":"kishore", "playing":["cricket","hockey"]},
                    {"name":"kishore", "playing":["volley ball"]},
                    {"name":"xyz","playing":["cricket"]}]
import itertools
import operator

by_name = operator.itemgetter('name')
result = []
for name, grp in itertools.groupby(sorted(input_dictionary, key=by_name), key=by_name):
    playing = set(itertools.chain.from_iterable(x['playing'] for x in grp))
    # If order of `playing` is important use `collections.OrderedDict`
    # playing = collections.OrderedDict.fromkeys(itertools.chain.from_iterable(x['playing'] for x in grp))
    result.append({'name': name, 'playing': list(playing)})

print(result)

5
投票
[{'playing': ['volley ball', 'basket ball', 'hockey', 'cricket'], 'name': 'kishore'}, {'playing': ['cricket'], 'name': 'xyz'}]

产品:

toutput = {}
for entry in input_dictionary:
    if entry['name'] not in toutput: toutput[entry['name']] = []
    for p in entry['playing']:
        if p not in toutput[entry['name']]:
            toutput[entry['name']].append(p)
output = list({'name':n, 'playing':l} for n,l in toutput.items())

或使用集合:

[{'name': 'kishore', 'playing': ['cricket', 'basket ball', 'volley ball', 'hockey']}, {'name': 'xyz', 'playing': ['cricket']}]

3
投票

这基本上是@perreal答案的细微变化(我是说,是在from collections import defaultdict toutput = defaultdict(set) for entry in input_dictionary: toutput[entry['name']].update(entry['playing']) output = list({'name':n, 'playing':list(l)} for n,l in toutput.items()) 版本添加之前的答案!)

defaultdict

0
投票
merged = {}
for d in input_dictionary:
    merged.setdefault(d["name"], set()).update(d["playing"])

output = [{"name": k, "playing": list(v)} for k,v in merged.items()]
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