我在python中有一个字典列表。现在我如何将这些词典合并到python中的单个实体中。 示例字典是
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"]}]
输出应该是:
[{"name":"kishore", "playing":["cricket","basket ball","volley ball","hockey"]},{"name":"xyz","playing":["cricket"]}]
答案 0 :(得分:14)
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)
输出:
[{'playing': ['volley ball', 'basket ball', 'hockey', 'cricket'], 'name': 'kishore'}, {'playing': ['cricket'], 'name': 'xyz'}]
答案 1 :(得分:5)
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())
产地:
[{'name': 'kishore', 'playing': ['cricket', 'basket ball', 'volley ball', 'hockey']}, {'name': 'xyz', 'playing': ['cricket']}]
或者,使用集合:
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())
答案 2 :(得分:3)
这基本上是@perreal的答案的一个轻微变体(我添加了defaultdict
版本之前的答案!)
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()]
答案 3 :(得分:0)
from collections import defaultdict
result = defaultdict(set)
[result[k[1]].update(v[1]) for k,v in [d.items() for d in input_dictionary]]
print [{'name':k, 'playing':v} for k,v in result.items()]