我遇到连接词典的问题。有这么多代码所以我在示例中显示了我的问题。
d1 = {'the':3, 'fine':4, 'word':2}
+
d2 = {'the':2, 'fine':4, 'word':1, 'knight':1, 'orange':1}
+
d3 = {'the':5, 'fine':8, 'word':3, 'sequel':1, 'jimbo':1}
=
finald = {'the':10, 'fine':16, 'word':6, 'knight':1, 'orange':1, 'sequel':1, 'jimbo':1}
它正在为wordcloud预备wordcounts。我不知道如何连接键的值,这对我来说很困惑。请帮忙。 最好的问候
答案 0 :(得分:5)
我会使用collections
中的Counter
进行此操作。
from collections import Counter
d1 = {'the':3, 'fine':4, 'word':2}
d2 = {'the':2, 'fine':4, 'word':1, 'knight':1, 'orange':1}
d3 = {'the':5, 'fine':8, 'word':3, 'sequel':1, 'jimbo':1}
c = Counter()
for d in (d1, d2, d3):
c.update(d)
print(c)
输出:
Counter({'fine': 16, 'the': 10, 'word': 6, 'orange': 1, 'jimbo': 1, 'sequel': 1, 'knight': 1})
答案 1 :(得分:2)
import itertools
d1 = {'the':3, 'fine':4, 'word':2}
d2 = {'the':2, 'fine':4, 'word':1, 'knight':1, 'orange':1}
d3 = {'the':5, 'fine':8, 'word':3, 'sequel':1, 'jimbo':1}
dicts = [d1, d2, d3]
In [31]: answer = {k:sum(d[k] if k in d else 0 for d in dicts) for k in itertools.chain.from_iterable(dicts)}
In [32]: answer
Out[32]:
{'sequel': 1,
'the': 10,
'fine': 16,
'jimbo': 1,
'word': 6,
'orange': 1,
'knight': 1}
答案 2 :(得分:2)
def sumDicts(*dicts):
summed = {}
for subdict in dicts:
for (key, value) in subdict.items():
summed[key] = summed.get(key, 0) + value
return summed
Shell示例:
>>> d1 = {'the':3, 'fine':4, 'word':2}
>>> d2 = {'the':2, 'fine':4, 'word':1, 'knight':1, 'orange':1}
>>> d3 = {'the':5, 'fine':8, 'word':3, 'sequel':1, 'jimbo':1}
>>> sumDicts(d1, d2, d3)
{'orange': 1, 'the': 10, 'fine': 16, 'jimbo': 1, 'word': 6, 'knight': 1, 'sequel': 1}