Python 如何对dict元素求和
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How to sum dict elements
提问by Nazmul Hasan
In Python, I have list of dicts:
在 Python 中，我有字典列表：
dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]
I want one final dict that will contain the sum of all dicts.
I.e the result will be: {'a':5, 'b':7}
我想要一个包含所有字典总和的最终字典。即结果将是：{'a':5, 'b':7}
N.B: every dict in the list will contain same number of key, value pairs.
注意：列表中的每个字典都将包含相同数量的键值对。
采纳答案by carl
A little ugly, but a one-liner:
有点丑，但单线：
dictf = reduce(lambda x, y: dict((k, v + y[k]) for k, v in x.iteritems()), dict1)
回答by paxdiablo
The following code shows one way to do it:
以下代码显示了一种方法：
dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]
final = {}
for k in dict1[0].keys(): # Init all elements to zero.
final[k] = 0
for d in dict1:
for k in d.keys():
final[k] = final[k] + d[k] # Update the element.
print final
This outputs:
这输出：
{'a': 5, 'b': 7}
as you desired.
如您所愿。
Or, as inspired by kriss, better but still readable:
或者，受 kriss 的启发，更好但仍然可读：
dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]
final = {}
for d in dict1:
for k in d.keys():
final[k] = final.get(k,0) + d[k]
print final
I pine for the days of the original, readable Python :-)
我怀念原始的、可读的 Python 的日子:-)
回答by Manoj Govindan
This might help:
这可能有帮助：
def sum_dict(d1, d2):
for key, value in d1.items():
d1[key] = value + d2.get(key, 0)
return d1
>>> dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]
>>> reduce(sum_dict, dict1)
{'a': 5, 'b': 7}
回答by John La Rooy
Leveraging sum()should get better performance when adding more than a few dicts
sum()添加多个 dicts 时，利用杠杆应该会获得更好的性能
>>> dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]
>>> from operator import itemgetter
>>> {k:sum(map(itemgetter(k), dict1)) for k in dict1[0]} # Python2.7+
{'a': 5, 'b': 7}
>>> dict((k,sum(map(itemgetter(k), dict1))) for k in dict1[0]) # Python2.6
{'a': 5, 'b': 7}
adding Stephan's suggestion
添加斯蒂芬的建议
>>> {k: sum(d[k] for d in dict1) for k in dict1[0]} # Python2.7+
{'a': 5, 'b': 7}
>>> dict((k, sum(d[k] for d in dict1)) for k in dict1[0]) # Python2.6
{'a': 5, 'b': 7}
I think Stephan's version of the Python2.7 code reads really nicely
我认为 Stephan 版本的 Python2.7 代码读起来非常好
回答by SiggyF
You can use the collections.Counter
您可以使用collections.Counter
counter = collections.Counter()
for d in dict1:
counter.update(d)
Or, if you prefer oneliners:
或者，如果您更喜欢单线：
functools.reduce(operator.add, map(collections.Counter, dict1))
回答by Dave Kirby
In Python 2.7 you can replace the dict with a collections.Counterobject. This supports addition and subtraction of Counters.
在 Python 2.7 中，您可以用collections.Counter对象替换 dict 。这支持计数器的加法和减法。
回答by trudolf
I was interested in the performance of the proposed Counter, reduce and sum methods for large lists. Maybe someone else is interested in this as well. You can have a look here: https://gist.github.com/torstenrudolf/277e98df296f23ff921c
我对针对大型列表的建议 Counter、reduce 和 sum 方法的性能感兴趣。也许其他人也对此感兴趣。你可以看看这里：https: //gist.github.com/torstenrudolf/277e98df296f23ff921c
I tested the three methods for this list of dictionaries:
我为此字典列表测试了三种方法：
dictList = [{'a': x, 'b': 2*x, 'c': x**2} for x in xrange(10000)]
the sum method showed the best performance, followed by reduce and Counter was the slowest. The time showed below is in seconds.
sum 方法表现出最好的性能，其次是reduce，而Counter 是最慢的。下面显示的时间以秒为单位。
In [34]: test(dictList)
Out[34]:
{'counter': 0.01955194902420044,
'reduce': 0.006518083095550537,
'sum': 0.0018319153785705566}
But this is dependent on the number of elements in the dictionaries. the sum method will slow down faster than the reduce.
但这取决于字典中的元素数量。sum 方法比reduce 慢得更快。
l = [{y: x*y for y in xrange(100)} for x in xrange(10000)]
In [37]: test(l, num=100)
Out[37]:
{'counter': 0.2401433277130127,
'reduce': 0.11110662937164306,
'sum': 0.2256883692741394}
回答by Kyan
Here is a reasonable beatiful one.
这是一个合理的美丽的。
final = {}
for k in dict1[0].Keys():
final[k] = sum(x[k] for x in dict1)
return final
回答by Aaron McMillin
One further one line solution
又一单线解决方案
dict(
functools.reduce(
lambda x, y: x.update(y) or x, # update, returns None, and we need to chain.
dict1,
collections.Counter())
)
This creates only one counter, uses it as an accumulator and finally converts back to a dict.
这仅创建一个计数器，将其用作累加器，最后转换回 dict。
回答by Ste
Here is another working solution (python3), quite general as it works for dict, lists, arrays. For non-common elements, the original value will be included in the output dict.
这是另一个工作解决方案（python3），非常通用，因为它适用于字典、列表、数组。对于非公共元素，原始值将包含在输出字典中。
def mergsum(a, b):
for k in b:
if k in a:
b[k] = b[k] + a[k]
c = {**a, **b}
return c
dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]
print(mergsum(dict1[0], dict1[1]))
