我有这样的字典,
data = {11L:[{'a':2,'b' :1},{'a':2,'b':3}], 22L:[{'a':3,'b':2},{'a':2,'b' :5},{'a':4,'b':2},{'a':1,'b':5},{'a':1,'b':0}], 33L:{'a':1,'b':2},{'a':3,'b':5},{'a':5,'b':2},{'a' :1,'b':3},{'a':1,'b':6},{'a':2,'b':0}], 44L:[{'a' :4,'b':2},{'a':4,'b':5},{'a':3,'b':1},{'a':3,'b' },{'a':2,'b':3},{'a':1,'b':2},{'a':1,'b':0}]}这里我将摆脱外键,并给出新的键值1,2,3等等,我想得到如下所示的结果,
result = {1:{'a':10,'b':7} 2:{ 'A':11, 'b':18},3:{ 'A':12, 'b':5},4:{ '一个':5, 'b':11},5: {'a':3,'b':9},6:{'a':3,'b':2},7:{'a':1,'b':0}}我尝试过这样的一些东西,但是我得到所需的结果,
d = defaultdict(int) for dct in data.values(): for k,v in dct.items():d [k] + = v print dict(d)我希望结果字典的键是动态的,就像上面的数据字典一样, 7个键值对,因此我们有结果字典有7个键等等
解决方案你想在这里使用一个列表,并且您想要使用 Counter()对象来使求和更容易:
from collections import counter from itertools import izip_longest fordcts在data.values()中:为i,dct in枚举(dcts):如果我> = len(结果): result.append(Counter(dct)) else: result [i] .update(dct)结果:
>>>结果 [Counter({'a':10,'b':7}),Counter({'b':18,'a':11}),Counter({'a':12, b':5}),Counter({'b':11,'a':5}),Counter({'b':9,'a':4}),Counter({'a' 'b':2}),Counter({'a':1,'b':0})]Counter()对象是 dict 的子类,所以它们以其他方式表现为字典。如果您有以 dict 值之后添加以下行:
result = [dict(r)for r in result]灵感来自Eric,您可以将以上内容转换为单行:
从集合import counter from itertools import izip_longest result = [sum(map,col),Counter()) for izip_longest中的col(* data.values(),fillvalue = {})]此版本与上述循环略有不同,该键在求和时从计数器中删除0。如果您想在上一个计数器中保留'b':0 ,请使用:
[reduce(lambda c,d:c.update(d)或c,col,Counter()) for izip_longest(* data.values(),fillvalue = {})]这再次使用 .update() / p>
I have a dictionary like this,
data={11L: [{'a': 2, 'b': 1},{'a': 2, 'b': 3}], 22L: [{'a': 3, 'b': 2},{'a': 2, 'b': 5},{'a': 4, 'b': 2},{'a': 1, 'b': 5}, {'a': 1, 'b': 0}], 33L: [{'a': 1, 'b': 2},{'a': 3, 'b': 5},{'a': 5, 'b': 2},{'a': 1, 'b': 3}, {'a': 1, 'b': 6},{'a':2,'b':0}], 44L: [{'a': 4, 'b': 2},{'a': 4, 'b': 5},{'a': 3, 'b': 1},{'a': 3, 'b': 3}, {'a': 2, 'b': 3},{'a':1,'b':2},{'a': 1, 'b': 0}]}Here i ll get rid of the outer keys, and give new key values 1, 2 , 3 so on, i want to get the result as shown below,
result={1:{'a':10,'b':7},2:{'a':11,'b':18},3:{'a':12,'b':5},4:{'a':5,'b':11},5:{'a':3,'b':9},6:{'a':3,'b':2},7:{'a':1,'b':0}}I tried some thing like this, but i dint get the required result,
d = defaultdict(int) for dct in data.values(): for k,v in dct.items(): d[k] += v print dict(d)I want the keys of result dictionary to be dynamic, like in the above data dictionary we have 44 which has highest with 7 key value pairs, hence we have the result dictionary with 7 keys and so on
解决方案You want to use a list here, and you want to perhaps use Counter() objects to make the summing that much easier:
from collections import Counter from itertools import izip_longest for dcts in data.values(): for i, dct in enumerate(dcts): if i >= len(result): result.append(Counter(dct)) else: result[i].update(dct)Result:
>>> result [Counter({'a': 10, 'b': 7}), Counter({'b': 18, 'a': 11}), Counter({'a': 12, 'b': 5}), Counter({'b': 11, 'a': 5}), Counter({'b': 9, 'a': 4}), Counter({'a': 3, 'b': 2}), Counter({'a': 1, 'b': 0})]Counter() objects are subclasses of dict, so they otherwise behave as dictionaries. If you have to have dict values afterwards, add the following line:
result = [dict(r) for r in result]Taking inspiration from Eric, you can transform the above into a one-liner:
from collections import Counter from itertools import izip_longest result = [sum(map(Counter, col), Counter()) for col in izip_longest(*data.values(), fillvalue={})]This version differs slightly from the loop above in that keys that are 0 are dropped from the counter when summing. If you want to keep 'b': 0 in the last counter, use:
[reduce(lambda c, d: c.update(d) or c, col, Counter()) for col in izip_longest(*data.values(), fillvalue={})]This uses .update() again.
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