问题
I want to create a finalDic which contains common keys and sum of their values
myDic = [{2:1, 3:1, 5:2}, {3:4, 6:4, 2:3}, {2:5, 3:6}, ...]
First find common keys
commonkey = [{2:1, 3:1}, {2:3, 3:4}, {2:5, 3:6}]
Then Sum and sort by their values
finalDic= {3:11, 2,9}
I've tried this and not even close what i want
import collections
myDic = [{2:1, 3:1, 5:2}, {3:4, 6:4, 2:3}, {2:5, 3:6}]
def commonKey(x):
i=0
allKeys = []
while i<len(x):
for key in x[0].keys():
allKeys.append(key)
i=i+1
commonKeys = collections.Counter(allKeys)
commonKeys = [i for i in commonKeys if commonKeys[i]>len(x)-1]
return commonKeys
print commonKey(myDic)
Thanks
回答1:
Here's how I'd do it:
my_dict = [{2:1, 3:1, 5:2}, {3:4, 6:4, 2:3}, {2:5, 3:6}]
# Finds the common keys
common_keys = set.intersection(*map(set, my_dict))
# Makes a new dict with only those keys and sums the values into another dict
summed_dict = {key: sum(d[key] for d in my_dict) for key in common_keys}
Or as a crazy one-liner:
{k: sum(d[k] for d in my_dict) for k in reduce(set.intersection, map(set, my_dict))}
回答2:
Only some pointers:
- obtain the keys from each directory in turn them into a set() and calculate the intersection() or all key sets. This will give you the common keys.
- now iterating over the original data and summing up the matching values from each dict is straight forward
The implementation is left to the OP as an exercise.
回答3:
l = [{2:1, 3:1, 5:2}, {3:4, 6:4, 2:3}, {2:5, 3:6}]
new_dict = {}
def unique_key_value(a,b):
return set(a).intersection(set(b))
def dict_sum(k, v):
if k not in new_dict.keys():
new_dict[k] = v
else:
new_dict[k] = new_dict[k] + v
for i in reduce(unique_key_value, l):
for k in l:
if i in k.keys():
dict_sum(i, k[i])
print new_dict
hope this helps. :)
回答4:
python 3.2
from collections import defaultdict
c=defaultdict(list)
for i in myDic:
for m,n in i.items():
c[m].append(n)
new_dic={i:sum(v) for i,v in c.items()if len(v)==len(myDic)}
print(new_dic)
来源:https://stackoverflow.com/questions/13985686/how-to-find-common-keys-in-a-list-of-dicts-and-sort-them-by-value