python中的字典操作列表-TypeError：不可哈希类型：'dict'

Question

I have a list of the folowing form:

oldlist = [{'x': {'a':1,'b':2}, 'y':2},{'x':{'a':6,'b':7}, 'y':2},{'x':{'a':1,'b':2}, 'y':3},{'x':{'a':1,'b':2}, 'y':2},{'x':{'a':10,'b':11}, 'y':4}]  

to be converted into

final = [{'x':{'a':1,'b':2},'y':[2,3,2],'count':3},{'x':{'a':6,'b':7},'y':[2],'count':1},{'x':{'a':10,'b':11},'y':[4],'count':1}]  

I have tried

oldlist = [{'x': {'a':1,'b':2}, 'y':2},{'x':{'a':6,'b':7}, 'y':2},{'x':{'a':1,'b':2}, 'y':3},{'x':{'a':1,'b':2}, 'y':2},{'x':{'a':10,'b':11}, 'y':4}]  
list1=[]  
list2=[]  
list3=[]  
s = set([d['x'] for d in oldlist])  
news=list(s)  
for item in oldlist:  
if item['x'] == news[0]:  
      list1.append(item['y'])  

if item['x'] == news[1]:  
      list2.append(item['y'])  

if item['x'] == news[2]:  
      list3.append(item['y'])  
final=[]  
dic1 = {'x':news[0],'y':list1,'count':len(list1)}  
dic2 = {'x':news[1],'y':list2,'count':len(list2)}  
dic3 = {'x':news[2],'y':list3,'count':len(list3)}  
final.append(dic1)  
final.append(dic2)  
final.append(dic3)  
print final

Getting

s = set([d['x'] for d in oldlist])
TypeError: unhashable type: 'dict'

Is there a simpler way to do it? Plus here I knew that x can have only three values so I created three variables list1, list2 and list3. What if x can have several other values and I have to find out a similar list of dictionaries like final! It should also work for strings!

EDIT:I tried this. But it all got messed up

s = list(frozenset(oldlist[0]['x'].items()))  
print s  
for item in oldlist:  
    s.append(frozenset(item['x'].items()))  

Answer 1

The set function can only handle hashable objects, like a string, number, tuple etc

Data types like List, dict are unhashable types, and hence the set function cannot handle them.

For some more clarity:

What do you mean by hashable in Python?

http://blog.lerner.co.il/is-it-hashable-fun-and-games-with-hashing-in-python/

A basic implementation of what you need:

for elem in oldlist:
    found = False
    for item in newlist:
        if elem['x'] == item['x']:
            y = item.get('y',[])
            item['y'] = t.append(elem['y'])
            found = True
            break
    if not found:
        newlist.append({'x':elem['x'], 'y':[elem['y']]})

This will give you the expected result

Answer 2

Set function of python does not allow dictionaries and you can not force it, try another method instead. (Take a closer look of the comment on the 5th and 6th line)

Try this code:

oldlist = [{'x': {'a':1,'b':2}, 'y':2},{'x':{'a':6,'b':7}, 'y':2},{'x':{'a':1,'b':2}, 'y':3},{'x':{'a':1,'b':2}, 'y':2},{'x':{'a':10,'b':11}, 'y':4}]  
list1=[]  
list2=[]  
list3=[]  
s = [d['x'] for d in oldlist] # Placed the dictionaries in a list
s = result = [dict(tupleized) for tupleized in set(tuple(item.items()) for item in s)] # This is the manual way on removing duplicates dictionaries in a list instead of using set
news=list(s)  
for item in oldlist:  
    if item['x'] == news[0]:  
        list1.append(item['y'])  

    if item['x'] == news[1]:  
        list2.append(item['y'])  

    if item['x'] == news[2]:  
        list3.append(item['y']) 

final=[]  
dic1 = {'x':news[0],'y':list1,'count':len(list1)}  
dic2 = {'x':news[1],'y':list2,'count':len(list2)}  
dic3 = {'x':news[2],'y':list3,'count':len(list3)}  
final.append(dic1)  
final.append(dic2)  
final.append(dic3)  
print final

Answer 3

You can use defaultdict where keys are frozenset objects created from value of x in the original dicts and values are list of relative y . Then you can construct the final result with list comprehension and turn frozensets back to dicts:

from collections import defaultdict

oldlist = [{'x': {'a':1,'b':2}, 'y':2},{'x':{'a':6,'b':7}, 'y':2},{'x':{'a':1,'b':2}, 'y':3},{'x':{'a':1,'b':2}, 'y':2},{'x':{'a':10,'b':11}, 'y':4}]
res = defaultdict(list)
for d in oldlist:
    res[frozenset(d['x'].items())].append(d['y'])

final = [{'x': dict(k), 'y': v, 'count': len(v)} for k, v in res.items()] # [{'y': [2, 3, 2], 'x': {'a': 1, 'b': 2}, 'count': 3}, {'y': [4], 'x': {'a': 10, 'b': 11}, 'count': 1}, {'y': [2], 'x': {'a': 6, 'b': 7}, 'count': 1}]