List of values for duplicate keys in dictionary Py

问题:

Apologies in advance if this question has already been explored here - I looked at different answers here but couldn't find what I need.

My goal is to create a dictionary like this -- {'a':[10, 9, 10, 10], 'b':[10, 9, 1, 0], 'c':[0, 5, 0, 1], and so on}

What I have is multiple dictionaries with duplicate keys (same keys in every other dictionary), something like this {'a':10, 'b': 0, 'c': 2} {'a':7, 'b': 4, 'c': 4} {'a':4, 'b': 5, 'c': 3}

I have no way of knowing the number of such dictionaries, or if there are keys continuing up to 'f', or a 'g' in them but I know that the keys are duplicated. I've tried defaultdict but what I get is--

defaultdict(<type 'list'>, {'a': [10]})
defaultdict(<type 'list'>, {'a': [10], 'b': [3]})
defaultdict(<type 'list'>, {'a': [10], 'b': [3], 'c': [0]})

and then the same thing for the next dictionary --

defaultdict(<type 'list'>, {'a': [4]})
defaultdict(<type 'list'>, {'a': [4], 'b': [5]})
defaultdict(<type 'list'>, {'a': [4], 'b': [5], 'c': [1]})

The code I have for the above output is --

d = collections.defaultdict(list)
    for k, v in z.iteritems():
        d[k].append(v)
        c = d.items()
        print d

If I do a print c instead (to print the d.items()) I get --

[('a', [10])]
[('a', [10]), ('b', [3])]
[('a', [10]), ('c', [0]), ('b', [3])]

which is again repeated for each dictionary. How do I get 1 dict holding all the keys, values --

{'a':[10,0,..], 'b':[4, 3, 4,..], etc.} ?

I should also add that the dicts I have are the result of a for loop and not stored individually in a unique variable.

回答1:

If I understand correctly your attetion, you are trying to merge various dictionaries. One way using built-ins (I am sure that soon someone will give you a numpy and collections answer) could look like this:

ds = [
    {'a':10, 'b': 0, 'c': 2},
    {'a':7, 'b': 4, 'c': 4},
    {'a':4, 'b': 5, 'c': 3} ]

merged = {}
for d in ds:
    for k, v in d.items ():
        if k not in merged: merged [k] = []
        merged [k].append (v)

print (merged)

(Quite verbose for clarity)

EDIT: After having read your comment stating "The result I want is a list of values/key.", you can use this on the resulting merged dictionary:

print ( [ (v, k) for k, v in merged.items () ] )

This yields:

[([10, 7, 4], 'a'), ([2, 4, 3], 'c'), ([0, 4, 5], 'b')]

回答2:

Is this what you need?

in_dicts = [{'a':10, 'b': 0, 'c': 2}, {'a':7, 'b': 4, 'c': 4}, {'a':4, 'b': 5, 'c': 3}]
out_dict = {}

for in_d in in_dicts:
  for k, v in in_d.iteritems():
    out_dict.setdefault(k, []).append(v)  
print out_dict

{'a': [10, 7, 4], 'c': [2, 4, 3], 'b': [0, 4, 5]}

回答3:

Here's one way of thinking about this:

Given a dictionary in Python, you can access the value by dictionary[key]. In your case, the values of the dictionaries are lists.

So you can come up with a function to take a key, see if it already exists in your dictionary, and if it does, just append the value to the dictionary. For example:

Say this is your current dictionary:

d = { 'a':[3, 17], 'b':[2, 4] }

and you have this data to parse:

data0 = { 'a':10, 'b':2, 'c':2 }

Since the key 'a' already exists, you want to take the value of data0['a'] and append to d by d['a'].append(data0['a']).

Let me know this helps / if you need any further clarification!

• mikeba

回答4:

Another way to do it is to initialise a dict and then populate it with unique keys across dictionaries, and their corresponding values without overwriting existing values. Steps: 1. Initialise an empty dict (say key_values) to store required key - values pairs 2. Iterate through the original list of dicts 3. Yield dicts one at a time 4. If a key has not already been added to the key_values dict, add the key and add its value to a list; otherwise, append value to the value of the existing key (which is a list).

list_of_dicts = [{'a':10, 'b': 0, 'c': 2}, {'a':7, 'b': 4, 'c': 4}, {'a':4, 'b': 5, 'c': 3}]
key_values = {}

for d in list_of_dicts:
    for k,v in d.iteritems():
        if k in key_values.keys():
            key_values[k].append(v)
        else:    
            key_values[k]=[v]

print key_values #prints out: {'a': [10, 7, 4], 'c': [2, 4, 3], 'b': [0, 4, 5]}