Python字典打印所有键的所有值

Question

I have two python lists:

keys=[1,2,2,3,2,3]
values=['apple','book','pen','soccer','paper','tennis']

The "keys" are cluster ID list for the corresponding words in "values" list. I wish to print key-value pairs using

keys=[1,2,2,3,2,3]
values=['apple','book','pen','soccer','paper','tennis']

dictionary = dict(zip(keys, values))
for key, value in dictionary.items() :
    print (key, value)

But it only prints

1 apple
2 paper
3 tennis

What I actually want is to get all values for all keys like this

1 [apple]
2 [book,pen,paper]
3 [soccer,tennis]

I know that my current code should logically print the first output as keys are unique. But how can I change it so that it will print all values for all keys? Thank you in advance!

Answer 1

from collections import defaultdict

keys=[1,2,2,3,2,3]
values=['apple','book','pen','soccer','paper','tennis']

d = defaultdict(list)
for k, v in zip(keys, values):
    d[k].append(v)

Answer 2

Looks like what you want is a mapping from one key to multiple values, one way to accomplish it would be:

from collections import defaultdict

d = defaultdict(list)

keys=[1,2,2,3,2,3]
values=['apple','book','pen','soccer','paper','tennis']

for tuple in zip(keys, values):
    d[tuple[0]].append(tuple[1])

print(d)  # defaultdict(<class 'list'>, {1: ['apple'], 2: ['book', 'pen', 'paper'], 3: ['soccer', 'tennis']})

Answer 3

You can use itertools :

import itertools
keys=[1,2,2,3,2,3]
values=['apple','book','pen','soccer','paper','tennis']
final_data = {a:[i[0] for i in b] for a, b in [(a, list(b)) for a, b in itertools.groupby(sorted(zip(values, keys), key=lambda x:x[-1]), key=lambda x:x[-1])]}

Output:

{1: ['apple'], 2: ['book', 'pen', 'paper'], 3: ['soccer', 'tennis']}

Answer 4

pure python also works

keys=[1,2,2,3,2,3]
values=['apple','book','pen','soccer','paper','tennis']

d = dict(zip(keys, [[] for _ in keys]))  # dict w keys, empty lists as values

for k, v in zip(keys, values):
    d[k].append(v)

d
Out[128]: {1: ['apple'], 2: ['book', 'pen', 'paper'], 3: ['soccer', 'tennis']}

Answer 5

Two method :

If you want you can use default dict as many already have been suggested :

Data is :

keys=[1,2,2,3,2,3]
values=['apple','book','pen','soccer','paper','tennis']

Method: 1

import collections


d=collections.defaultdict(list)

for i in zip(keys,values):
    d[i[0]].append(i[1])

print(d)

output:

{1: ['apple'], 2: ['book', 'pen', 'paper'], 3: ['soccer', 'tennis']}

Or if you want to develop your own logic without importing any external module then you can try:

result={}
for i in zip(keys,values):
    if i[0] not in result:
        result[i[0]]=[i[1]]
    else:
        result[i[0]].append(i[1])

print(result)

output:

{1: ['apple'], 2: ['book', 'pen', 'paper'], 3: ['soccer', 'tennis']}