Python字典打印所有键的所有值

Question

我有两个 python 列表：

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

“keys”是“values”列表中对应单词的集群 ID 列表。我希望使用

打印键值对

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)

但它只打印

1 apple
2 paper
3 tennis

我真正想要的是像这样获取所有键的所有值

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

我知道我当前的代码应该在逻辑上打印第一个输出，因为键是唯一的。但是我怎样才能改变它，让它打印所有键的所有值呢？提前谢谢！

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

看起来您想要的是从一个键到多个值的映射，实现它的一种方法是：

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

你可以使用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])]}

输出：

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

Answer 4

纯python也可以

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

两种方法：

如果你愿意，你可以使用已经建议的默认字典：

数据是：

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

方法：1

import collections


d=collections.defaultdict(list)

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

print(d)

输出：

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

或者如果您想开发自己的逻辑而不导入任何外部模块，那么您可以尝试：

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)

输出：

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