如何获得 o/p 作为 {"Chennai": ["Ram", "stephen"], "Mumbai": "Laxman"} 对于给定的 i/p = {"Ram": "Chennai", "Laxman": "Mumbai", "stephen": "Chennai"}

Question

我在这里尝试使用 defaultdict 函数，但它似乎不起作用

s = {"Ram": "Chennai", "Laxman": "Mumbai", "stephen": "Chennai"} 
d = defaultdict(list)
for k, v in s:
    d[k].append(v)
sorted(d.items())
print(d)

Answer 1

这应该适合你：

from collections import defaultdict


s = {"Ram": "Chennai", "Laxman": "Mumbai", "stephen": "Chennai"} 
d = defaultdict(list)
for k, v in s.items():
    d[v].append(k)

sorted(d.items())
print(d)

输出：

defaultdict(<class 'list'>, {'Chennai': ['Ram', 'stephen'], 'Mumbai': ['Laxman']})

Answer 2

您正在尝试同时迭代键和值，因此您必须使用 s.items() 将两个键值对作为元组获取：[('Ram', 'Chennai'), ('Laxman', 'Mumbai'), ('stephen', 'Chennai')]

from collections import defaultdict
s = {"Ram": "Chennai", "Laxman": "Mumbai", "stephen": "Chennai"} 
d = defaultdict(list)
for k, v in s.items():
    d[v].append(k) # Append k instead of v
sorted(d.items())
print(d) # {'Chennai': ['Ram', 'stephen'], 'Mumbai': ['Laxman']}

Answer 3

您正在遍历 dict 的 key 和 val 而没有调用 items()，而且新的 dict 应该有 keys 与城市名称因此必须在 for 循环中的 v 和 k 中互换。这是代码：

from collections import defaultdict
s = {"Ram": "Chennai", "Laxman": "Mumbai", "stephen": "Chennai"} 
d = defaultdict(list)
for k, v in s.items():
    d[v].append(k)
sorted(d.items())
print(d)

输出：

defaultdict(<class 'list'>, {'Chennai': ['Ram', 'stephen'], 'Mumbai': ['Laxman']})

如何获得 o/p 作为 {"Chennai": ["Ram", "stephen"], "Mumbai": "Laxman"} 对于给定的 i/p = {"Ram": "Chennai", "Laxman": "Mumbai", "stephen": "Chennai"}

how to get o/p as {"Chennai": ["Ram", "stephen"], "Mumbai": "Laxman"} for given i/p = {"Ram": "Chennai", "Laxman": "Mumbai", "stephen": "Chennai"}

python

dictionary

defaultdict