如何使用 Python 制作具有重复键的字典？

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You can't have a dict with duplicate keys for definition! Instead you can use a single key and, as the value, a list of elements that had that key.

So you can follow these steps:

See if the current element's key (of your initial set) is in the final dict. If it is, go to step 3
Update dict with key
Append the new value to the dict[key] list
Repeat [1-3]

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Python dictionaries don't support duplicate keys. One way around is to store lists or sets inside the dictionary.

One easy way to achieve this is by using defaultdict:

from collections import defaultdict


data_dict = defaultdict(list)

All you have to do is replace

data_dict[regNumber] = details

with

data_dict[regNumber].append(details)

and you'll get a dictionary of lists.

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You can't have duplicated keys in a dictionary. Use a dict of lists:

for line in data_list:
regNumber = line[0]
name = line[1]
phoneExtn = line[2]
carpark = line[3].strip()
details = (name,phoneExtn,carpark)
if not data_dict.has_key(regNumber):
data_dict[regNumber] = [details]
else:
data_dict[regNumber].append(details)

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You can change the behavior of the built in types in Python. For your case it's really easy to create a dict subclass that will store duplicated values in lists under the same key automatically:

class Dictlist(dict):
def __setitem__(self, key, value):
try:
self[key]
except KeyError:
super(Dictlist, self).__setitem__(key, [])
self[key].append(value)

Output example:

>>> d = dictlist.Dictlist()
>>> d['test'] = 1
>>> d['test'] = 2
>>> d['test'] = 3
>>> d
{'test': [1, 2, 3]}
>>> d['other'] = 100
>>> d
{'test': [1, 2, 3], 'other': [100]}

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If you want to have lists only when they are necessary, and values in any other cases, then you can do this:

class DictList(dict):
def __setitem__(self, key, value):
try:
# Assumes there is a list on the key
self[key].append(value)
except KeyError: # If it fails, because there is no key
super(DictList, self).__setitem__(key, value)
except AttributeError: # If it fails because it is not a list
super(DictList, self).__setitem__(key, [self[key], value])

You can then do the following:

dl = DictList()
dl['a']  = 1
dl['b']  = 2
dl['b'] = 3

Which will store the following {'a': 1, 'b': [2, 3]}.

I tend to use this implementation when I want to have reverse/inverse dictionaries, in which case I simply do:

my_dict = {1: 'a', 2: 'b', 3: 'b'}
rev = DictList()
for k, v in my_dict.items():
rev_med[v] = k

Which will generate the same output as above: {'a': 1, 'b': [2, 3]}.

CAVEAT: This implementation relies on the non-existence of the append method (in the values you are storing). This might produce unexpected results if the values you are storing are lists. For example,

dl = DictList()
dl['a']  = 1
dl['b']  = [2]
dl['b'] = 3

would produce the same result as before {'a': 1, 'b': [2, 3]}, but one might expected the following: {'a': 1, 'b': [[2], 3]}.

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You can refer to the following article: http://www.wellho.net/mouth/3934_Multiple-identical-keys-in-a-Python-dict-yes-you-can-.html

In a dict, if a key is an object, there are no duplicate problems.

For example:

class p(object):
def __init__(self, name):
self.name = name
def __repr__(self):
return self.name
def __str__(self):
return self.name
d = {p('k'): 1, p('k'): 2}

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I just posted an answer to a question that was subsequently closed as a duplicate of this one (for good reasons I think), but I'm surprised to see that my proposed solution is not included in any of the answers here.

Rather than using a defaultdict or messing around with membership tests or manual exception handling, you can easily append values onto lists within a dictionary using the setdefault method:

results = {}                              # use a normal dictionary for our output
for k, v in some_data:                    # the keys may be duplicates
results.setdefault(k, []).append(v)   # magic happens here!

This is a lot like using a defaultdict, but you don't need a special data type. When you call setdefault, it checks to see if the first argument (the key) is already in the dictionary. If doesn't find anything, it assigns the second argument (the default value, an empty list in this case) as a new value for the key. If the key does exist, nothing special is done (the default goes unused). In either case though, the value (whether old or new) gets returned, so we can unconditionally call append on it, knowing it should always be a list.

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Dictionary does not support duplicate key, instead you can use defaultdict
Below is the example of how to use defaultdict in python3x to solve your problem

from collections import defaultdict


sdict = defaultdict(list)
keys_bucket = list()


data_list = [lines.split(",") for lines in contents.split("\n")]
for data in data_list:
key = data.pop(0)
detail = data
    

keys_bucket.append(key)
if key in keys_bucket:
sdict[key].append(detail)
else:
sdict[key] = detail


print("\n", dict(sdict))

Above code would produce output as follow:

{'EDF768': [[' Bill Meyer', ' 2456', ' Vet_Parking'], [' Jenny Meyer', ' 9987', ' Vet_Parking']], 'TY5678': [[' Jane Miller', ' 8987', ' AgHort_Parking'], [' Jo King', ' 8987', ' AgHort_Parking']], 'GEF123': [[' Jill Black', ' 3456', ' Creche_Parking']], 'ABC234': [[' Fred Greenside', ' 2345', ' AgHort_Parking']], 'GH7682': [[' Clara Hill', ' 7689', ' AgHort_Parking']], 'JU9807': [[' Jacky Blair', ' 7867', ' Vet_Parking'], [' Mike Green', ' 3212', ' Vet_Parking']], 'KLOI98': [[' Martha Miller', ' 4563', ' Vet_Parking']], 'ADF645': [[' Cloe Freckle', ' 6789', ' Vet_Parking']], 'DF7800': [[' Jacko Frizzle', ' 4532', ' Creche_Parking']], 'WER546': [[' Olga Grey', ' 9898', ' Creche_Parking']], 'HUY768': [[' Wilbur Matty', ' 8912', ' Creche_Parking']]}

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It's pertty old question but maybe my solution help someone.

by overriding __hash__ magic method, you can save same objects in dict.

Example:

from random import choices


class DictStr(str):
"""
This class behave exacly like str class but
can be duplicated in dict
"""
def __new__(cls, value='', custom_id='', id_length=64):
# If you want know why I use __new__ instead of __init__
# SEE: https://stackoverflow.com/a/2673863/9917276
obj = str.__new__(cls, value)
if custom_id:
obj.id = custom_id
else:
# Make a string with length of 64
choice_str = "abcdefghijklmopqrstuvwxyzABCDEFJHIJKLMNOPQRSTUVWXYZ1234567890"
obj.id = ''.join(choices(choice_str, k=id_length))
return obj


def __hash__(self) -> int:
return self.id.__hash__()

Now lets create a dict:

>>> a_1 = DictStr('a')
>>> a_2 = DictStr('a')
>>> a_3 = 'a'
>>> a_1
a
>>> a_2
a
>>> a_1 == a_2 == a_3
True
>>> d = dict()
>>> d[a_1] = 'some_data'
>>> d[a_2] = 'other'
>>> print(d)
{'a': 'some_data', 'a': 'other'}

NOTE: This solution can apply to any basic data structure like (int, float,...)

EXPLANATION :

We can use almost any object as key in dict class (or mostly known as HashMap or HashTable in other languages) but there should be a way to distinguish between keys because dict have no idea about objects.

For this purpose objects that want to add to dictionary as key somehow have to provide a unique identifier number(I name it uniq_id, it's actually a number somehow created with hash algorithm) for themself.

Because dictionary structure widely use in most of solutions, most of programming languages hide object uniq_id generation inside a hash name buildin method that feed dict in key search

So if you manipulate hash method of your class you can change behaviour of your class as dictionary key