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Question

Let's say I have a list of dictionaries:

[
    {'id': 1, 'name': 'john', 'age': 34},
    {'id': 1, 'name': 'john', 'age': 34},
    {'id': 2, 'name': 'hanna', 'age': 30},
]

How can I obtain a list of unique dictionaries (removing the duplicates)?

[
    {'id': 1, 'name': 'john', 'age': 34},
    {'id': 2, 'name': 'hanna', 'age': 30},
]

Answer 1

So make a temporary dict with the key being the id . This filters out the duplicates. The values() of the dict will be the list

In Python2.7

>>> L=[
... {'id':1,'name':'john', 'age':34},
... {'id':1,'name':'john', 'age':34},
... {'id':2,'name':'hanna', 'age':30},
... ]
>>> {v['id']:v for v in L}.values()
[{'age': 34, 'id': 1, 'name': 'john'}, {'age': 30, 'id': 2, 'name': 'hanna'}]

In Python3

>>> L=[
... {'id':1,'name':'john', 'age':34},
... {'id':1,'name':'john', 'age':34},
... {'id':2,'name':'hanna', 'age':30},
... ] 
>>> list({v['id']:v for v in L}.values())
[{'age': 34, 'id': 1, 'name': 'john'}, {'age': 30, 'id': 2, 'name': 'hanna'}]

In Python2.5/2.6

>>> L=[
... {'id':1,'name':'john', 'age':34},
... {'id':1,'name':'john', 'age':34},
... {'id':2,'name':'hanna', 'age':30},
... ] 
>>> dict((v['id'],v) for v in L).values()
[{'age': 34, 'id': 1, 'name': 'john'}, {'age': 30, 'id': 2, 'name': 'hanna'}]

Answer 2

The usual way to find just the common elements in a set is to use Python's set class. Just add all the elements to the set, then convert the set to a list , and bam the duplicates are gone.

The problem, of course, is that a set() can only contain hashable entries, and a dict is not hashable.

If I had this problem, my solution would be to convert each dict into a string that represents the dict , then add all the strings to a set() then read out the string values as a list() and convert back to dict .

A good representation of a dict in string form is JSON format. And Python has a built-in module for JSON (called json of course).

The remaining problem is that the elements in a dict are not ordered, and when Python converts the dict to a JSON string, you might get two JSON strings that represent equivalent dictionaries but are not identical strings. The easy solution is to pass the argument sort_keys=True when you call json.dumps() .

EDIT: This solution was assuming that a given dict could have any part different. If we can assume that every dict with the same "id" value will match every other dict with the same "id" value, then this is overkill; @gnibbler's solution would be faster and easier.

EDIT: Now there is a comment from André Lima explicitly saying that if the ID is a duplicate, it's safe to assume that the whole dict is a duplicate. So this answer is overkill and I recommend @gnibbler's answer.

Answer 3

In case the dictionaries are only uniquely identified by all items (ID is not available) you can use the answer using JSON. The following is an alternative that does not use JSON, and will work as long as all dictionary values are immutable

[dict(s) for s in set(frozenset(d.items()) for d in L)]

Answer 4

Here's a reasonably compact solution, though I suspect not particularly efficient (to put it mildly):

>>> ds = [{'id':1,'name':'john', 'age':34},
...       {'id':1,'name':'john', 'age':34},
...       {'id':2,'name':'hanna', 'age':30}
...       ]
>>> map(dict, set(tuple(sorted(d.items())) for d in ds))
[{'age': 30, 'id': 2, 'name': 'hanna'}, {'age': 34, 'id': 1, 'name': 'john'}]

Answer 5

You can use numpy library (works for Python2.x only):

   import numpy as np 

   list_of_unique_dicts=list(np.unique(np.array(list_of_dicts)))

To get it worked with Python 3.x (and recent versions of numpy), you need to convert array of dicts to numpy array of strings, eg

list_of_unique_dicts=list(np.unique(np.array(list_of_dicts).astype(str)))

Answer 6

a = [
{'id':1,'name':'john', 'age':34},
{'id':1,'name':'john', 'age':34},
{'id':2,'name':'hanna', 'age':30},
]

b = {x['id']:x for x in a}.values()

print(b)

outputs:

[{'age': 34, 'id': 1, 'name': 'john'}, {'age': 30, 'id': 2, 'name': 'hanna'}]

Answer 7

Since the id is sufficient for detecting duplicates, and the id is hashable: run 'em through a dictionary that has the id as the key. The value for each key is the original dictionary.

deduped_dicts = dict((item["id"], item) for item in list_of_dicts).values()

In Python 3, values() doesn't return a list; you'll need to wrap the whole right-hand-side of that expression in list() , and you can write the meat of the expression more economically as a dict comprehension:

deduped_dicts = list({item["id"]: item for item in list_of_dicts}.values())

Note that the result likely will not be in the same order as the original. If that's a requirement, you could use a Collections.OrderedDict instead of a dict .

As an aside, it may make a good deal of sense to just keep the data in a dictionary that uses the id as key to begin with.

Answer 8

We can do with pandas

import pandas as pd
yourdict=pd.DataFrame(L).drop_duplicates().to_dict('r')
Out[293]: [{'age': 34, 'id': 1, 'name': 'john'}, {'age': 30, 'id': 2, 'name': 'hanna'}]

Notice slightly different from the accept answer.

drop_duplicates will check all column in pandas , if all same then the row will be dropped .

For example :

If we change the 2nd dict name from john to peter

L=[
    {'id': 1, 'name': 'john', 'age': 34},
    {'id': 1, 'name': 'peter', 'age': 34},
    {'id': 2, 'name': 'hanna', 'age': 30},
]
pd.DataFrame(L).drop_duplicates().to_dict('r')
Out[295]: 
[{'age': 34, 'id': 1, 'name': 'john'},
 {'age': 34, 'id': 1, 'name': 'peter'},# here will still keeping the dict in the out put 
 {'age': 30, 'id': 2, 'name': 'hanna'}]

Answer 9

Expanding on John La Rooy ( Python - List of unique dictionaries ) answer, making it a bit more flexible:

def dedup_dict_list(list_of_dicts: list, columns: list) -> list:
    return list({''.join(row[column] for column in columns): row
                for row in list_of_dicts}.values())

Calling Function:

sorted_list_of_dicts = dedup_dict_list(
    unsorted_list_of_dicts, ['id', 'name'])

Answer 10

I don't know if you only want the id of your dicts in the list to be unique, but if the goal is to have a set of dict where the unicity is on all keys' values.. you should use tuples key like this in your comprehension :

>>> L=[
...     {'id':1,'name':'john', 'age':34},
...    {'id':1,'name':'john', 'age':34}, 
...    {'id':2,'name':'hanna', 'age':30},
...    {'id':2,'name':'hanna', 'age':50}
...    ]
>>> len(L)
4
>>> L=list({(v['id'], v['age'], v['name']):v for v in L}.values())
>>>L
[{'id': 1, 'name': 'john', 'age': 34}, {'id': 2, 'name': 'hanna', 'age': 30}, {'id': 2, 'name': 'hanna', 'age': 50}]
>>>len(L)
3

Hope it helps you or another person having the concern....

Answer 11

There are a lot of answers here, so let me add another:

import json
from typing import List

def dedup_dicts(items: List[dict]):
    dedupped = [ json.loads(i) for i in set(json.dumps(item, sort_keys=True) for item in items)]
    return dedupped

items = [
    {'id': 1, 'name': 'john', 'age': 34},
    {'id': 1, 'name': 'john', 'age': 34},
    {'id': 2, 'name': 'hanna', 'age': 30},
]
dedup_dicts(items)

Answer 12

I have summarized my favorites to try out:

https://repl.it/@SmaMa/Python-List-of-unique-dictionaries

# ----------------------------------------------
# Setup
# ----------------------------------------------

myList = [
  {"id":"1", "lala": "value_1"},
  {"id": "2", "lala": "value_2"}, 
  {"id": "2", "lala": "value_2"}, 
  {"id": "3", "lala": "value_3"}
]
print("myList:", myList)

# -----------------------------------------------
# Option 1 if objects has an unique identifier
# -----------------------------------------------

myUniqueList = list({myObject['id']:myObject for myObject in myList}.values())
print("myUniqueList:", myUniqueList)

# -----------------------------------------------
# Option 2 if uniquely identified by whole object
# -----------------------------------------------

myUniqueSet = [dict(s) for s in set(frozenset(myObject.items()) for myObject in myList)]
print("myUniqueSet:", myUniqueSet)

# -----------------------------------------------
# Option 3 for hashable objects (not dicts)
# -----------------------------------------------

myHashableObjects = list(set(["1", "2", "2", "3"]))
print("myHashAbleList:", myHashableObjects)

Answer 13

In python 3, simple trick, but based on unique field (id):

data = [ {'id': 1}, {'id': 1}]

list({ item['id'] : item for item in data}.values())

Answer 14

If there is not a unique id in the dictionaries, then I'd keep it simple and define a function as follows:

def unique(sequence):
    result = []
    for item in sequence:
        if item not in result:
            result.append(item)
    return result

The advantage with this approach, is that you can reuse this function for any comparable objects. It makes your code very readable, works in all modern versions of Python, preserves the order in the dictionaries, and is fast too compared to its alternatives.

>>> L = [
... {'id': 1, 'name': 'john', 'age': 34},
... {'id': 1, 'name': 'john', 'age': 34},
... {'id': 2, 'name': 'hanna', 'age': 30},
... ] 
>>> unique(L)
[{'id': 1, 'name': 'john', 'age': 34}, {'id': 2, 'name': 'hanna', 'age': 30}]

Answer 15

In python 3.6+ (what I've tested), just use:

import json

#Toy example, but will also work for your case 
myListOfDicts = [{'a':1,'b':2},{'a':1,'b':2},{'a':1,'b':3}]
#Start by sorting each dictionary by keys
myListOfDictsSorted = [sorted(d.items()) for d in myListOfDicts]

#Using json methods with set() to get unique dict
myListOfUniqueDicts = list(map(json.loads,set(map(json.dumps, myListOfDictsSorted))))

print(myListOfUniqueDicts)

Explanation: we're mapping the json.dumps to encode the dictionaries as json objects, which are immutable. set can then be used to produce an iterable of unique immutables. Finally, we convert back to our dictionary representation using json.loads . Note that initially, one must sort by keys to arrange the dictionaries in a unique form. This is valid for Python 3.6+ since dictionaries are ordered by default.

Answer 16

A quick-and-dirty solution is just by generating a new list.

sortedlist = []

for item in listwhichneedssorting:
    if item not in sortedlist:
        sortedlist.append(item)

Answer 17

Well all the answers mentioned here are good, but in some answers one can face error if the dictionary items have nested list or dictionary, so I propose simple answer

a = [str(i) for i in a]
a = list(set(a))
a = [eval(i) for i in a]

Answer 18

Objects can fit into sets. You can work with objects instead of dicts and if needed after all set insertions convert back to a list of dicts. Example

class Person:
    def __init__(self, id, age, name):
        self.id = id
        self.age = age
        self.name = name

my_set = {Person(id=2, age=3, name='Jhon')}

my_set.add(Person(id=3, age=34, name='Guy'))

my_set.add({Person(id=2, age=3, name='Jhon')})

# if needed convert to list of dicts
list_of_dict = [{'id': obj.id,
                 'name': obj.name,
                 'age': obj.age} for obj in my_set]

Answer 19

Let me add mine.

1. sort target dict so that {'a' : 1, 'b': 2} and {'b': 2, 'a': 1} are not treated differently

2. make it as json

3. deduplicate via set (as set does not apply to dicts)

4. again, turn it into dict via json.loads

import json

[json.loads(i) for i in set([json.dumps(i) for i in [dict(sorted(i.items())) for i in target_dict]])]

Answer 20

Pretty straightforward option:

L = [
    {'id':1,'name':'john', 'age':34},
    {'id':1,'name':'john', 'age':34},
    {'id':2,'name':'hanna', 'age':30},
    ]


D = dict()
for l in L: D[l['id']] = l
output = list(D.values())
print output

Answer 21

Heres an implementation with little memory overhead at the cost of not being as compact as the rest.

values = [ {'id':2,'name':'hanna', 'age':30},
           {'id':1,'name':'john', 'age':34},
           {'id':1,'name':'john', 'age':34},
           {'id':2,'name':'hanna', 'age':30},
           {'id':1,'name':'john', 'age':34},]
count = {}
index = 0
while index < len(values):
    if values[index]['id'] in count:
        del values[index]
    else:
        count[values[index]['id']] = 1
        index += 1

output:

[{'age': 30, 'id': 2, 'name': 'hanna'}, {'age': 34, 'id': 1, 'name': 'john'}]

Answer 22

This is the solution I found:

usedID = []

x = [
{'id':1,'name':'john', 'age':34},
{'id':1,'name':'john', 'age':34},
{'id':2,'name':'hanna', 'age':30},
]

for each in x:
    if each['id'] in usedID:
        x.remove(each)
    else:
        usedID.append(each['id'])

print x

Basically you check if the ID is present in the list, if it is, delete the dictionary, if not, append the ID to the list