从字典列表中删除重复项 python

Question

I am trying to remove the duplicates from following list

 distinct_cur = [{'rtc': 0, 'vf': 0, 'mtc': 0, 'doc': 'good job', 'foc': 195, 'st': 0.0, 'htc': 2, '_id': ObjectId('58e86a550a0aeff4e14ca6bb'), 'ftc': 0}, 
 {'rtc': 0, 'vf': 0, 'mtc': 0, 'doc': 'good job', 'foc': 454, 'st': 0.8, 'htc': 1, '_id': ObjectId('58e8d03958ae6d179c2b4413'), 'ftc': 1},
 {'rtc': 0, 'vf': 2, 'mtc': 1, 'doc': 'test', 'foc': 45, 'st': 0.8, 'htc': 12, '_id': ObjectId('58e8d03958ae6d180c2b4446'), 'ftc': 0}] 

of dictionaries based on condition that if 'doc' key value text is same then one of the dictionary should be removed. I have tried the following solution

distinct_cur = [dict(y) for y in set(tuple(x.items()) for x in cur)] 

but duplicates are still present in the final list.

Below is the desired output as in 1st and 2nd distinct_cur text of key 'doc' value is same (good job):

[{'rtc': 0, 'vf': 0, 'mtc': 0, 'doc': 'good job', 'foc': 195, 'st': 0.0, 'htc': 2, '_id': ObjectId('58e86a550a0aeff4e14ca6bb'), 'ftc': 0}, 
 {'rtc': 0, 'vf': 2, 'mtc': 1, 'doc': 'test', 'foc': 45, 'st': 0.8, 'htc': 12, '_id': ObjectId('58e8d03958ae6d180c2b4446'), 'ftc': 0}] 

Thanks in advance!

Answer 1

You're creating a set out of different elements and expect that it will remove the duplicates based on a criterion that only you know.

You have to iterate through your list, and add to the result list only if doc has a different value than the previous ones: for instance like this:

done = set()
result = []
for d in distinct_cur:
    if d['doc'] not in done:
        done.add(d['doc'])  # note it down for further iterations
        result.append(d)

that will keep only the first occurrence(s) of the dictionaries which have the same doc key by registering the known keys in an aux set.

Another possibility is to use a dictionary with the key as the "doc" key of the dictionary, iterating backwards in the list so the first items overwrite the last ones in the list:

result = {i['doc']:i for i in reversed(distinct_cur)}.values()

Answer 2

I see 2 similar solutions that depend on your domain problem: do you want to keep the first instance of a key or the last instance?

Using the last (so as to overwrite the previous matches) is simpler:

d = {r['doc']: r for r in distinct_cur}.values()

Answer 3

一个用于对doc的 primary_key 上的字典distinct_cur列表进行重复数据删除的衬垫

[i for n, i in enumerate(distinct_cur) if i.get('doc') not in [y.get('doc') for y in distinct_cur[n + 1:]]]

Answer 4

Try this:

distinct_cur  =[dict(t) for t in set([tuple(d.items()) for d in distinct_cur])]

Worked for me...