remove duplicates from nested dictionaries in list

Question

Hello,

quick and very basic newbie question.

If i have list of dictionaries looking like this:

L = []
L.append({"value1": value1, "value2": value2, "value3": value3, "value4": value4})

Let's say there exists multiple entries where value3 and value4 are identical to other nested dictionaries. How can i quick and easy find and remove those duplicate dictionaries.

Preserving order is of no importance.

Thanks.

EDIT:

If there are five inputs, like this:

L = [{"value1": fssd, "value2": dsfds, "value3": abcd, "value4": gk},
    {"value1": asdasd, "value2": asdas, "value3": dafdd, "value4": sdfsdf},
    {"value1": sdfsf, "value2": sdfsdf, "value3": abcd, "value4": gk},
    {"value1": asddas, "value2": asdsa, "value3": abcd, "value4": gk},
    {"value1": asdasd, "value2": dskksks, "value3": ldlsld, "value4": sdlsld}]

The output shoud look like this:

L = [{"value1": fssd, "value2": dsfds, "value3": abcd, "value4": gk},
    {"value1": asdasd, "value2": asdas, "value3": dafdd, "value4": sdfsdf},
    {"value1": asdasd, "value2": dskksks, "value3": ldlsld, "value4": sdlsld}

Answer 1

for dic in list: 
  for anotherdic in list:
    if dic != anotherdic:
      if dic["value3"] == anotherdic["value3"] or dic["value4"] == anotherdic["value4"]:
        list.remove(anotherdic)

Tested with

list = [{"value1": 'fssd', "value2": 'dsfds', "value3": 'abcd', "value4": 'gk'},
{"value1": 'asdasd', "value2": 'asdas', "value3": 'dafdd', "value4": 'sdfsdf'},
{"value1": 'sdfsf', "value2": 'sdfsdf', "value3": 'abcd', "value4": 'gk'},
{"value1": 'asddas', "value2": 'asdsa', "value3": 'abcd', "value4": 'gk'},
{"value1": 'asdasd', "value2": 'dskksks', "value3": 'ldlsld', "value4": 'sdlsld'}]

worked fine for me :)

Answer 2

That's a list of one dictionary and but, assuming there are more dictionaries in the list l:

l = [ldict for ldict in l if ldict.get("value3") != value3 or ldict.get("value4") != value4]

But is that what you really want to do? Perhaps you need to refine your description.

BTW, don't use list as a name since it is the name of a Python built-in.

EDIT: Assuming you started with a list of dictionaries, rather than a list of lists of 1 dictionary each that should work with your example. It wouldn't work if either of the values were None, so better something like:

l = [ldict for ldict in l if not ( ("value3" in ldict and ldict["value3"] == value3) and ("value4" in ldict and ldict["value4"] == value4) )]

But it still seems like an unusual data structure.

EDIT: no need to use explicit gets.

Also, there are always tradeoffs in solutions. Without more info and without actually measuring, it's hard to know which performance tradeoffs are most important for the problem. But, as the Zen sez: "Simple is better than complex".

Answer 3

You can use a temporary array to store an items dict. The previous code was bugged for removing items in the for loop.

(v,r) = ([],[])
for i in l:
    if ('value4', i['value4']) not in v and ('value3', i['value3']) not in v:
        r.append(i)
    v.extend(i.items())
l = r

Your test:

l = [{"value1": 'fssd', "value2": 'dsfds', "value3": 'abcd', "value4": 'gk'},
    {"value1": 'asdasd', "value2": 'asdas', "value3": 'dafdd', "value4": 'sdfsdf'},
    {"value1": 'sdfsf', "value2": 'sdfsdf', "value3": 'abcd', "value4": 'gk'},
    {"value1": 'asddas', "value2": 'asdsa', "value3": 'abcd', "value4": 'gk'},
    {"value1": 'asdasd', "value2": 'dskksks', "value3": 'ldlsld', "value4": 'sdlsld'}]

ouputs

{'value4': 'gk', 'value3': 'abcd', 'value2': 'dsfds', 'value1': 'fssd'}
{'value4': 'sdfsdf', 'value3': 'dafdd', 'value2': 'asdas', 'value1': 'asdasd'}
{'value4': 'sdlsld', 'value3': 'ldlsld', 'value2': 'dskksks', 'value1': 'asdasd'}

Answer 4

Here's one way:

keyfunc = lambda d: (d['value3'], d['value4'])

from itertools import groupby
giter = groupby(sorted(L, key=keyfunc), keyfunc)

L2 = [g[1].next() for g in giter]
print L2

Answer 5

If I understand correctly, you want to discard matches that come later in the original list but do not care about the order of the resulting list, so:

(Tested with 2.5.2)

tempDict = {}
for d in L[::-1]:
    tempDict[(d["value3"],d["value4"])] = d
L[:] = tempDict.itervalues()
tempDict = None

Answer 6

In Python 2.6 or 3.*:

import itertools
import pprint

L = [{"value1": "fssd", "value2": "dsfds", "value3": "abcd", "value4": "gk"},
    {"value1": "asdasd", "value2": "asdas", "value3": "dafdd", "value4": "sdfsdf"},
    {"value1": "sdfsf", "value2": "sdfsdf", "value3": "abcd", "value4": "gk"},
    {"value1": "asddas", "value2": "asdsa", "value3": "abcd", "value4": "gk"},
    {"value1": "asdasd", "value2": "dskksks", "value3": "ldlsld", "value4": "sdlsld"}]

getvals = operator.itemgetter('value3', 'value4')

L.sort(key=getvals)

result = []
for k, g in itertools.groupby(L, getvals):
    result.append(g.next())

L[:] = result
pprint.pprint(L)

Almost the same in Python 2.5, except you have to use g.next() instead of next(g) in the append.