CODEWITHSUNDEEP
python
kevin
kevin

Reputation: 2014

Python lists with irregular format

I have the data (the results of LDA using Gensim), which looks like this:

[(1, 0.97456828373415116)]
[(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)]
[(0, 0.93783520386426555), (1, 0.015481826214088806), (2, 0.015545735781026492), (3, 0.015535246185968628), (4, 0.015601987954650424)]
[(2, 0.98493696818505228)]
[(3, 0.99067359305252778)]
[(0, 0.73578249201070511), (3, 0.25197028613750805)]

I would like to convert to the following format:

[(0, 0), (1, 0.97456828373415116), (2, 0), (3, 0), (4, 0)]
[(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)]
[(0, 0.93783520386426555), (1, 0.015481826214088806), (2, 0.015545735781026492), (3, 0.015535246185968628), (4, 0.015601987954650424)]
[(0, 0), (1, 0), (2, 0.98493696818505228), (3, 0), (4, 0)]
[(0, 0), (1, 0), (2, 0), (3, 0.96747728928637211), (4, 0)]
[(0, 0), (1, 0), (2, 0), (3, 0.99067359305252778), (4, 0)]
[(0, 0.73578249201070511), (1, 0), (2, 0), (3, 0.25197028613750805), (4, 0)]

Upvotes: 1

Views: 57

Answers (2)

zxq9
zxq9

Reputation: 13164

One very simple way to do this is to use a constructed dict with your defaults, and then update it:

>>> d = dict([(0,0),(1,0),(2,0),(3,0)])
>>> print(d)
{0: 0, 1: 0, 2: 0, 3: 0}
>>> d.update([(0, 0.73578249201070511), (3, 0.25197028613750805)])
>>> print(d)
{0: 0.7357824920107051, 1: 0, 2: 0, 3: 0.25197028613750805}

Edit

Incorporating hgwell's suggestion to output a list of tuples, here is a complete function (which could probably be done better somehow, but this works anyway):

def listify(l):
    res = []
    for j in l:
        d = dict([(0,0),(1,0),(2,0),(3,0),(4,0)])
        d.update(j)
        res.append(list(d.items()))
    return res

and in action...

>>> z = listify([[(1, 0.97456828373415116)],
                 [(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)],
                 [(2, 0.98493696818505228)]])
>>> pprint(z)
[[(0, 0), (1, 0.9745682837341512), (2, 0), (3, 0), (4, 0)],
 [(0, 0.9188312525648973),
  (1, 0.020225186991467976),
  (2, 0.020314851937259213),
  (3, 0.0203822948891845),
  (4, 0.020246413617191008)],
 [(0, 0), (1, 0), (2, 0.9849369681850523), (3, 0), (4, 0)]]

Upvotes: 1

TigerhawkT3
TigerhawkT3

Reputation: 49330

You can change each sublist into a dict with the map() function:

data = [[(1, 0.97456828373415116)],
[(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)],
[(0, 0.93783520386426555), (1, 0.015481826214088806), (2, 0.015545735781026492), (3, 0.015535246185968628), (4, 0.015601987954650424)],
[(2, 0.98493696818505228)],
[(3, 0.99067359305252778)],
[(0, 0.73578249201070511), (3, 0.25197028613750805)]]

results = list(map(dict, data))

Then use the dict.get method and specify a default of 0 for keys that are not present in the dictionary:

for i in range(5):
    print(results[0].get(i, 0))

Result of the above:

0
0.9745682837341512
0
0
0

Upvotes: 1

Related Questions