CODEWITHSUNDEEP
pythonpython-3.x
user92654
user92654

Reputation: 39

Python: reconstruct a nested list

I have a nested list:

  output= [('the', 'B', 'NNP'), ('wall', 'I', 'NNP'), ('street', 'I', 'NNP'), ('journal', 'I', 'NNP'), ('reported', 'O', 'VB'), ('today', 'O', 'NNP'), ('that', 'O', 'NNP'), ('apple', 'B', 'NNP'), ('corporation', 'I', 'NNP'), ('made', 'O', 'VB'), ('money', 'O', 'NNP'), ('.', 'O', '.'), ('georgia', 'B', 'NNP'), ('tech', 'I', 'NNP'), ('is', 'O', 'NNP'), ('a', 'O', '.'), ('university', 'O', 'NNP'), ('in', 'O', 'NNP'), ('georgia', 'B', 'NNP'),('.', 'O', '.')]

I want to re-format it to the following expected format:

new_output= [(['the', 'wall', 'street', 'journal', 'reported', 'today', 'that', 'apple', 'corporation', 'made', 'money'], ['B', 'I', 'I', 'I', 'O', 'O', 'O', 'B', 'I', 'O', 'O']), (['georgia', 'tech', 'is', 'a', 'university', 'in', 'georgia'], ['B', 'I', 'O', 'O', 'O', 'O', 'B'])]

My attempt is:

import string
word = []
token = []
result_word = []
result_token = []

result = []
for i in output[0]:
    for every_word in i:
        word.append(every_word)
result_word = " ".join(" ".join(word).split()[::3])

How can I get my expected format?

Upvotes: 0

Views: 91

Answers (3)

ingvar
ingvar

Reputation: 4375

output = [('the', 'B', 'NNP'), ('wall', 'I', 'NNP'), ('street', 'I', 'NNP'), ('journal', 'I', 'NNP'), ('reported', 'O', 'VB'), ('today', 'O', 'NNP'), ('that', 'O', 'NNP'), ('apple', 'B', 'NNP'), ('corporation', 'I', 'NNP'), ('made', 'O', 'VB'), ('money', 'O', 'NNP'), ('.', 'O', '.'), ('georgia', 'B', 'NNP'), ('tech', 'I', 'NNP'), ('is', 'O', 'NNP'), ('a', 'O', '.'), ('university', 'O', 'NNP'), ('in', 'O', 'NNP'), ('georgia', 'B', 'NNP'),('.', 'O', '.')]
result, words, tokens = [], [], []
for word, token, _ in output:  # this is tuple like ('the', 'B', 'NNP')
    if word == '.':   # end of sentence, save current and start new one
        result.append((words, tokens))
        words, tokens = [], []
    else:   # add new word to current sentence
        words.append(word)
        tokens.append(token)

print(result)

Output:

[(['the', 'wall', 'street', 'journal', 'reported', 'today', 'that', 'apple', 'corporation', 'made', 'money'], ['B', 'I', 'I', 'I', 'O', 'O', 'O', 'B', 'I', 'O', 'O']), (['georgia', 'tech', 'is', 'a', 'university', 'in', 'georgia'], ['B', 'I', 'O', 'O', 'O', 'O', 'B'])]

Upvotes: 1

Dani Mesejo
Dani Mesejo

Reputation: 61910

You could do something like this:

from itertools import groupby
from operator import itemgetter

output = [('the', 'B', 'NNP'), ('wall', 'I', 'NNP'), ('street', 'I', 'NNP'), ('journal', 'I', 'NNP'),
          ('reported', 'O', 'VB'), ('today', 'O', 'NNP'), ('that', 'O', 'NNP'), ('apple', 'B', 'NNP'),
          ('corporation', 'I', 'NNP'), ('made', 'O', 'VB'), ('money', 'O', 'NNP'), ('.', 'O', '.'),
          ('georgia', 'B', 'NNP'), ('tech', 'I', 'NNP'), ('is', 'O', 'NNP'), ('a', 'O', '.'),
          ('university', 'O', 'NNP'), ('in', 'O', 'NNP'), ('georgia', 'B', 'NNP'), ('.', 'O', '.')]

sentences = [list(group) for k, group in groupby(output, lambda x: x[0] == ".") if not k]
result = [tuple(map(list, zip(*map(itemgetter(0, 1), sentence)))) for sentence in sentences]
print(result)

Output

[(['the', 'wall', 'street', 'journal', 'reported', 'today', 'that', 'apple', 'corporation', 'made', 'money'], ['B', 'I', 'I', 'I', 'O', 'O', 'O', 'B', 'I', 'O', 'O']), (['georgia', 'tech', 'is', 'a', 'university', 'in', 'georgia'], ['B', 'I', 'O', 'O', 'O', 'O', 'B'])]

Explanation

As far as I understood you want to unpack the first and the last element of each of the sentences.

The line:

sentences = [list(group) for k, group in groupby(output, lambda x: x[0] == ".") if not k]

splits output into sentences by each ., the second line just unpacks each sentence:

result = [tuple(map(list, zip(*map(itemgetter(0, 1), sentence)))) for sentence in sentences]

As you want to a list of tuple of lists and zip returns a list of tuples you have to map each tuple with list and then convert the result of map to a tuple.

Upvotes: 2

Patrick Haugh
Patrick Haugh

Reputation: 61014

You can use groupby to group the non-period items into sentences, then use zip to split the words from the parts of speech indicators:

from itertools import groupby

l =   output= [('the', 'B', 'NNP'), ('wall', 'I', 'NNP'), ('street', 'I', 'NNP'), ('journal', 'I', 'NNP'), ('reported', 'O', 'VB'), ('today', 'O', 'NNP'), ('that', 'O', 'NNP'), ('apple', 'B', 'NNP'), ('corporation', 'I', 'NNP'), ('made', 'O', 'VB'), ('money', 'O', 'NNP'), ('.', 'O', '.'), ('georgia', 'B', 'NNP'), ('tech', 'I', 'NNP'), ('is', 'O', 'NNP'), ('a', 'O', '.'), ('university', 'O', 'NNP'), ('in', 'O', 'NNP'), ('georgia', 'B', 'NNP'),('.', 'O', '.')]


groups = (g for k, g in groupby(l, lambda x: x[0] != '.') if k)
zs = (zip(*g) for g in groups)
res = [(next(z), next(z)) for z in zs]

res is then

[(('the', 'wall', 'street', 'journal', 'reported', 'today', 'that', 'apple', 'corporation', 'made', 'money'), 
  ('B', 'I', 'I', 'I', 'O', 'O', 'O', 'B', 'I', 'O', 'O')), 
 (('georgia', 'tech', 'is', 'a', 'university', 'in', 'georgia'), 
  ('B', 'I', 'O', 'O', 'O', 'O', 'B'))
]

Upvotes: 2

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