stackit stackit - 1 year ago 218
Python Question

How to compute skipgrams in python?

A k skipgram is an ngram which is a superset of all ngrams and each (k-i )skipgram till (k-i)==0 (which includes 0 skip grams). So how to efficiently compute these skipgrams in python?

Following is the code i tried but it is not doing as expected:

input_list = ['all', 'this', 'happened', 'more', 'or', 'less']
def find_skipgrams(input_list, N,K):
bigram_list = []

for k in range(K+1):
for i in range(len(input_list)-1):
if i+k+1<len(input_list):
for j in range(N+1):
if i+k+j+1<len(input_list):

return bigram_list


The above code is not rendering correctly, but print
find_skipgrams(['all', 'this', 'happened', 'more', 'or', 'less'],2,1)
gives following output

[['this', 'happened', 'more'], ['happened', 'more', 'or'], ['more',
'or', 'less'], ['or', 'less'], ['less'], ['happened', 'more', 'or'],
['more', 'or', 'less'], ['or', 'less'], ['less'], ['less']]

The code listed here also does not give correct output:

print skipgram_ndarray("What is your name") gives:
['What,is', 'is,your', 'your,name', 'name,', 'What,your', 'is,name']

name is a unigram!

Answer Source

From the paper that OP links, the following string:

Insurgents killed in ongoing fighting


2-skip-bi-grams = {insurgents killed, insurgents in, insurgents ongoing, killed in, killed ongoing, killed fighting, in ongoing, in fighting, ongoing fighting}

2-skip-tri-grams = {insurgents killed in, insurgents killed ongoing, insurgents killed fighting, insurgents in ongoing, insurgents in fighting, insurgents ongoing fighting, killed in ongoing, killed in fighting, killed ongoing fighting, in ongoing fighting}.

With slight modification to NLTK's ngrams code (

from itertools import chain, combinations
import copy
from nltk.util import ngrams

def pad_sequence(sequence, n, pad_left=False, pad_right=False, pad_symbol=None):
    if pad_left:
        sequence = chain((pad_symbol,) * (n-1), sequence)
    if pad_right:
        sequence = chain(sequence, (pad_symbol,) * (n-1))
    return sequence

def skipgrams(sequence, n, k, pad_left=False, pad_right=False, pad_symbol=None):
    sequence_length = len(sequence)
    sequence = iter(sequence)
    sequence = pad_sequence(sequence, n, pad_left, pad_right, pad_symbol)

    if sequence_length + pad_left + pad_right < k:
        raise Exception("The length of sentence + padding(s) < skip")

    if n < k:
        raise Exception("Degree of Ngrams (n) needs to be bigger than skip (k)")    

    history = []
    nk = n+k

    # Return point for recursion.
    if nk < 1: 
    # If n+k longer than sequence, reduce k by 1 and recur
    elif nk > sequence_length: 
        for ng in skipgrams(list(sequence), n, k-1):
            yield ng

    while nk > 1: # Collects the first instance of n+k length history
        nk -= 1

    # Iterative drop first item in history and picks up the next
    # while yielding skipgrams for each iteration.
    for item in sequence:
        current_token = history.pop(0)      
        # Iterates through the rest of the history and 
        # pick out all combinations the n-1grams
        for idx in list(combinations(range(len(history)), n-1)):
            ng = [current_token]
            for _id in idx:
            yield tuple(ng)

    # Recursively yield the skigrams for the rest of seqeunce where
    # len(sequence) < n+k
    for ng in list(skipgrams(history, n, k-1)):
        yield ng

Let's do some doctest to match the example in the paper:

>>> two_skip_bigrams = list(skipgrams(text, n=2, k=2))
[('Insurgents', 'killed'), ('Insurgents', 'in'), ('Insurgents', 'ongoing'), ('killed', 'in'), ('killed', 'ongoing'), ('killed', 'fighting'), ('in', 'ongoing'), ('in', 'fighting'), ('ongoing', 'fighting')]
>>> two_skip_trigrams = list(skipgrams(text, n=3, k=2))
[('Insurgents', 'killed', 'in'), ('Insurgents', 'killed', 'ongoing'), ('Insurgents', 'killed', 'fighting'), ('Insurgents', 'in', 'ongoing'), ('Insurgents', 'in', 'fighting'), ('Insurgents', 'ongoing', 'fighting'), ('killed', 'in', 'ongoing'), ('killed', 'in', 'fighting'), ('killed', 'ongoing', 'fighting'), ('in', 'ongoing', 'fighting')]

But do note that if n+k > len(sequence), it will yield the same effects as skipgrams(sequence, n, k-1) (this is not a bug, it's a fail safe feature), e.g.

>>> three_skip_trigrams = list(skipgrams(text, n=3, k=3))
>>> three_skip_fourgrams = list(skipgrams(text, n=4, k=3))
>>> four_skip_fourgrams  = list(skipgrams(text, n=4, k=4))
>>> four_skip_fivegrams  = list(skipgrams(text, n=5, k=4))
>>> print len(three_skip_trigrams), three_skip_trigrams
10 [('Insurgents', 'killed', 'in'), ('Insurgents', 'killed', 'ongoing'), ('Insurgents', 'killed', 'fighting'), ('Insurgents', 'in', 'ongoing'), ('Insurgents', 'in', 'fighting'), ('Insurgents', 'ongoing', 'fighting'), ('killed', 'in', 'ongoing'), ('killed', 'in', 'fighting'), ('killed', 'ongoing', 'fighting'), ('in', 'ongoing', 'fighting')]
>>> print len(three_skip_fourgrams), three_skip_fourgrams 
5 [('Insurgents', 'killed', 'in', 'ongoing'), ('Insurgents', 'killed', 'in', 'fighting'), ('Insurgents', 'killed', 'ongoing', 'fighting'), ('Insurgents', 'in', 'ongoing', 'fighting'), ('killed', 'in', 'ongoing', 'fighting')]
>>> print len(four_skip_fourgrams), four_skip_fourgrams 
5 [('Insurgents', 'killed', 'in', 'ongoing'), ('Insurgents', 'killed', 'in', 'fighting'), ('Insurgents', 'killed', 'ongoing', 'fighting'), ('Insurgents', 'in', 'ongoing', 'fighting'), ('killed', 'in', 'ongoing', 'fighting')]
>>> print len(four_skip_fivegrams), four_skip_fivegrams 
1 [('Insurgents', 'killed', 'in', 'ongoing', 'fighting')]

This allows n == k but it disallow n > k as shown in the lines :

if n < k:
        raise Exception("Degree of Ngrams (n) needs to be bigger than skip (k)")    

For understanding sake, let's try to understand the "mystical" line:

for idx in list(combinations(range(len(history)), n-1)):
    pass # Do something

Given a list of unique items, combinations produce this:

>>> from itertools import combinations
>>> x = [0,1,2,3,4,5]
>>> list(combinations(x,2))
[(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]

And since the indices of a list of tokens is always unique, e.g.

>>> sent = ['this', 'is', 'a', 'foo', 'bar']
>>> current_token = sent.pop(0) # i.e. 'this'
>>> range(len(sent))

It's possible to compute the possible combinations (without replacement) of the range:

>>> n = 3
>>> list(combinations(range(len(sent)), n-1))
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]

If we map the indices back to the list of tokens:

>>> [tuple(sent[id] for id in idx) for idx in combinations(range(len(sent)), 2)
[('is', 'a'), ('is', 'foo'), ('is', 'bar'), ('a', 'foo'), ('a', 'bar'), ('foo', 'bar')]

Then we concatenate with the current_token, we get the skipgrams for the current token and context+skip window:

>>> [tuple([current_token]) + tuple(sent[id] for id in idx) for idx in combinations(range(len(sent)), 2)]
[('this', 'is', 'a'), ('this', 'is', 'foo'), ('this', 'is', 'bar'), ('this', 'a', 'foo'), ('this', 'a', 'bar'), ('this', 'foo', 'bar')]

So after that we move on to the next word.

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