Remove elements from list which are just before some specific element

Question

Let's say I am having a list as:

a = ['no', 'no', 'no', 'yes', 'no', 'yes', 'no']

Here I want to remove every 'no' which is preceded by every 'yes'. So my resultant list should be like:

['no', 'no', 'yes', 'yes', 'no']

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I've found that in order to remove an element from a list by its value, we may use list.remove(..) as:

a = ['no', 'no', 'no', 'yes', 'no', 'yes', 'no']
a.remove('no')
print a

But it gives me result with only removing first occurrence of 'no' as:

['no', 'no', 'yes', 'no', 'yes', 'no']

How can I achieve the desired result by removing all the occurrence of 'no' which are preceded by all 'yes' in my list?

Answer 1

For removing all the occurrence of 'no' which are present just before the 'yes' in your list, you may use list comprehension with itertools.zip_longest(...) in Python 3.x (which is equivalent of iterools.izip_longest(..) in Python 2.x) (having default fillvalue as None) to achieve this as :

>>> a = ['no', 'no', 'no', 'yes', 'no', 'yes', 'no']

# Python 3.x solution
>>> from itertools import zip_longest
>>> [x for x, y in zip_longest(a, a[1:]) if not(x=='no' and y=='yes')]
['no', 'no', 'yes', 'yes', 'no']

# Python 2.x solution
>>> from itertools import izip_longest
>>> [x for x, y in izip_longest(a, a[1:]) if not(x=='no' and y=='yes')]
['no', 'no', 'yes', 'yes', 'no']

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You might be interested in taking a look at the zip_longest document which says:

Make an iterator that aggregates elements from each of the iterables. If the iterables are of uneven length, missing values are filled-in with fillvalue. Iteration continues until the longest iterable is exhausted.

Answer 2

Try this:

a = ['no', 'no', 'no', 'yes', 'no', 'yes', 'no']
a = ' '.join(a)  
print(a.replace('no yes', 'yes').split(' '))

What it is doing is: 1. merging the list into a string with ' '.join() 2. replacing all the occurrencies of 'no yes' with 'yes' by a.replace() 3. splitting it back into a list with a.split(' ')

Answer 3

Iterate with the condition and append last item:

[i for i, j in zip(a, a[1:]) if (i == 'yes' or j == 'no')] + a[-1:]

Answer 4

An interestingly roundabout way, using regex with a look-ahead:

>>> import re
>>> s = ' '.join(a)                          # convert it into string
>>> out = re.sub('no (?=yes)', '', s)        # remove
>>> out.split()                              # get back the list
=> ['no', 'no', 'yes', 'yes', 'no']