CODEWITHSUNDEEP
pythonlist
Kevin
Kevin

Reputation: 597

Python group by adjacent items in a list with same attributes

Please see the simplified example:

A = [(721,'a'),(765,'a'),(421,'a'),(422,'a'),(106,'b'),(784,'a'),(201,'a'),(206,'b'),(207,'b')]

I want group adjacent tuples with attribute 'a', every two pair wise and leave tuples with 'b' alone.

So the desired tuple would looks like:

A = [[(721,'a'),(765,'a')],
     [(421,'a'),(422,'a')],
     [(106,'b')],
     [(784,'a'),(201,'a')],
     [(206,'b')],
     [(207,'b')]]

What I can do is to build two separated lists contains tuples with a and b.

Then pair tuples in a, and add back. But it seems not very efficient. Any faster and simple solutions?

Upvotes: 3

Views: 1077

Answers (3)

Nikaido
Nikaido

Reputation: 4629

No need to use two lists. Edit: If the 'a' are not assumed to come always as pairs/adjacent

A = [(721,'a'),(765,'a'),(421,'a'),(422,'a'),(106,'b'),(784,'a'),(201,'a'),(206,'b'),(207,'b')]
new_list = []
i = 0
while i < len(A):
    if i == len(A)-1:
        new_list.append([A[i]])
        i+=1
    elif (A[i][1]==A[i+1][1]=='a') :
        new_list.append([A[i], A[i+1]])
        i += 2
    else:
        new_list.append([A[i]])
        i += 1
print(new_list)

Upvotes: 0

Ajax1234
Ajax1234

Reputation: 71451

You can use itertools.groupby:

import itertools
A=[(721,'a'),(765,'a'),(421,'a'),(422,'a'),(106,'b'),(784,'a'),(201,'a'),(206,'b'),(207,'b')]
def split(s):
  return [s[i:i+2] for i in range(0, len(s), 2)]

new_data = [i if isinstance(i, list) else [i] for i in list(itertools.chain(*[split(list(b)) if a == 'a' else list(b) for a, b in itertools.groupby(A, key=lambda x:x[-1])]))

Output:

[[(721, 'a'), (765, 'a')], [(421, 'a'), (422, 'a')], [(106, 'b')], [(784, 'a'), (201, 'a')], [(206, 'b')], [(207, 'b')]]

Upvotes: 2

match
match

Reputation: 11060

Assuming a items are always in pairs, a simple approach would be as follows.

Look at the first item - if it's an a, use it and the next item as a pair. Otherwise, just use the single item. Then 'jump' forward by 1 or 2, as appropriate:

A=[(721,'a'),(765,'a'),(421,'a'),(422,'a'),(106,'b'),(784,'a'),(201,'a'),(206,'b'),(207,'b')]

result = []
count = 0
while count <= len(A)-1:
    if A[count][1] == 'a':
        result.append([A[count], A[count+1]])
        count += 2
    else:
        result.append([A[count]])
        count += 1

print(result)

Upvotes: 2

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