CODEWITHSUNDEEP
python
minion
minion

Reputation: 571

how to print this pattern using python iterator

L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
head = 'head'
tail = 'tail'

suppose we can and can only get the iterator of some iterable(L). and we can not know the length of L. Is that possible to print the iterable as:

'head123tail'
'head456tail' 
'head789tail' 
'head10tail'

My try at it is as follows.

L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
head = 'head'
tail = 'tail'
slice_size = 3

i = iter(L)
try:
    while True:
        counter = 0
        while counter < slice_size:
            e = next(i)
            if counter == 0:
                print(head, end='')
            print(e, end='')
            counter += 1
        else:
            print(tail)
except StopIteration:
    if counter > 0:
        print(tail)

Upvotes: 0

Views: 235

Answers (2)

tobias_k
tobias_k

Reputation: 82899

You can split the iterator into chunks of three, using chain and slice from itertools and a for loop, and then join them. The for loop will do most of what your try/while True/except construct is doing.

L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
head = 'head'
tail = 'tail'
slice_size = 3
iterator = iter(L)

from itertools import chain, islice
for first in iterator:
    chunk = chain([first], islice(iterator, slice_size - 1))
    print(head, ''.join(str(x) for x in chunk), tail)

However, if your iterator is just a list, you can just use range with a step parameter:

for start in range(0, len(L), slice_size):
    chunk = L[start : start + slice_size]
    print(head, ''.join(str(x) for x in chunk), tail)

Upvotes: 1

Moses Koledoye
Moses Koledoye

Reputation: 78546

Here's one way to do it with itertools.groupby and itertools.count.

groupby on the key function lambda _: next(c)//3 groups the items in the iterables in threes in successions. The logic uses the integer division of the next object in the count item on 3:

from itertools import groupby, count

L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
head = 'head'
tail = 'tail'

c = count()    
for _, g in groupby(L, lambda _: next(c)//3):
    item = head + ''.join(map(str, g)) + tail
    print(item)

Output:

head123tail
head456tail
head789tail
head10tail

Upvotes: 2

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