CODEWITHSUNDEEP
python
user308827
user308827

Reputation: 21971

Itertools partial lists python

I have the foll. list in python:

[1, 2, 3, 4]

Is there an python itertools function that results in foll:

[1]
[1, 2]
[1, 2, 3]
[1, 2, 3, 4]

Upvotes: 1

Views: 498

Answers (6)

Jose Ricardo Bustos M.
Jose Ricardo Bustos M.

Reputation: 8164

One line solution, using partial and islice,

from itertools import islice
from functools import partial
my_list = [1, 2, 3, 4]

[list(l) for l in map(partial(islice, my_list), range(1,len(my_list)+1))]

you get,

[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4]]

in other words,

from itertools import islice
from functools import partial
my_list = [1, 2, 3, 4]
p = partial(islice, my_list)
for i in range(1,5):
    print(list(p(i)))

you get,

[1]
[1, 2]
[1, 2, 3]
[1, 2, 3, 4]

Upvotes: 1

Moinuddin Quadri
Moinuddin Quadri

Reputation: 48077

If it is must to use itertools, you may use itertools.islice as:

from itertools import islice
my_list = [1, 2, 3, 4]

for i in range(1, len(my_list)+1):
    print list(islice(my_list, i))

However there is absolutely no need to use itertools here. You may achieve this via simple list slicing as:

for i in range(len(my_list)):
    print my_list[:i+1]

Both of the above solutions will print the result as:

[1]
[1, 2]
[1, 2, 3]
[1, 2, 3, 4]

Upvotes: 2

georg
georg

Reputation: 214959

This is trivial without itertools:

def fall(it):
    ls = []
    for x in it:
        ls.append(x)
        yield ls

for x in fall(xrange(20)):
    print x

Note that this works with any iterable, not just a list.

If you still want itertools, something like this should work (py3):

for x in itertools.accumulate(map(lambda x: [x], it)):
     print(x)

Again, it's lazy and works with any iterable.

Upvotes: 6

Maurice Meyer
Maurice Meyer

Reputation: 18106

This can be written as one-liner using list comprehension:

>>> [ list[:x+1] for x in range(len(list)) ]
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4]]

Upvotes: 2

rtkaleta
rtkaleta

Reputation: 691

You don't need itertools, just use a map:

>>> l = [1, 2, 3, 4]
>>> for sub_list in map(lambda index: l[:index + 1], range(len(l))):
...     print sub_list

Upvotes: 0

inspectorG4dget
inspectorG4dget

Reputation: 113975

There isn't anything in itertools, that I can think of, but this should work:

def incremental(L):
    for i in range(1, len(L)+1):
        yield L[:i]

Output:

In [53]: print(*incremental([1, 2, 3, 4]), sep='\n')
[1]
[1, 2]
[1, 2, 3]
[1, 2, 3, 4]

Upvotes: 3

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