Im trying to write a function that creates set of dynamic sublists each containing 5 elements from a list passed to it.Here's my attempt at the code
def sublists(seq):
i=0
x=[]
while i<len(seq)-1:
j=0
while j<5:
X.append(seq[i]) # How do I change X after it reaches size 5?
#return set of sublists
EDIT:
Sample input: [1,2,3,4,5,6,7,8,9,10]
Expected output: [[1,2,3,4,5],[6,7,8,9,10]]
Well, for starters, you'll need to (or at least should) have two lists, a temporary one and a permanent one that you return (Also you will need to increase j and i or, more practically, use a for loop, but I assume you just forgot to post that).
EDIT removed first code as the style given doesn't match easily with the expected results, see other two possibilities.
Or, more sensibly:
def sublists(seq):
x=[]
for i in range(0,len(seq),5):
x.append(seq[i:i+5])
return x
Or, more sensibly again, a simple list comprehension:
def sublists(seq):
return [seq[i:i+5] for i in range(0,len(seq),5)]
When given the list:
l = [1,2,3,4,5,6,7,8,9,10]
They will return
[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
By "dynamic sublists", do you mean break up the list into groups of five elements? This is similar to your approach:
def sublists(lst, n):
ret = []
i = 0
while i < len(lst):
ret.append(seq[i:i+n])
i += n
return ret
Or, using iterators:
def sublists(seq, n):
it = iter(seq)
while True:
r = list(itertools.islice(it, 5))
if not r:
break
yield r
which will return an iterator of lists over list of length up to five. (If you took out the list call, weird things would happen if you didn't access the iterators in the same order.)
Have you considered using itertools.combinations(...)?
For example:
>>> from itertools import combinations
>>> l = [1,2,3,4,5,6]
>>> list(combinations(l, 5))
[(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6), (1, 3, 4, 5, 6), (2, 3, 4, 5, 6)]
