python - Creating sublists -

the opposite of list flattening.

given list , length n return list of sub lists of length n.

def sublist(lst, n):     sub=[] ; result=[]     in lst:         sub+=[i]         if len(sub)==n: result+=[sub] ; sub=[]     if sub: result+=[sub]     return result 

an example:

if list is:

[1,2,3,4,5,6,7,8] 

and n is:

3 

return:

[[1, 2, 3], [4, 5, 6], [7, 8]] 

is there more eloquent / concise way?

an aside, preferred when appending lists lists (in context above):

list1+=[list2] 

or:

list1.append(list2) 

given (according summerfeild's 'programming in python 3') same?

thanks.

such list of lists constructed using list comprehension:

in [17]: seq=[1,2,3,4,5,6,7,8] in [18]: [seq[i:i+3] in range(0,len(seq),3)] out[18]: [[1, 2, 3], [4, 5, 6], [7, 8]] 

there grouper idiom:

in [19]: import itertools in [20]: list(itertools.izip_longest(*[iter(seq)]*3)) out[20]: [(1, 2, 3), (4, 5, 6), (7, 8, none)] 

but note missing elements filled value none. izip_longest can take fillvalue parameter if other none desired.

---

list1+=[list2] -- noting brackets time -- equivalent list1.append(list2). highest priority when writing code readability, not speed. reason, go list1.append(list2). readability subjective, however, , influenced idioms you're familiar with.

happily, in case, readability , speed seem coincide:

in [41]: %timeit list1=[1,2,3]; list1.append(list2) 1000000 loops, best of 3: 612 ns per loop  in [42]: %timeit list1=[1,2,3]; list1+=[list2] 1000000 loops, best of 3: 847 ns per loop