The opposite of list flattening.
Given a list and a length n return a list of sub lists of length n.
def sublist(lst, n):
sub=[] ; result=[]
for i in lst:
sub+=[i]
if len(sub)==n: result+=[sub] ; sub=[]
if sub: result+=[sub]
return result
An example:
If the 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 a more eloquent / concise way?
An aside, what is preferred when appending lists to lists (in the context above):
list1+=[list2]
Or:
list1.append(list2)
Given that (according to Summerfeild's 'Programming in Python 3') they are the same?
Thanks.
Answer
Such a list of lists could be constructed using a list comprehension:
In [17]: seq=[1,2,3,4,5,6,7,8]
In [18]: [seq[i:i+3] for i in range(0,len(seq),3)]
Out[18]: [[1, 2, 3], [4, 5, 6], [7, 8]]
There is also the 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 that missing elements are filled with the value None. izip_longest can take a fillvalue parameter as well if something other than None is desired.
list1+=[list2] -- noting the brackets this time -- is equivalent to list1.append(list2). My highest priority when writing code is readability,
not speed. For this reason, I would go with list1.append(list2). Readability is subjective, however, and probably is influenced greatly by what idioms you're familiar with.
Happily, in this case, readability and speed seem to 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