Creating sublists [duplicate]

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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

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