getting the last and first few elements in a list in python

Question

I'm trying to divide a list, according to user input. for example, I can get a list of

(["A","B","C","D"],num_of_groups)
>>>["AB","BC","CD","DA"]

my problem is with the last element. because like in the example, I need to take the last and first elements in a list (could be also last two and first three .. )

I tried doing so by adding all the element twice and then slicing, but then I have a problem that when the user wants a group of all elements, I'm returning two groups of the elements instead of only the one. I'll explain:

this is what I'm doing

list_copy = list(char_list)*2
#get the slice in the size of n.
slices.append(list_copy[i:i+n])

this is what happaning when user eners a size same as the list size

(["A","B","C"],3)

>>>["ABC","ACB","BAC","BCA","CAB","CBA","ABC","ACB","BAC","BCA","CAB","CBA"]
*instead of ["ABC","ACB","BAC","BCA","CAB","CBA"]

is there any other way of doing so? I would love some help! thanks!

Answer 1

Use the modulus operator % to start at the beginning when necessary:

>>> l = ['a', 'b', 'c', 'd']
>>> [l[i%len(l)]+l[(i+1)%len(l)] for i in range(len(l))]
['ab', 'bc', 'cd', 'da']
>>> [l[i%len(l)]+l[(i+1)%len(l)] for i in range(2*len(l))]
['ab', 'bc', 'cd', 'da', 'ab', 'bc', 'cd', 'da']

Answer 2

using the wonderful power of zip :

>>> l = ["A","B","C","D"]        
>>> for i in zip(l,l[1:]+l[:1]): print(''.join(i))


AB
BC
CD
DA

This may be confusing but what is essentially happening is iterating through 2 lists, your original one and a copy of your original where the each element is moved over one. You could also use this for a list comprehension:

>>> [''.join(i) for i in zip(l,l[1:]+l[:1])]
['AB', 'BC', 'CD', 'DA']

Answer 3

It's a little ambiguous as to what you're trying to achieve. In your first example, you wrap around to get DA , however in your last example you only want ['ABC'] and not perform the same wrapping to get ['ABC', 'BCA', 'CAB'] ?

If you're looking to get the wrapping effect, then here's something you can do:

from itertools import islice, cycle

def group(l, size):
  return [ ''.join(islice(cycle(l), i, i + size)) for i in xrange(len(l)) ]

Using itertools.cycle is a little more convenient than doubling the list. Since you're working off an iterator instead of a list, you have to pair it up with itertools.islice to do the slicing.

Here's the example output:

>>> group(['A', 'B', 'C', 'D'], 2)
['AB', 'BC', 'CD', 'DA']

>>> group(['A', 'B', 'C'], 1)
['A', 'B', 'C']

>>> group(['A', 'B', 'C'], 2)
['AB', 'BC', 'CA']

>>> group(['A', 'B', 'C'], 3)
['ABC', 'BCA', 'CAB']

>>> group(['A', 'B', 'C'], 4)
['ABCA', 'BCAB', 'CABC']