Insert element in Python list after every nth element

Question

Say I have a Python list like this:

letters = ['a','b','c','d','e','f','g','h','i','j']

I want to insert an 'x' after every nth element, let's say three characters in that list. The result should be:

letters = ['a','b','c','x','d','e','f','x','g','h','i','x','j']

I understand that I can do that with looping and inserting. What I'm actually looking for is a Pythonish-way, a one-liner maybe?

Solution 1

I've got two one liners.

Given:

>>> letters = ['a','b','c','d','e','f','g','h','i','j']

1. Use enumerate to get index, add 'x' every 3^rd letter, eg: mod(n, 3) == 2, then concatenate into string and list() it.

   >>> list(''.join(l + 'x' * (n % 3 == 2) for n, l in enumerate(letters)))
   
   ['a', 'b', 'c', 'x', 'd', 'e', 'f', 'x', 'g', 'h', 'i', 'x', 'j']
   

   But as @sancho.s points out this doesn't work if any of the elements have more than one letter.

2. Use nested comprehensions to flatten a list of lists^(a), sliced in groups of 3 with 'x' added if less than 3 from end of list.

   >>> [x for y in (letters[i:i+3] + ['x'] * (i < len(letters) - 2) for
        i in xrange(0, len(letters), 3)) for x in y]
   
   ['a', 'b', 'c', 'x', 'd', 'e', 'f', 'x', 'g', 'h', 'i', 'x', 'j']
   

(a) [item for subgroup in groups for item in subgroup] flattens a jagged list of lists.

Solution 2

Try this

i = n
while i < len(letters):
    letters.insert(i, 'x')
    i += (n+1)

where n is after how many elements you want to insert 'x'.

This works by initializing a variable i and setting it equal to n. You then set up a while loop that runs while i is less then the length of letters. You then insert 'x' at the index i in letters. Then you must add the value of n+1 to i. The reason you must do n+1 instead of just n is because when you insert an element to letters, it expands the length of the list by one.

Trying this with your example where n is 3 and you want to insert 'x', it would look like this

letters = ['a','b','c','d','e','f','g','h','i','j']
i = 3
while i < len(letters):
    letters.insert(i, 'x')
    i += 4

print letters

which would print out

['a', 'b', 'c', 'x', 'd', 'e', 'f', 'x', 'g', 'h', 'i', 'x', 'j']

which is your expected result.

Solution 3

Although using list.insert() in a for loop seems to be more memory efficient, in order to do it in one-line, you can also append the given value at the end of every equally divided chunks split on every nth index of the list.

>>> from itertools import chain

>>> n = 2
>>> ele = 'x'
>>> lst = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

>>> list(chain(*[lst[i:i+n] + [ele] if len(lst[i:i+n]) == n else lst[i:i+n] for i in xrange(0, len(lst), n)]))
[0, 1, 'x', 2, 3, 'x', 4, 5, 'x', 6, 7, 'x', 8, 9, 'x', 10]

Solution 4

I want to add a new element per item.

How about this ?

a=[2,4,6]
for b in range (0,len(a)):
    a.insert(b*2,1)

a is now

[1, 2, 1, 4, 1, 6]

Solution 5

A pretty straightforward method:

>>> letters = ['a','b','c','d','e','f','g','h','i','j']
>>> new_list = []
>>> n = 3
>>> for start_index in range(0, len(letters), n):
...     new_list.extend(letters[start_index:start_index+n])
...     new_list.append('x')
... 
>>> new_list.pop()
'x'
>>> new_list
['a', 'b', 'c', 'x', 'd', 'e', 'f', 'x', 'g', 'h', 'i', 'x', 'j']

You can also use the grouper recipe from the itertools documentation for the chunking.

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