Insert element in Python list after every nth element

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.

Try this

i = nwhile 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 = 3while i < len(letters):    letters.insert(i, 'x')    i += 4print letters

which would print out

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

which is your expected result.

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]