assigning every Nth element in a list in python

Question

I want to set every Nth element in a list to something else.

(Like this question which is for Matlab.)

Here's an attempt with N=2:

>>> x=['#%d' % i for i in range(10)]
>>> x
['#0', '#1', '#2', '#3', '#4', '#5', '#6', '#7', '#8', '#9']
>>> x[0::2] = 'Hey!'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: attempt to assign sequence of size 4 to extended slice of size 5

How can I fix this? Slicing seems to expect an iterable, not just a single value.

This isn't an array, it's a list. That matters, because there is an array type in the standard library, as well as a different one in the common library numpy, and with a numpy array, x[::2] = 'Hey!' would actually have worked.. — DSM
– DSM, Commented Apr 9, 2013 at 18:46

ecatmur · Accepted Answer · 2013-04-09 18:31:31Z

5

The right side of a slice assignment has to be a sequence of an appropriate length; Python is treating 'Hey!' as a sequence of the characters 'H', 'e', 'y', '!'.

You can be clever and create a sequence of the appropriate length:

x[::2] = ['Hey!'] * len(x[::2])

However, the clearest way to do this is a for loop:

for i in range(0, len(x), 2):
    x[i] = 'Hey!'

answered Apr 9, 2013 at 18:31

ecatmur

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2 Comments

ecatmur Over a year ago

@JasonS there is a one-liner (added above) but it's significantly less readable.

cwallenpoole Over a year ago

I think the first solution is quite Pythonic, personally.

Finn · Accepted Answer · 2013-04-09 18:42:26Z

1

It is also possible with list comprehension:

x=['#%d' % i for i in range(10)]
['Hey!' if i%3 == 0 else b for  i,b in enumerate(x)]

This is appearently with n=3.

edited Apr 9, 2013 at 18:42

answered Apr 9, 2013 at 18:33

Finn

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