Edit the value of every Nth item in a list

Question

What's the most pythonic way of performing an arithmetic operation on every nth value in a list? For example, if I start with list1:

list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

I would like to add 1 to every second item, which would give:

list2 = [1, 3, 3, 5, 5, 7, 7, 9, 9, 11]

I've tried:

list1[::2]+1

and also:

for x in list1:
    x=2        
    list2 = list1[::x] + 1

Answer 1

You could use slicing with a list comprehension as follows:

In [26]: list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

In [27]: list1[1::2] = [x+1 for x in list1[1::2]]

In [28]: list1
Out[28]: [1, 3, 3, 5, 5, 7, 7, 9, 9, 11]

Answer 2

numpy allows you to use += operation with slices too:

In [15]: import numpy as np

In [16]: l = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

In [17]: l[1::2] += 1

In [18]: l
Out[18]: array([ 1,  3,  3,  5,  5,  7,  7,  9,  9, 11])

Answer 3

Use enumerate and a list comprehension

>>> list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> [v+1 if i%2!=0 else v for i,v in enumerate(list1)]
[1, 3, 3, 5, 5, 7, 7, 9, 9, 11]

Answer 4

list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for i in range(1, len(list1), 2):
    list1[i] +=1
print(list1)

using i%2 seems not very efficient

Answer 5

Try this:

list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for i in range(1,len(list1),2):
    list1[i] += 1

Answer 6

You can create an iterator representing the delta ( itertools.cycle([0, 1]) and then add its elements to your existing list.

>>> list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> [a + b for a,b in zip(list1, itertools.cycle([0,1]))]
[1, 3, 3, 5, 5, 7, 7, 9, 9, 11]
>>>

Answer 7

a = [i for i in range(1,11)] 
#a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
b = [a[i]+1 if i%2==1 else a[i] for i in range(len(a))] 
#b = [1, 3, 3, 5, 5, 7, 7, 9, 9, 11]