how to grab internal index of current loop in python when iter a list?

Question

i got a listx =[1,2,3,4,5,6,7,8,9]

i want to alter every Nth item of the list. For example i want to modify for every 2 item step, let say i want to modify by +1 . so i want to get result = [1+1,2,3+1,4,5+1,6,7+1,8,9+1] =[2,2,4,4,6,6,8,8,9]

i can do this by using for-loop , by adding counter variable , then check the counter by counter%2==0. But this time i 'm just curious using single line statement. Here what i want =

newlistx=[i+1 for i in listx] <- this will modify all items, so i'm expecting i can use some internal indexing use in this iteration process, become like this :

newlistx=[i+1 if (__indexing__%step==0) else i for i in listx] where step=2.

Actually, i can use list.index() function , like this :

newlistx=[i+1 if listx.index(i)%2==0 else i for i in listx]

the problem this me thod only works if all the item is unique, if i got items which have same value then index() will return wrong value.

Again, i'm just curious if i can grab some internal indexing or counter , if exist.

Dietrich Epp · Accepted Answer · 2012-06-19 20:31:20Z

7

You can use the enumerate function.

newlist = [x + 1 if n % step == 0 else x
           for (n, x) in enumerate(oldlist)]

The enumerate function iterates over a sequence and yields the objects with their indexes.

edited Jun 19, 2012 at 20:31

answered Jun 19, 2012 at 9:20

Dietrich Epp

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2 Comments

Markus Over a year ago

Your use of n for the index and i for the list item makes me cringe.

Dietrich Epp Over a year ago

@Markus: Indeed, I never would have written code like that on my own. Fixed.

Rich Tier · Accepted Answer · 2012-06-19 11:32:31Z

1

new_list = [n + 1 if i & 1 else n for i, n in enumerate(listx)]

answered Jun 19, 2012 at 11:32

Rich Tier

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