list index out of range ,how to stop at nonexisting value

Question

if i+1 is non existing value then stopn the loop and value of i+1 where i is end of list should not execute.

my_list = [1,2,3,4,5]
for i in range(len(my_list)):
if my_list[i] == my_list[i+1]:
    print('ok')

#IndexError: list index out of range

how about `range(len(my_list) - 1)` instead of `range(len(my_list))`? — L.Grozinger, Sep 24 '20 at 10:59
if you want to use len(list) then use len(list)-1 as its length — Swaroop Maddu, Sep 24 '20 at 11:00

score 1 · Answer 1 · answered Sep 24 '20 at 10:59

1

Try this,

my_list = [1, 2, 3, 4, 5]
for i in range(len(my_list) - 1):
if my_list[i] == my_list[i+1]:
    print('ok')

answered Sep 24 '20 at 10:59

Maran Sowthri

829
6
14

score 1 · Answer 2 · answered Sep 24 '20 at 11:00

Error occurs because my_list[i+1] tries to access 6th element of list containing 5 elements.

Not sure what is your goal here, but this is possible fix - since you are comparing current and next element all the time, you can loop from first to 5th element:

my_list = [1,2,3,4,5]
for i in range(len(my_list) - 1):
if my_list[i] == my_list[i+1]:
    print('ok')

score 1 · Answer 3 · answered Sep 24 '20 at 11:05

1

my_list = [1,2,3,4,5]
for i in range(len(my_list)-1):
    if my_list[i] == my_list[i+1]:
        print('ok')

If the length of the list is n we have elements from 0 to n-1 only since list indices start from 0.

answered Sep 24 '20 at 11:05

Swaroop Maddu

4,289
2
26
38

score 0 · Answer 4 · answered Sep 24 '20 at 11:01

0

#hope this code helps you

my_list = [1,2,3,4,5]
for i in range(len(my_list)):
    if i in my_list:
        print("ok")
        i = i +1

answered Sep 24 '20 at 11:01

Arbaz

11
1
4

no,i doesn;t helps – Shah Vipul Sep 24 '20 at 12:18

list index out of range ,how to stop at nonexisting value

4 Answers4