Checking if the next element in a python list is empty

Question

So what I am trying to accomplish is to check whether an element is empty by using a counter + 1 but I keep getting index out of range which essentially means the next element doesnt exist, but instead of throwing an exception I want the program to return a boolean to my if statement is that possible..? In essence I want to peek forward to the next element of a tuple within a dictionary actually and see if it is empty.

>>> counter = 1
>>> list = 1,2,3,4
>>> print list
>>> (1, 23, 34, 46)
>>> >>> list[counter]
23
>>> list[counter + 1]
34
>>> list[counter + 2]
46

>>> if list[counter + 3]:
...     print hello
... else:
...     print bye
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: tuple index out of range

possible duplicate of [If list index exists, do X](http://stackoverflow.com/questions/11786157/if-list-index-exists-do-x) — Mel, Jun 29 '15 at 12:19

The6thSense · Accepted Answer · 2015-06-29T12:27:19.837

4

You could use try/catch to catch error if you index a not available index of a list

And the main thing it is a bad practice to name variable with keywords i.e. list,set etc

try:
    if list[counter + 3]:
        print "yes"
except IndexError:
    print 'bye'

edited Jun 29 '15 at 12:27

answered Jun 29 '15 at 11:58

The6thSense

8,103
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2

"Easier to ask forgiveness than permission." – nkorth Jun 29 '15 at 11:59

score 3 · Answer 2 · answered Jun 29 '15 at 11:57

3

You can use len to check for whether you are within the range.

For example:

>>> l = 1,2,3,4
>>> len(l)
4

Also, a tuple is not a list. It is generally considered bad practice to name things as list or array etc.

answered Jun 29 '15 at 11:57

Jay Bosamiya

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score 1 · Answer 3 · answered Jun 29 '15 at 11:59

1

The easiest way to check presence of index in tuple or list is to compare given index to length of it.

if index + 1 > len(my_list):
    print "Index is to big"
else:
    print "Index is present"

answered Jun 29 '15 at 11:59

Vladyslav Lubenskyi

527
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This is wrong for python 3 at least. See my answer. – Saelyth Apr 05 '20 at 08:07

score 1 · Answer 4 · answered Apr 05 '20 at 08:09

Python 3 code without exceptions:

my_list = [1, 2, 3]
print(f"Lenght of list: {len(my_list)}")
for index, item in enumerate(my_list):
    print(f"We are on element: {index}")
    next_index = index + 1
    if next_index > len(my_list) - 1:
        print(f"Next index ({next_index}) doesn't exists")
    else:
        print(f"Next index exists: {next_index}")

Prints this:

>>> Lenght of list: 3
>>> We are on element: 0
>>> Next index exists: 1
>>> We are on element: 1
>>> Next index exists: 2
>>> We are on element: 2
>>> Next index (3) doesn't exists

Checking if the next element in a python list is empty

4 Answers4