Continually catch exception in python

Question

I need to keep catching an exception while indexing a list throws a IndexError exception, for example:

l = []
while((l[3]) throws IndexError):
    //Add more data to the list
    l += [3]

How can I keep checking to see if the method call has thrown an exception without having nested a nested try/catch block?

Why would you not want to use a try:... except: block? It's Pythonic. — anon582847382
– anon582847382, Commented Mar 17, 2014 at 10:18
@BurhanKhalid The number being indexed changes each run, this is dependant because it is reading in data from the USB port. — Dean
– Dean, Commented Mar 17, 2014 at 10:30
Doing this with an exception or any sort of "until we have enough" loop seems silly. Why not just compare the length you need to the length you have and add the amount of data you need? if len(l) < needed: l += [3] * (needed - len(l)) — user2357112
– user2357112, Commented Mar 17, 2014 at 10:38

skamsie · Accepted Answer · 2014-03-17 10:14:02Z

4

It depends what you would like to extend your list with. Assuming 'None' you could do it like this:

l = []
while True:
  try:
    l[3] = 'item'
    break
  except IndexError:
    l.extend([None])

print l # [None, None, None, 'item']

answered Mar 17, 2014 at 10:14

skamsie

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2 Comments

Dean Over a year ago

I hadn't thought of wrapping this in a infinite loop, partly because in my application its already wrapped in an infinite loop and don't want it to become uncontrollable.

skamsie Over a year ago

You could do: while len(l) <= 4: (4 not 3 because indexing starts with 0) or even wrap this into a function and have the index as argument

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