How to handle error in a for loop (python) and continue execution

Question

I m trying to get my script to still print action 2 "print (list1[3])" and skip action 1 "print (list1[9])" if it can not execute it. I aplozise in advnace if my question is not clear enough I m trying to do my best to explain the issue. I m a beginner.

list1 = ['a','b','c','d','f']

try:
  for i in list1:
    #action 1
    print (list1[9])
    #action 2
    print (list1[3])
    break
except:
  pass

Just wrap the line in question in try/except rather than the whole loop. — Mark
– Mark, Commented May 23, 2020 at 2:15
@Sushanth the code can not print (print (list1[9])) because the list only contain 5 elements therefore it can not print the 9 element because it does not excite I want to skip that an still print the third one using print (list1[9]. right now my code can not do that. I m trying to fix it. I hope you understand what I mean ? — secatt
– secatt, Commented May 23, 2020 at 2:21

Chems Eddine · Accepted Answer · 2020-05-23 02:43:25Z

2

Try this cleaner way

  l = ['a','b','c','d','f']

  # execute the function in try-except
  def try_to_do(fun, index):
     try:
        fun(l[index])
     except:
        pass

  for j in l:
     try_to_do(print, 11)
     try_to_do(print, 1)

  print('Done')

answered May 23, 2020 at 2:43

Chems Eddine

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Comments

CircuitSacul · Accepted Answer · 2020-05-23 02:44:35Z

2

Just put a try for each action instead of both actions together, like this

list1 = ['a','b','c','d','f']


for i in list1:
  try:
    #action 1
    print (list1[9])
  except IndexError:
    pass
  try:
    #action 2
    print (list1[3])
  except IndexError:
    pass
  break

edited May 23, 2020 at 2:44

answered May 23, 2020 at 2:15

CircuitSacul

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7 Comments

sushanth Over a year ago

Wouldn't adding single try..except block enough.. rather than adding to each condition.

CircuitSacul Over a year ago

@Sushanth Not to solve the problem, because if action1 fails then action2 won't execute

Mark Over a year ago

In general, you should try to catch specific errors Since you know you are expecting an IndexError you should catch that specifically with except IndexError:

CircuitSacul Over a year ago

@MarkMeyer I was focusing on answering his question without adding more detail which would make it harder to understand the answer, but you are correct.

Mark Over a year ago

@secatt just replace except: in the above answer with except IndexError: and it will only catch that specific error.

|

bigbounty · Accepted Answer · 2020-05-23 04:17:39Z

0

Use try except inside the loop

list1 = ['a','b','c','d','f']

for i in list1:
    try:
        #action 1
        print (list1[9])
    except:
        print(e.with_traceback())
    try:
        #action 2
        print (list1[3])
    except Exception as e:
        print(e.with_traceback())

edited May 23, 2020 at 4:17

answered May 23, 2020 at 2:16

bigbounty

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1 Comment

bigbounty Over a year ago

@MarkMeyer Changed accordingly

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