Continuing for loop when exception occurs

Question

I would like for the loop to keep going even exception is generated at first iteration. How to do this?

mydict = {}
wl = ["test", "test1", "test2"]
    
try:
  for i in wl:
   a = mydict['sdf']
   print(i)
            
except:
       # I want the loop to continue and print all elements of list, instead of exiting it after exception
       # exception will occur because mydict doesn't have 'sdf' key
    pass

Move `try` / `except` inside loop body – awesoon Aug 03 '20 at 04:51 — awesoon, Aug 03 '20 at 04:51

score 1 · Answer 1 · answered Aug 03 '20 at 04:53

1

You can use dict.get(). It will return None if the key not exist. You can also specify default value in dict.get(key, default_value)

for i in wl:
    a = mydict.get('sdf')
    print(i)

answered Aug 03 '20 at 04:53

deadshot

8,881
4
20
39

score 0 · Answer 2 · answered Aug 03 '20 at 04:54

0

The best I can advice is to move the try inside the loop as below:

mydict = {}
wl = ["test", "test1", "test2"]
for i in wl:
    try:
        a = mydict['sdf']
        print(i)

    except:
        continue

answered Aug 03 '20 at 04:54

Albert Alberto

801
8
15

1

This is incorrect. The `print` will be skipped. – user Aug 03 '20 at 05:16
Those two lines of code are independent, so you could just move the `print` to be first. – John Gordon Aug 03 '20 at 05:18
1

Also, using a bare `except` that catches all exceptions is bad practice. Only catch the exceptions that you know might be raised. – Thierry Lathuille Aug 03 '20 at 05:18

score 0 · Answer 3 · answered Aug 03 '20 at 04:59

0

This is my approach for your problem
Hope its work for you as you want

mydict = {}
wl = ["test", "test1", "test2"]

for i in wl:
    try:
        a = mydict['sdf']
    except:
        pass
    print(i)

answered Aug 03 '20 at 04:59

Amirul Akmal

401
6
13

Continuing for loop when exception occurs

3 Answers3