Python: How to manage ZeroDivisionError

Question

I'm very new to Python and I'm struggling with this error, here is my code:

new_dictionary = dict(((i,f), abs(int(-d/-(-first_dictionary[i, f]//second_dictionary[i, f])))) for i in I for f in F)

I get in return this: ZeroDivisionError: float division by zero

I know that this error occurs because some values of first_dictionary and of second_dictionary are equal to 0.0, but I don't know how to manage it.

I tried to define this function:

def my_function(dict1, dict2):
    first_dictionary = dict1
    second_dictionary = dict2
    for i in I:
        for f in F:
            if first_dictionary[i, f] == 0 or second_dictionary[i, f] == 0.0:
                new_value = 0
            else:
                new_value = abs(int(-d/-(-first_dictionary[i, f]//second_dictionary[i, f])))
    return new_value


new_dictionary = dict(((i, f), my_function(first_dictionary, second_dictionary)) for i in I for f in F)

But the output is different from what I expected, it gives me the same value for all key in the dictionary.

How can I resolve this error? Thanks in advance!

can you share your dictionary values? the question will become more clear.. — Aditya, May 20 '17 at 10:52
Are you suppose to override the "new_value" always in the loops? Why bother with the for loops then? — Danfoa, May 20 '17 at 10:57

score 1 · Answer 1 · edited May 23 '17 at 12:26

1

Source - Reference

>>> try:
    ...     f(0)
    ... except ZeroDivisionError:
    ...     import traceback
    ...     traceback.print_last()
    ... 
    Traceback (most recent call last):
      File "<stdin>", line 1, in <module>

You can use this to check the first instance of what is causing the error

edited May 23 '17 at 12:26

Community

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1

answered May 20 '17 at 10:47

Aditya

2,380
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zwer · Accepted Answer · 2017-05-20T11:09:38.517

1

You don't need to do two loops, if you want explicit check instead of error capture:

def safe_division(a, b):
    if not b:
        return 0
    return a / b

new_dictionary = dict(((i,f), abs(int(safe_division(-d, safe_division(-first_dictionary[i, f], second_dictionary[i, f]))))) for i in I for f in F)

Actually, let's make it a tad bit better with dict comprehension:

new_dictionary = {(i, f): abs(int(safe_division(-d, safe_division(-first_dictionary[i, f], second_dictionary[i, f])))) for i in I for f in F}

edited May 20 '17 at 11:09

answered May 20 '17 at 11:02

zwer

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Thank you for the answer, but this return me another error: ZeroDivisionError: integer division or modulo by zero – Zeno May 20 '17 at 11:06
I had a typo, the above should never divide by zero. – zwer May 20 '17 at 11:11
Thank you for your answer, now is working! Very clever! – Zeno May 20 '17 at 11:12

Python: How to manage ZeroDivisionError

2 Answers2