Why Does Python Return 0 for Simple Division Calculation

Why does Python return 0 for simple division calculation?

In Python 2, 25/100 is zero when performing an integer divison. since the result is less than 1.

You can "fix" this by adding from __future__ import division to your script. This will always perform a float division when using the / operator and use // for integer division.

Another option would be making at least one of the operands a float, e.g. 25.0/100.

In Python 3, 25/100 is always 0.25.

Make division by zero equal to zero

Check if the denominator is zero before dividing. This avoids the overhead of catching the exception, which may be more efficient if you expect to be dividing by zero a lot.

def weird_division(n, d):
    return n / d if d else 0

Why does the division get rounded to an integer?

You're using Python 2.x, where integer divisions will truncate instead of becoming a floating point number.

>>> 1 / 2
0

You should make one of them a float:

>>> float(10 - 20) / (100 - 10)
-0.1111111111111111

or from __future__ import division, which the forces / to adopt Python 3.x's behavior that always returns a float.

>>> from __future__ import division
>>> (10 - 20) / (100 - 10)
-0.1111111111111111

BMI calculator always return 0. (Python)

Type cast it to float. / will do integer division always.

That is

bmi = float(weight) / float(length * length)

It will print

Weight in kg: 10
Length in cm: 20
0.025

Or do

from __future__ import division

How to return 0 with divide by zero

In numpy v1.7+, you can take advantage of the "where" option for ufuncs. You can do things in one line and you don't have to deal with the errstate context manager.

>>> a = np.array([-1, 0, 1, 2, 3], dtype=float)
>>> b = np.array([ 0, 0, 0, 2, 2], dtype=float)

# If you don't pass `out` the indices where (b == 0) will be uninitialized!
>>> c = np.divide(a, b, out=np.zeros_like(a), where=b!=0)
>>> print(c)
[ 0.   0.   0.   1.   1.5]

In this case, it does the divide calculation anywhere 'where' b does not equal zero. When b does equal zero, then it remains unchanged from whatever value you originally gave it in the 'out' argument.

How to return 0 with divide by 0 from string?

you can try

try:
  #Your operation here
except ZeroDivisionError:
  # return 0 here 

More information here: https://en.wikibooks.org/wiki/Python_Programming/Exceptions

Returning math error (division by 0 etc) that otherwise crashes method

Wrap your statements with an try: except: else: block

Fx = input("what is the desired function of x?" )
listY = []

try:
    for i in range(0,1001):
        x = i/100
        y = eval(Fx)
        listY.append(y)
except ZeroDivisionError:
    print("Divide by zero error for f({})".format(x))
except ValueError:
    print("Invalid value for f({})".format(x))
else:
    print("This function's minimum value is", min(listY))
    print("This function's maximum value is", max(listY))

I'm not sure how you can define there was a min or a max when you have undefined results. You could use limits around the undefined result(s) to understand which direction you approached the undefined value.

But if you want to simply return the min or max of the values that were defined you can move the try block inside the for loop, e.g:

Fx = input("what is the desired function of x?" )
listY = []

undefined = False
for i in range(0,1001):
    try:
        x = i/100
        y = eval(Fx)
    except ZeroDivisionError:
        print("Divide by zero error for f({})".format(x))
        undefined = True
    except ValueError:
        print("Invalid value for f({})".format(x))
        undefined = True
    else:
        listY.append(y)

if undefined:
    print("Ignoring undefined results")
print("This function's minimum value is", min(listY))
print("This function's maximum value is", max(listY))

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