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Home/ Questions/Stabilizing the error with the best answers: invalid value encountered in true_divide
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Zoé Bertrand
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Zoé Bertrand
Asked: May 12, 20222022-05-12T11:53:23+00:00 2022-05-12T11:53:23+00:00In: python

Stabilizing the error with the best answers: invalid value encountered in true_divide

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I’m building a new program, but when I run it, an error pops up. The error displayed is as follows:

RuntimeWarning: invalid value encountered in true_divide

I have tried several workarounds, but they still do not get the desired results. If you have come across this situation and have a solution for the “invalid value encountered in true_divide” problem, pls let me know. This is what I do:

# import necessary packages
import numpy as np
  
# Create 2 Numpy arrays
Array1 = np.array([6, 2, 0])
Array2 = np.array([3, 2, 0])
  
# divide the values in Array1 by the
# values in Array2
np.divide(Array1, Array2)

Thanks.

invalid value encountered
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    lyytutoria Expert
    2022-06-01T05:20:49+00:00Added an answer on June 1, 2022 at 5:20 am

    The cause: This error happens because  you are dividing the elements of Array1 by the elements of Array2. And it returns the quotient value.

    6/3=2 (Valid Operation)
    2/2=1 (Valid Operation)
    0/0 which is an invalid operation so a Warning is shown and the result is not a Number (nan).

    Solution: Using seterr method which takes invalid as a parameter and assign ignore as a value to it. By this way, the warning message which contains invalid in can be hiden.

    Syntax: numpy.seterr(invalid=’ignore’)

    # import necessary packages
    import numpy as np
      
    # Create 2 Numpy arrays
    Array1 = np.array([6, 2, 0])
    Array2 = np.array([3, 2, 0])
      
    # Supress/hide the warning
    np.seterr(invalid='ignore')
      
    # divide the values in Array1 by the 
    # values in Array2
    np.divide(Array1, Array2)

    Output:

    array([ 2., 1., nan])
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  2. Zoé Breton
    2022-05-25T19:48:54+00:00Added an answer on May 25, 2022 at 7:48 pm

    This is why you will get the runtime warning.

    log_norm_images = np.where(b_0 > 0, np.divide(diff_images, b_0), 0)

    Is that the inner expression?

    np.divide(diff_images, b_0)

    Gets evaluated first and is run on all diff_images and b_0 elements (even if you ignore elements that involve division by zero). The warning occurs before any code that ignores these elements. This is why it is a warning, not an error. There are legitimate cases such as this where division-by-zero doesn’t pose a problem since it is being dealt with in a later operation.

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