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I am creating a calculator application for all types of mathematical algorithms. However, I want to identify if a root is complex and then have an exception for it. I came up with this:

if x == complex():
    print("Error 05: Complex Root")

However, nothing is identified or printed when I run the app, knowing that x is a complex root.

  • Are the indents exactly as in your question? Is there any error? Could you add `else` part of `if` statement? What about `complex()`? Shouldn't you pass an argument to it and return `True` or `False`? – Tadeck Mar 21 '12 at 23:57
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    Wait a second - what if a complex root is not an error? Sometimes that's the right answer. Are you sure it should be flagged as an error? – duffymo Mar 22 '12 at 00:02

5 Answers5

28

I'm not 100% sure what you're asking, but if you want to check if a variable is of complex type you can use isinstance. For example,

x = 5j
if isinstance(x, complex):
    print 'X is complex'

prints

X is complex
Adam Mihalcin
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>>> isinstance(1j, complex)
True
Ignacio Vazquez-Abrams
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9

Try this:

if isinstance(x, complex):
    print("Error 05: Complex Root")

This prints error for 2 + 0j, 3j, but does not print anything for 2, 2.12 etc.

Also think about throwing an error (ValueError or TypeError) when the variable is complex.

Tadeck
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5

In NumPy v1.15, a function is included: numpy.iscomplex(x)

where x is the number, that is to be identified.

Soham Bhattacharyya
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0

One way to do it could be to do,

if type(x) == complex():
    print("Error 05: Complex Root")

As others have pointed out, isinstance works too

Evelyn
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