How to declare a sympy Piecewise with multiple limits for multiple variables

Question

In sympy, how do I declare a Piecewise function with multiple limits for multiple variables in a sub-function?

Here is my context and attempt:

from sympy import Piecewise, Symbol, exp
from sympy.abc import z
x1 = Symbol('x1')
x2 = Symbol('x2')
f = 2*pow(z,2)*exp(-z*(x1 + x2 + 2))
p = Piecewise((f, z > 0 and x1 > 0 and x2 > 0), (0, True))

And the error I receive is:

TypeError                                 Traceback (most recent call last)
<ipython-input-47-5e3db02fe3dc> in <module>()
----> 1 p = Piecewise((f, z > 0 and x1 > 0 and x2 > 0), (0, True))

C:\Anaconda3\lib\site-packages\sympy\core\relational.py in __nonzero__(self)
    193 
    194     def __nonzero__(self):
--> 195         raise TypeError("cannot determine truth value of Relational")
    196 
    197     __bool__ = __nonzero__

TypeError: cannot determine truth value of Relational

Hanshan · Accepted Answer · 2016-08-19 05:17:18Z

5

Ah, there is a sympy And function for this:

from sympy import Piecewise, Symbol, exp, And
from sympy.abc import z
x1 = Symbol('x1')
x2 = Symbol('x2')
f = 2*pow(z,2)*exp(-z*(x1 + x2 + 2))
p = Piecewise((f, And(z > 0, x1 > 0, x2 > 0)), (0, True))

answered Aug 19, 2016 at 5:17

Hanshan

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jiedxu · Accepted Answer · 2022-01-13 08:54:21Z

2

You can use & as well:

from sympy import Piecewise, Symbol, exp
from sympy.abc import z
x1 = Symbol('x1')
x2 = Symbol('x2')
f = 2 * pow(z,2) * exp(-z * (x1 + x2 + 2))
p = Piecewise((f, (z > 0) & (x1 > 0) & (x2 > 0)), (0, True))

There are more examples in their documentation.

answered Jan 13, 2022 at 8:54

jiedxu

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