CODEWITHSUNDEEP
roundingsympylambdify
arsenis
arsenis

Reputation: 33

rounding part of a sympy expression

I have a sympy symbolic expression that contains sin and cos. At the end of the code I use lambdify((Phi),function(Phi))(phi) but I'm getting many non-zero values because of the angle dependence which consists of several pi values, since sin(pi) doesn't return exactly 0. What I would like to do if possible, is to round while still in symbolic form. Something like sin(phi).round(5) where phi is just a symbol, so that when you lambdify the expression and sub in a value for phi the result of sin is automatically rounded to 5d.p. Is there any way to do so?

Thanks

Upvotes: 0

Views: 192

Answers (1)

Oscar Benjamin
Oscar Benjamin

Reputation: 14480

You can use the floor function to round to 5 decimal places symbolically:

In [14]: f = floor(100000*x) / 100000

In [15]: lambdify(x, f)(0.123456789)
Out[15]: 0.12345

Upvotes: 1

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