How to define periodic function in python?

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how can you define a periodic function in python, e.g. the "sawtooth" function?

f(x) = x, for –π < x < π and 2-pi periodically continued on IR

Can you do this with a lambda function?

You can make use of decorators:

def periodically_continued(a, b):
    interval = b - a
    return lambda f: lambda x: f((x - a) % interval + a)

@periodically_continued(-1, 1)
def f(x):
    return x

g = periodically_continued(0, 1)(lambda x: -x)

assert f(2.5) == 0.5
assert g(2.5) == -0.5

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as a normal function you can use modulo (%) also with float:

from math import pi

def f(x):
    return (x+pi) % (2*pi) - pi

this easily translates to a lambda expression:

lambda x: (x+pi) % (2*pi) - pi

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You could write a function that takes a function and a period, and returns a function:

import math

def periodic_function(func, period, offset):
    return lambda x: func( ((x - offset) % period ) + offset )

and use that then:

sawtooth = periodic_function(lambda x: x, 2*math.pi, math.pi)

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A very simple way is to limit the inputs to the first period. Iteratively remove one period until the input falls within the defined values. For example the following example gives a square wave with period 2*pi.

def f(t):
    while t>2*np.pi:
            t=t-2*np.pi
    if 0.0 <= t <=np.pi:
        return 1.0
    else:
        return -1.0

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Comments
  • plz explain 2-pi periodically continued on IR
  • probably would be tricky with a lambda but I'd be tempted to use fmod from the mathmodule
  • 2-pi periodically continued on IR (the set of all real numbers) means f(x) is defined for all real numbers x, and furthermore f(x) is identically equal to f(x + 2pi)
  • @AhsanulHaque: I believe "IR" is serving as an ASCII approximation to "ℝ" here.
  • docs.scipy.org/doc/scipy-0.13.0/reference/generated/…
  • +1 Nice idea! expense is an odd word to use, though. Did you mean expanse? extent? (I'd probably just call it periodic, or perhaps periodic_on.)
  • You also want to replace x % interval with (x - a) % interval, I think.
  • @MarkDickinson Thanks for correction. Oh my poor English skill >_<
  • oha, now there is a missing ly :D But nice use of decorators!
  • You don't need the Python 3 disclaimer: % works for floating-point numbers in both Python 2 and Python 3. And I think you want (x + pi) % (2*pi) - pi to get the right result.
  • @Mark Dickinson: you are right. just made a quick plot... thanks & updated.
  • What is the offset?