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There is a good way to compute rotation number of a circle homeomorphism (this was the way Poincaré thinked of it): you calculate the rotation number buy its continued fraction in a direct way.

You start from a point $x$ and $f(x)$: this gives you a decomposition of the circle into points that are on the right side of $x$ (in $]x,f(x)[$) and points which are on its left side (in $]f(x),x[$). You look at $f^2(x)$ and you write $R$ if it is on the right side of $x$, $L$ otherwise. Iteranting $f$ you find a sequence of $R$'s and $L$'s. If you get $LLLLR$, for example, you record 4 (this is the number of $L$'s) and you approximate the rotation number of $f$ by $1/4$.

Renormalizing $f$, you iterate this process finding $\rho=[0,a_1,a_2,\ldots,a_k]$.

I won't be more precise here.

Every detail is very well explained in de Melo & van Strien's One-Dimensional Dynamics, section I.1.

You can find a paper by Bruin (Numerical determination of the continued fraction expansion of the rotation number) in which he compares different methods on Arnold tongues.

EDIT: Recently, I wrote for myself some sage lines implementing the algorithm I described you. I was interested in irrational rotation numbers, so these lines do not work properly for rational numbers, but I think you can adapt it very easily (comments are welcome to improve it!).

L=9 #length for cf-expansion

def partfrac(x):
    return x-floor(x)

#computing rotation number
def rotation(f):
    a=[0]
    orbit=[]
    orbit.append(partfrac(f(0)))

    def shift(x):  #set f(0) as the origin + 1
        if partfrac(x)>orbit[0]:
            return partfrac(x)-1
        return partfrac(x)

    def first_return(p,pre_p,y):
        x=shift(f(y))
        while x<pre_p or x>p:
            x=shift(f(x))
        return x

    a.append(1)
    x=orbit[0]

    if shift(f(orbit[0]))==0:
        return 'map with fixed point'

    if shift(f(orbit[0]))<0:
        while shift(f(x))<0:
            a[1]=a[1]+1
            x = shift(f(x))
            if a[1]>10000:
                return 'approximatively 0'
        orbit.append(shift(x))
        z = shift(f(x))
        a.append(0)
        while z>0:
            y = z
            z = first_return(shift(orbit[0]),shift(orbit[1]),z)
            a[2]=a[2]+1
            if a[2]>10000:
                return 'approximatively 0'
        orbit.append(y)

    if shift(f(orbit[0]))>0:
        def shift(y):  #set f(0) as the origin
            if partfrac(y)>=orbit[0]:
                return partfrac(y)-1
            return partfrac(y)
        orbit.append(orbit[0]-1)
        a.append(0)
        while shift(f(x))>0:
            a[2] = a[2] + 1
            x = shift(f(x))
            if a[2]>10000:
                return 'approximatively rational'
        orbit.append(shift(x))
        z = shift(f(x))

    for i in range(1,L):
        a.append(0)
        if shift(orbit[i+1])<shift(orbit[i]):
            while z>0:
                y = z
                z = first_return(shift(orbit[i]),shift(orbit[i+1]),z)
                a[i+2]=a[i+2]+1
                if a[i+2]>10000:
                    return 'approximatively rational'
        if shift(orbit[i+1])>shift(orbit[i]):
            while z<0:
                y = z
                z = first_return(shift(orbit[i+1]),shift(orbit[i]),z)
                a[i+2]=a[i+2]+1
                if a[i+2]>10000:
                    return 'approximatively rational'
        orbit.append(y)

    print a  #shows the cf expansion

    #computing rational approximations
    p=[0,1]
    q=[1,a[1]]
    for i in range(1,L+1):
        p.append(a[i+1]*p[i]+p[i-1])
        q.append(a[i+1]*q[i]+q[i-1])
    return simplify(p[L+1]/q[L+1]) #returns the rational approximation