How to make this rounding function faster?

Question

I am trying to write a function to round values to the nearest valid odds in a list from here: https://api.developer.betfair.com/services/webapps/docs/display/1smk3cen4v3lu3yomq5qye0ni/Betfair+Price+Increments

My code is here:

def nclosest_valid_odds( x ):
    """
    https://api.developer.betfair.com/services/webapps/docs/display/1smk3cen4v3lu3yomq5qye0ni/Betfair+Price+Increments
    """

    r = np.empty_like( x )

    r[ x<1.0 ] = np.nan

    bidx      = (1.0<=x) & (x<=2.0)
    r[ bidx ] = 0.01 * np.round( x[ bidx ] / 0.01 )

    bidx      = (2.0<x) & (x<=3.0 )
    r[ bidx ] = 0.02 * np.round( x[ bidx ] / 0.02 )    

    bidx      = (3.0<x) & (x<=4.0)
    r[ bidx ] = 0.05 * np.round( x[ bidx ] / 0.05 )

    bidx      = (4.0<x) & (x<=6.0)
    r[ bidx ] = 0.1 * np.round( x[ bidx ] / 0.1 )

    bidx      = (6.0<x) & (x<=10.0)
    r[ bidx ] = 0.2 * np.round( x[ bidx ] / 0.2 )

    bidx      = (10.0<x) & (x<=20.0)
    r[ bidx ] = 0.5 * np.round( x[ bidx ] / 0.5 )

    bidx      = (20.0<x) & (x<=30.0)
    r[ bidx ] = np.round( x[ bidx ] )

    bidx      = (30.0<x) & (x<=50.0)
    r[ bidx ] = 2.0 * np.round( x[ bidx ] / 2.0 )

    bidx      = (50.0<x) & (x<=100.0)
    r[ bidx ] = 5.0 * np.round( x[ bidx ] / 5.0 )

    bidx      = (100.0<x) & (x<=1000)
    r[ bidx ] = 10.0 * np.round( x[ bidx ] / 10.0 )

    return r

A floor version using numba is here:

def floor_closest_valid_odds( x ):

    r = np.zeros_like( x )

    for i in range( len( r ) ):
        if x[i]<=1.0:
            r[i] = np.nan
        elif x[i]<=2.0:            
            r[i] = 0.01 * np.floor( x[i] / 0.01 )            
        elif x[i]<=3.0:                        
            r[i] = 0.02 * np.floor( x[i] / 0.02 )            
        elif x[i]<=4.0:            
            r[i] = 0.05 * np.floor( x[i] / 0.05 )
        elif x[i]<=6.0:            
            r[i] = 0.1 * np.floor( x[i] / 0.1 )
        elif x[i]<=10.0:            
            r[i] = 0.5 * np.floor( x[i] / 0.5 )
        elif x[i]<=20.0:            
            r[i] = 1.0 * np.floor( x[i] / 1.0 )
        elif x[i]<=30.0:            
            r[i] = 2.0 * np.floor( x[i] / 2.0 )
        elif x[i]<=50.0:            
            r[i] = 2.0 * np.floor( x[i] / 2.0 )
        elif x[i]<=100.0:            
            r[i] = 5.0 * np.floor( x[i] / 5.0 )
        elif x[i]<=1000.0:            
            r[i] = 2.0 * np.floor( x[i] / 2.0 )
        else:            
            r[i] = 1000.0
    return r

jfloor_closest_valid_odds = autojit( floor_closest_valid_odds )

And I time the code using this:

    x = np.random.randn( 1000000 )

    with Timer( 'nclosest_odds' ):
        r = nclosest_valid_odds( x )

    r =helpers.jfloor_closest_valid_odds( x )    
    with Timer( 'jfloor_closest_valid_odds' ):
        r = helpers.jfloor_closest_valid_odds( x )                 

Timings on my machine:

nclosest odds : 0.06 seconds
jfloor_closest_odds : 0.01 seconds

How could I speed up the code using numpy and/or numba?

Solution:

I discovered that the multi-threading example from http://numba.pydata.org/numba-doc/dev/examples.html can be turned into a vectorizing function. Using this results in the best performance on my 2 core laptop.

Numba's Vectorize function is OK too, though not as good.

I have uploaded the vectorizer code to github: https://github.com/jsphon/MTVectorizer

A chart of timing comparisons is below. The x-axis represents the size of the input array. The y-axis represents time.

This chart is from a dual core laptop.

This chart is from a desktop with i7 920 CPU. It has 8 cores.

Answer 1

You can get a speed increase in numpy by creating a magnitude array and then doing the rounding all at the end with the magnitude array.

def nclosest_valid_odds_3( x ):

    magnitudes = np.empty_like(x)

    magnitudes[x < 1] = np.nan
    magnitudes[(1 <= x) & (x <= 2)] = 0.01

    v = [2, 3, 4, 6, 10, 20, 30, 50, 100, 1000]
    m = [0.02, 0.05, 0.1, 0.2, 0.5, 1, 2, 5, 10]
    for low, high, mag in zip(v, v[1:], m):
        magnitudes[(low < x) & (x <= high)] = mag

    return magnitudes * np.round(x / magnitudes)