It is worth noting that the accepted answer will round small floats down to zero as demonstrated below:

>>> import numpy as np 
>>> arr = np.asarray([2.92290007e+00, -1.57376965e-03, 4.82011728e-08, 1.92896977e-12])
>>> print(arr)
[ 2.92290007e+00 -1.57376965e-03  4.82011728e-08  1.92896977e-12]
>>> np.round(arr, 2)
array([ 2.92, -0.  ,  0.  ,  0.  ]) 

You can use set_printoptions and a custom formatter to fix this and get a more numpy-esque printout with fewer decimal places:

>>> np.set_printoptions(formatter={'float': "{0:0.2e}".format})
>>> print(arr)
[2.92e+00 -1.57e-03 4.82e-08 1.93e-12]  

This way, you get the full versatility of format and maintain the precision of numpy's datatypes.

Also note that this only affects printing, not the actual precision of the stored values used for computation.

Summary: This is a correct way to do it, the top answer has cases that do not work:

def round_step_size(quantity: Union[float, Decimal], step_size: Union[float, Decimal]) -> float:
    """Rounds a given quantity to a specific step size
    :param quantity: required
    :param step_size: required
    :return: decimal
    """
    precision: int = int(round(-math.log(step_size, 10), 0))
    return float(round(quantity, precision))

My reputation is too low to post a comment on the top answer from Ruggero Turra and point out the issue. However it has cases which did not work for example:

def getRoundedThresholdv1(a, MinClip):
    return round(float(a) / MinClip) * MinClip

getRoundedThresholdv1(quantity=13.200000000000001, step_size=0.0001)

Returns 13.200000000000001 right back whether using numpy or the standard library round. I didn't even find this by stress testing the function. It just came up when using it in production code and spat an error.

Note full credit for this answer comes out of an open source github repo which is not mine found here