Numpy argmax - random tie breaking

Use np.random.choice -

np.random.choice(np.flatnonzero(b == b.max()))

Let's verify for an array with three max candidates -

In [298]: b
Out[298]: array([0, 5, 2, 5, 4, 5])

In [299]: c=[np.random.choice(np.flatnonzero(b == b.max())) for i in range(100000)]

In [300]: np.bincount(c)
Out[300]: array([    0, 33180,     0, 33611,     0, 33209])

In the case of a multi-dimensional array, choice won't work.

An alternative is

def randargmax(b,**kw):
  """ a random tie-breaking argmax"""
  return np.argmax(np.random.random(b.shape) * (b==b.max()), **kw)

If for some reason generating random floats is slower than some other method, random.random can be replaced with that other method.

Easiest way is

np.random.choice(np.where(b == b.max())[0])