Using Additional kwargs with a Custom Function for Scipy's cdist (or pdist)?

Question

I am using a custom metric function with scipy's cdist function. The custom function is something like

def cust_metric(u,v):
  dist = np.cumsum(np.gcd(u,v) * k) 
  return dist

where k is an arbitrary coefficient.

Ideally, I was hoping to pass k as an argument when calling cdist like so: d_ar = scipy.spatial.distance.cdist(arr1, arr2, metric=cust_metric(k=7)) However, this throws an error.

I was wondering if there is a simple solution that I may be missing? A quick but non-elegant fix is to declare k as a global variable and adjust it when needed.

Are `arr1` and `arr2` 1D arrays? If so, you can just use `broadcasting`. — Quang Hoang, May 19 '20 at 14:07

fuglede · Accepted Answer · 2020-05-31T11:28:14.290

According to its documentation, the value for metric should be a callable (or a string for a particular fixed collection). In your case you could obtain that through

def cust_metric(k):
    return lambda u, v: np.cumsum(np.gcd(u, v) * k)

I do imagine your actual callable would look somewhat different since the moment u and v are 2D arrays, the np.cumsum returns an array, while the callable is supposed to produce a scalar. For example:

In [25]: arr1 = np.array([[5, 7], [6, 1]])

In [26]: arr2 = np.array([[6, 7], [6, 1]])

In [28]: def cust_metric(k):
    ...:     return lambda u, v: np.sqrt(np.sum((k*u - v)**2))
    ...:

In [29]: scipy.spatial.distance.cdist(arr1, arr2, metric=cust_metric(k=7))
Out[29]:
array([[51.03920062, 56.08029957],
       [36.        , 36.49657518]])

In [30]: scipy.spatial.distance.cdist(arr1, arr2, metric=cust_metric(k=1))
Out[30]:
array([[1.        , 6.08276253],
       [6.        , 0.        ]])

Using Additional kwargs with a Custom Function for Scipy's cdist (or pdist)?

1 Answers1