Clash Royale CLAN TAG#URR8PPP

NumPy setdiff1d with tolerance - Comparing a numpy array to another and saving only the unique values - outside of a tolerance

I have two numpy arrays:

A= [ 3.8357 3.2450]

B= [ 5.6132 3.2415 3.6086 3.5666 3.8769 4.3587]


I want to compare A to B and only keep the value in A that is unique - outside of a +/-0.04 tolerance (i.e. A=[3.8357]).

Any ideas as to how I can do this?

1 Answer
1

Approach #1

We could use broadcasting -

broadcasting

A[(np.abs(np.subtract.outer(A,B)) > 0.04).all(1)]


Approach #2

We could leverage searchsorted to have a generic numpy.isin with tolerance specifier for use in generic problems, like so -

searchsorted

numpy.isin

def isin_tolerance(A, B, tol):
A = np.asarray(A)
B = np.asarray(B)

Bs = np.sort(B) # skip if already sorted
idx = np.searchsorted(Bs, A)

linvalid_mask = idx==len(B)
idx[linvalid_mask] = len(B)-1
lval = Bs[idx] - A
lval[linvalid_mask] *=-1

rinvalid_mask = idx==0
idx1 = idx-1
idx1[rinvalid_mask] = 0
rval = A - Bs[idx1]
rval[rinvalid_mask] *=-1
return np.minimum(lval, rval) <= tol


Hence, to solve our case -

out = A[~isin_tolerance(A, B, tol=0.04)]


Sample run -

In [294]: A
Out[294]: array([13.8357, 3.245 , 3.8357])

In [295]: B
Out[295]: array([5.6132, 3.2415, 3.6086, 3.5666, 3.8769, 4.3587])

In [296]: A[~isin_tolerance(A, B, tol=0.04)]
Out[296]: array([13.8357, 3.8357])


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.