Let's say I have a 2d numpy array A:
A = [[0.3, 0.2],
[1.0, 0.1],
[0.3, 0.1],
[1.0, 0.1]]
What I would like is something that maps rows of A to their empirical distribution:
f([0.3, 0.2]) = 0.25
f([1.0, 0.1]) = 0.50
f([-12, 140]) = 0.00
Is there a nice way to do this?
I suggest using numpy.allclose. You can choose a tolerance, here I put 1.e-10 :
import numpy as np
A = np.array([[0.3, 0.2],[1.0, 0.1],[0.3, 0.1], [1.0, 0.1]])
def f(x,tol=1.e-10):
l = [np.allclose(x,row,tol) for row in A]
return l.count(True)/float(A.shape[0])
print f(np.array([0.3,0.2]))
print f(np.array([1.0, 0.1]))
print f(np.array([-12, 140]))
Here's my hacky closure-based method:
def pdf_maker(A, round_place=10):
counts = {}
for i in range(A.shape[0]):
key = tuple([round(a,round_place) for a in A[i]])
try:
counts[key] += 1.0
except KeyError:
counts[key] = 1.0
pdf = {}
for key in counts:
pdf[key] = counts[key] / A.shape[0]
def f_pdf(row):
key = tuple([round(a,round_place) for a in row])
try:
return pdf[key]
except KeyError:
return 0.0
return f_pdf
I'm sure there's a cleaner way to do it, though.
f?