搜索两个 3 列的 numpy 数组并找到在 Python 中符合条件的位置

Question

I have two numpy array like

Population A

Score A Score B Answer
  1       0.3      1    
  0       0.5      0
  1       0.6      1     
  0       0.7      1
  1       0.9      1

Population B

Score A Score B Answer
  1       0.3      1    
  0       0.5      0
  1       0.6      1     
  1       0.7      1
  0       0.9      1

 Sample Results are
 ScoreB     Ratio 
  0.3        1
  0.5        1
  0.6        1
  0.7        1

I have to find score/value of score B in each population, any value above that value becomes 1 else 0, for example if you pick 0.5 in population A then first value is 0 rest are 1, similarly if you pick 0.6 in population B for example then first two value are 0 rest are 1.

I have to do this iteratively/ algorithmic possibly in while loop I guess and without creating or replacing scoreB such that

Ratio = (counts(scoreA=1&scoreb=1&Answer=1) in population A/ counts(scoreA=1&scoreb=1&Answer=1) in population B) == 1

Note: Score B is sorted so not to worry about that

Answer 1

You can use logical_and or simply the * operator on numpy array which gives element wise multiplication

import numpy as np
def count_match (v, A, B, C):

  cond = np.logical_and(A == 1, B >= v)
  cond = np.logical_and(cond, C == 1)
  # alternatively, use an element wise product:
  cond = (A == 1) * (B >= v) * (C == 1)

  # counting the number of 1
  count = np.sum(cond)
  return count

A_a = np.array([1, 0, 1, 0, 1])
B_a = np.array([0.3, 0.5, 0.6, 0.7, 0.9])
C_a = np.array([1, 0, 1, 1, 1])

A_b = np.array([1, 0, 1, 1, 0])
B_b = np.array([0.3, 0.5, 0.6, 0.7, 0.9])
C_b = np.array([1, 0, 1, 1, 1])
for v in [0.3, 0.5, 0.6, 0.7, 0.9]:
  print(v, count_match(v, A_a, B_a, C_a) / count_match(v, A_b, B_b, C_b)) # divides by zero brr

note btw that for popA and popB, the condition on array A and C do not change, so you may precompute them already (I doubt that it changes time that much)

def count_match2(v, pre_cond, B):
  return np.sum(pre_cond * (B >= v))

for v in [0.3, 0.5, 0.6, 0.7, 0.9]:
  print(v, count_match2(v, (A_a == 1) * (C_a == 1), B_a) / count_match2(v, (A_b == 1) * (C_b == 1), B_b)) # divides by zero brr