在整个参数空间/二维数组 Numpy 上应用 function

Question

I'm in one of those weird places, where I know exactly what I want to do. I could easily code it up using for loops. but I'm trying to learn Numpy and I can't formulate how to solve this in Numpy.

I want to have a 2d array or parameter space. All values between 1200 and 1800, and all combinations therein. So [1200, 1200], [1200, 1201], [1200, 1202].... [1201, 1200], [1201, 1201] etc.

I want to apply a function across this entire parameter space. The function uses a further 2 arrays, which are also values in 1200-1800 range. But they are random values, so these 2 extra arrays are random values in the 1200-1800 range, so [1356, 1689, 1436, ...] and [1768, 1495, 1358, ...] etc. check_array1 and check_array2.

The function needs to move through the parameter space checking a condition, which is basically if x < check_array1 and y < check_array2 then 1 else 0 . Where x and y are the each specific point in the 2d parameter space. It needs to check against every value combination in the check arrays. Sum the total, do a comparison to another static value, and return the difference.

Each unique combination in the parameter space grid will then have a unique value associated with it based on how those specific x and y values from the parameter space compare to the 2 check arrays.

Hopefully the above makes, I just can't figure out how to work this into a Numpy friendly problem. Sorry for the wall of text.

Edit: I've written it in more basic Python to better illustrate what I'm trying to do.

check1 = np.random.randint(1200, 1801, 300)
check2 = np.random.randint(1200, 1801, 300)

def check_this_double(i, j, check1, check2):
   total = 0
   for num in range(0, len(check1)):
       if ((i < check1[num]) or (j < check2[num])):
           total += 1           
   return total

outputs = {}
for i in range(1200, 1801):
   for j in range(1200, 1801):
    outputs[i,j] = check_this_double(i, j, check1, check2)

Edit 2: I believe I have it. Following from Mountains code creating the p_space and then using np.vectorize on a normal Python fuction.

check1 = np.random.randint(1200, 1801, 300)
check2 = np.random.randint(1200, 1801, 300)

def calc(i, j):    
   total = np.where(np.logical_or(check1 < i, checks2 < j), 1, 0)
   return total.sum()

rate_calv_v = np.vectorize(rate_calc)

final = rate_calv_v(p_space[:, 0], p_space[:, 1])

Feels kind of like cheating:), there must be way to do it without np.vectorize. But this works for me I believe.

Answer 1

I don't fully understand the problem you are trying to solve. I hope the following will give you a starting point on how numpy can be used. I recommend going through a numpy introductory tutorial.

numpy boolean indexing and vector math can improve speed and reduce the need for loops. Here is my understanding of the first part of your questions.

import numpy as np

xv, yv = np.meshgrid(np.arange(1200, 1801), np.arange(1200, 1801))
p_space = np.stack((xv, yv), axis=-1)  # the 2d array described

# print original values
print(p_space[0,:10,0])
print(p_space[0,-10:,0])

old_shape = p_space.shape
p_space = p_space.reshape(-1, 2)  # flatten the array for the compare

check1 = np.random.randint(1200, 1801, len(p_space))
check2 = np.random.randint(1200, 1801, len(p_space))

# you can used this to access and modify values that meet the condition
index_array = np.logical_and(p_space[:, 0] < check1, p_space[:, 1] < check2)

# do some sort of complex math
p_space[index_array] = p_space[index_array] / 2 + 10

# get the sum across the two columns
print(np.sum(p_space, axis=0))

p_space = p_space.reshape(old_shape)  # return to the grid shape

# print modified values
print(p_space[0,:10,0])  # likely to be changed based on checks
print(p_space[0,-10:,0])  # unlikely to be changed