гибкий запрос на основе двух массивов numpy

Question

Я хотел бы создать несколько динамический запрос на основе массива numpy. В идеале, индексы одного массива (uu) должны быть возвращены с учетом условий для каждого столбца во втором массиве (cond).

Образец ниже демонстрирует то, что я имею в виду, и он работает с использованием цикла. Мне интересно, есть ли более эффективный метод. Спасибо за вашу помощь.

import numpy as np 
# create an array that has four columns, each contain a vector (here: identical vectors) 
n = 101 
u = np.linspace(0,100,n) 
uu = np.ones((4, n)) * u 
# ideally I would like the indices of uu 
#  (so I could access another array of the same shape as uu) 
#  that meets the following conditions: 
#  the condition based on which values in uu should be selected 
#   in the first column (index 0) all values <= 10. should be selected 
#   in the second column (index 1) all values <= 20. should be selected 
cond = np.array([10,20]) 


# this gives the correct indices, but in a series of 1D solutions 
#  this would work as a work-around 
for i in range(cond.size): 
    ix = np.where(uu[i,:] <= cond[i]) 
    print(ix) 


# this seems like a work-around using True/False 
# but here I am not sure how to best convert this to indices 
for i in range(cond.size): 
    uu[i,:] = uu[i,:] <= cond[i] 
print(uu) 

Answer 1

Numpy позволяет сравнивать массивы непосредственно:

import numpy as np 
# just making my own uu with random numbers 
n = 101 
uu = np.random.rand(n,4) 

# then if you have an array or a list of values... 
cond = [.2,.5,.7,.8] 

# ... you can directly compare the array to it 
comparison = uu <= cond 

# comparison now has True/False values, and the shape of uu 
# you can directly use comparison to get the desired values in uu 
uu[comparison] # gives the values of uu where comparison ir True 

# but if you really need the indices you can do 
np.where(comparison) 
#returns 2 arrays containing i-indices and j-indices where comparison is True