Creating new numpy arrays based on condition

I have 2 numpy arrays:

aa = np.random.rand(5,5)
bb = np.random.rand(5,5)

How can I create a new array which has a value of 1 when both aa and bb exceed 0.5?

回答1:

With focus on performance and using two methods few aproaches could be added. One method would be to get the boolean array of valid ones and converting to int datatype with .astype() method. Another way could involve using np.where that lets us select between 0 and 1 based on the same boolean array. Thus, essentially we would have two methods, one that harnesses efficient datatype conversion and another that uses selection criteria. Now, the boolean array could be obtained in two ways - One using simple comparison and another using np.logical_and. So, with two ways to get the boolean array and two methods to convert the boolean array to int array, we would end up with four implementations as listed below -

out1 = ((aa>0.5) & (bb>0.5)).astype(int)
out2 = np.logical_and(aa>0.5, bb>0.5).astype(int)
out3 = np.where((aa>0.5) & (bb>0.5),1,0)
out4 = np.where(np.logical_and(aa>0.5, bb>0.5), 1, 0)

You can play around with the datatypes to use less precision types, which shouldn't hurt as we are setting the values to 0 and 1 anyway. The benefit should be noticeable speedup as it leverages memory efficiency. We could use int8, uint8, np.int8, np.uint8 types. Thus, the variants of the earlier listed approaches using the new int datatypes would be -

out5 = ((aa>0.5) & (bb>0.5)).astype('int8')
out6 = np.logical_and(aa>0.5, bb>0.5).astype('int8')
out7 = ((aa>0.5) & (bb>0.5)).astype('uint8')
out8 = np.logical_and(aa>0.5, bb>0.5).astype('uint8')

out9 = ((aa>0.5) & (bb>0.5)).astype(np.int8)
out10 = np.logical_and(aa>0.5, bb>0.5).astype(np.int8)
out11 = ((aa>0.5) & (bb>0.5)).astype(np.uint8)
out12 = np.logical_and(aa>0.5, bb>0.5).astype(np.uint8)

Runtime test (as we are focusing on performance with this post) -

In [17]: # Input arrays
    ...: aa = np.random.rand(1000,1000)
    ...: bb = np.random.rand(1000,1000)
    ...: 

In [18]: %timeit ((aa>0.5) & (bb>0.5)).astype(int)
    ...: %timeit np.logical_and(aa>0.5, bb>0.5).astype(int)
    ...: %timeit np.where((aa>0.5) & (bb>0.5),1,0)
    ...: %timeit np.where(np.logical_and(aa>0.5, bb>0.5), 1, 0)
    ...: 
100 loops, best of 3: 9.13 ms per loop
100 loops, best of 3: 9.16 ms per loop
100 loops, best of 3: 10.4 ms per loop
100 loops, best of 3: 10.4 ms per loop

In [19]: %timeit ((aa>0.5) & (bb>0.5)).astype('int8')
    ...: %timeit np.logical_and(aa>0.5, bb>0.5).astype('int8')
    ...: %timeit ((aa>0.5) & (bb>0.5)).astype('uint8')
    ...: %timeit np.logical_and(aa>0.5, bb>0.5).astype('uint8')
    ...: 
    ...: %timeit ((aa>0.5) & (bb>0.5)).astype(np.int8)
    ...: %timeit np.logical_and(aa>0.5, bb>0.5).astype(np.int8)
    ...: %timeit ((aa>0.5) & (bb>0.5)).astype(np.uint8)
    ...: %timeit np.logical_and(aa>0.5, bb>0.5).astype(np.uint8)
    ...: 
100 loops, best of 3: 5.6 ms per loop
100 loops, best of 3: 5.61 ms per loop
100 loops, best of 3: 5.63 ms per loop
100 loops, best of 3: 5.63 ms per loop
100 loops, best of 3: 5.62 ms per loop
100 loops, best of 3: 5.62 ms per loop
100 loops, best of 3: 5.62 ms per loop
100 loops, best of 3: 5.61 ms per loop

In [20]: %timeit 1 * ((aa > 0.5) & (bb > 0.5)) #@BPL's vectorized soln
100 loops, best of 3: 10.2 ms per loop

回答2:

What about this?

import numpy as np

aa = np.random.rand(5, 5)
bb = np.random.rand(5, 5)

print aa
print bb

cc = 1 * ((aa > 0.5) & (bb > 0.5))
print cc

回答3:

when element of aa and bb at index i is exceed than 0.5 then new array have 1 at index i

aa = np.random.rand(5,5)
bb = np.random.rand(5,5)
new_arr = []
for i in range(5):
    for j in range(5):
        if aa[i] >0.5 and bb[i]>0.5:
              new_arr[i] = 1
        else:
              new_arr[i] = "any Value You want