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A = np.array([[0.94366988, 0.86095311, 0.88896715, 0.93630641, 0.74075403, 0.52849619 , 0.03094677, 0.85707681, 0.88457925, 0.67279696, 0.26601085, 0.4823794 , 0.74741157, 0.78575729, 0.00978911, 0.9203284, 0.02453695, 0.84884703 , 0.2050248, 0.03703224, 0.92931392, 0.11930532, 0.01411064, 0.7832698 , 0.58188015, 0.66897565, 0.75119007, 0.01323558, 0.03402649, 0.99735115 , 0.21031727, 0.78123225, 0.6815842, 0.46647604, 0.66323375, 0.03424828 , 0.08031627, 0.76570656, 0.34760863, 0.06177743, 0.6987531, 0.4106426 , 0.6648871, 0.02776868, 0.93053125, 0.46395717, 0.23971605, 0.9771735 , 0.66202407, 0.10482388]])

Convert the entries of a into 0 (if activation <= 0.5) or 1 (if activation > 0.5)

for i in range(A.shape[1]): if A[i]>0.5: Y_prediction[i] = 1 else: Y_prediction[i] = 0

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

And how to use vectorize this thx

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1

7 Answers 7

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71

I think you need vectorized function np.where:

B = np.where(A > 0.5, 1, 0) print (B) [[1 1 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 0 0 1 1 1 1 0 0 1 0 1 1 0 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0]] 
B = np.where(A <= 0.5, 0, 1) print (B) [[1 1 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 0 0 1 1 1 1 0 0 1 0 1 1 0 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0]]

But better is holdenweb solution if need convert to 0 and 1 only.

np.where is better if need convert to another scalars like 5 and 10 or a and b:

C = np.where(A > 0.5, 5, 10) print (C) [[ 5 5 5 5 5 5 10 5 5 5 10 10 5 5 10 5 10 5 10 10 5 10 10 5 5 5 5 10 10 5 10 5 5 10 5 10 10 5 10 10 5 10 5 10 5 10 10 5 5 10]] D = np.where(A > 0.5, 'a', 'b') print (D) [['a' 'a' 'a' 'a' 'a' 'a' 'b' 'a' 'a' 'a' 'b' 'b' 'a' 'a' 'b' 'a' 'b' 'a' 'b' 'b' 'a' 'b' 'b' 'a' 'a' 'a' 'a' 'b' 'b' 'a' 'b' 'a' 'a' 'b' 'a' 'b' 'b' 'a' 'b' 'b' 'a' 'b' 'a' 'b' 'a' 'b' 'b' 'a' 'a' 'b']]

Timings:

np.random.seed(223) A = np.random.rand(1,1000000) #jez In [64]: %timeit np.where(A > 0.5, 1, 0) 100 loops, best of 3: 7.58 ms per loop #holdenweb In [65]: %timeit (A > 0.5).astype(int) 100 loops, best of 3: 3.47 ms per loop #stamaimer In [66]: %timeit element_wise_round(A) 1 loop, best of 3: 318 ms per loop
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Good answer. np.where is my goto method for this. I prefer it over broadcasting to bools and converting to int, it's just more readable imo and applicable to other scalars and datatypes as you show.
34

Standard numpy broadcasting can be used to compare each element with a scalar value, yielding a Boolean for each element. The ndarray.astype method then converts the True values to 1 and the False values to zero.

In [16]: (A > 0.5).astype(int) Out[16]: array([[1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0]])
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Doesn't seem to work.
I assure you it does. Just checked with Python 3.9 and numpy 1.21.2. Not a terribly information-packed bug report, so I regret I'm unable to help further.
While checking I also checked the timing, and this method appears (on my computer!) to run about twice as fast as the most-upvoted answer. Not that speed is everything.
I've checked it both within my local Python install (also running 3.9) and here, and it didn't work on a numpy array for me.
Again, I have no information that might help me to help you solve my problem. This is like "Help. My car is broken. Tell me what I need to do to fix it." I used the numpy array given in the question. I'm sure there are a million things it wouldn't work on, but none of them will be numpy arrays of real numbers.
3 
np.random.seed(223) A = np.random.rand(1000000) A = [0 if i <=0.5 else 1 for i in A]
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Thx！can you tell me how to implement use for loop?
this is a list comprehension, which is much faster than using for loops in python. You don't need to loop to get the result. @jezrael 's answer is also a good method.
@Tkanno - I cannot add your solution for timeit, because ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() :(
@jezrael that's because you've declared the array A shape (1,1000000). my method works on a flattened array, another limitation. if you flatten A you can timeit on my method.
@Tkanno - I try set input by data similar OP. But in my opinion here list comprehension is slowiest, because not vectorized. Is possible update your solution working with A? Then I can add timings too (if want)
2

The problem with your code is the dimension of A[i] in your code. A is initialised as matrix[1,n] (1 row and n columns - you have double [[]] brackets in np.array([[]])), so you reference of type A[i] actually means "the whole i'th row, or an array(n) in this case). I'm quite new to python and numpy, but to access individual cells and set each to 1 or 0 could be coded this way (of course, only if you are forced to use FOR loops here):

 for i in range(A.shape[1]): if A[0,i]>0.5: Y_prediction[0,i] = 1 else: Y_prediction[0,i] = 0

But,definitely, the best solution, as others described above, is to avoid for loop.

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1

this may solve your issue

a=1 b=0 predict=list() for i in A: if i>0.5: predict.append(a) else: predict.append(b) print(predict)

we first assign value a to 0 and b to 1 but you can directly use 1 and 0 on append function 'A' contains the list in form of probability from which we assign the new list name predict and after comparing each items in list we put value into list by using append methord predict list contain your ans

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0

You can make the built-in round function element wise use np.vectorize.

import numpy as np element_wise_round = np.vectorize(round, otypes=[np.int]) print element_wise_round(A)
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0 
A1=A-0.5 A2=np.sign(A1)

Now you have 1 and -1. If you do not have 0 in the original array, you can convert the -1 to 0 to have only 1 and 0 array:

indices = A2 == -1 A2[indices] = 0

The result:

A2

array([[1., 1., 1., 1., 1., 1., 0., 1., 1., 1., 0., 0., 1., 1., 0., 1., 0., 1., 0., 0., 1., 0., 0., 1., 1., 1., 1., 0., 0., 1., 0., 1., 1., 0., 1., 0., 0., 1., 0., 0., 1., 0., 1., 0., 1., 0., 0., 1., 1., 0.]])

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