Convert Numpy Array To 0 Or 1

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

Solution 1:

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

Solution 2:

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]])

Solution 3:

np.random.seed(223)
A = np.random.rand(1000000)
A = [0if i <=0.5else1for i in A]

Solution 4:

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] =1else:
    Y_prediction[0,i] = 0

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

Solution 5:

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)