# 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
, 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

``````A = [0 if i <=0.5 else 1 for i in A]
``````

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
``````

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]])
``````

You can make the built-in `round` function element wise use `np.vectorize`.
``````import numpy as np