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
4 answers

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
and1
only.np.where
is better if need convert to another scalars like5
and10
ora
andb
: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. Thendarray.astype
method then converts theTrue
values to 1 and theFalse
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 builtin
round
function element wise usenp.vectorize
.import numpy as np element_wise_round = np.vectorize(round, otypes=[np.int]) print element_wise_round(A)