One Hot Encoding using numpy
Question:
If the input is zero I want to make an array which looks like this:
[1,0,0,0,0,0,0,0,0,0]
and if the input is 5:
[0,0,0,0,0,1,0,0,0,0]
For the above I wrote:
np.put(np.zeros(10),5,1)
but it did not work.
Is there any way in which, this can be implemented in one line?
Answers:
Something like :
np.array([int(i == 5) for i in range(10)])
Should do the trick.
But I suppose there exist other solutions using numpy.
edit : the reason why your formula does not work : np.put does not return anything, it just modifies the element given in first parameter. The good answer while using np.put() is :
a = np.zeros(10)
np.put(a,5,1)
The problem is that it can’t be done in one line, as you need to define the array before passing it to np.put()
The problem here is that you save your array nowhere. The put function works in place on the array and returns nothing. Since you never give your array a name you can not address it later. So this
one_pos = 5
x = np.zeros(10)
np.put(x, one_pos, 1)
would work, but then you could just use indexing:
one_pos = 5
x = np.zeros(10)
x[one_pos] = 1
In my opinion that would be the correct way to do this if no special reason exists to do this as a one liner. This might also be easier to read and readable code is good code.
Taking a quick look at the manual, you will see that np.put does not return a value. While your technique is fine, you are accessing None instead of your result array.
For a 1-D array it is better to just use direct indexing, especially for such a simple case.
Here is how to rewrite your code with minimal modification:
arr = np.zeros(10)
np.put(arr, 5, 1)
Here is how to do the second line with indexing instead of put:
arr[5] = 1
The np.put mutates its array arg in-place. It’s conventional in Python for functions / methods that perform in-place mutation to return None; np.put adheres to that convention. So if a is a 1D array and you do
a = np.put(a, 5, 1)
then a will get replaced by None.
Your code is similar to that, but it passes an un-named array to np.put.
A compact & efficient way to do what you want is with a simple function, eg:
import numpy as np
def one_hot(i):
a = np.zeros(10, 'uint8')
a[i] = 1
return a
a = one_hot(5)
print(a)
output
[0 0 0 0 0 1 0 0 0 0]
import time
start_time = time.time()
z=[]
for l in [1,2,3,4,5,6,1,2,3,4,4,6,]:
a= np.repeat(0,10)
np.put(a,l,1)
z.append(a)
print("--- %s seconds ---" % (time.time() - start_time))
#--- 0.00174784660339 seconds ---
import time
start_time = time.time()
z=[]
for l in [1,2,3,4,5,6,1,2,3,4,4,6,]:
z.append(np.array([int(i == l) for i in range(10)]))
print("--- %s seconds ---" % (time.time() - start_time))
#--- 0.000400066375732 seconds ---
Use np.identity or np.eye. You can try something like this with your input i, and the array size s:
np.identity(s)[i:i+1]
For example, print(np.identity(5)[0:1]) will result:
[[ 1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]
If you are using TensorFlow, you can use tf.one_hot: https://www.tensorflow.org/api_docs/python/array_ops/slicing_and_joining#one_hot
Usually, when you want to get a one-hot encoding for classification in machine learning, you have an array of indices.
import numpy as np
nb_classes = 6
targets = np.array([[2, 3, 4, 0]]).reshape(-1)
one_hot_targets = np.eye(nb_classes)[targets]
The one_hot_targets is now
array([[[ 0., 0., 1., 0., 0., 0.],
[ 0., 0., 0., 1., 0., 0.],
[ 0., 0., 0., 0., 1., 0.],
[ 1., 0., 0., 0., 0., 0.]]])
The .reshape(-1) is there to make sure you have the right labels format (you might also have [[2], [3], [4], [0]]). The -1 is a special value which means “put all remaining stuff in this dimension”. As there is only one, it flattens the array.
Copy-Paste solution
def get_one_hot(targets, nb_classes):
res = np.eye(nb_classes)[np.array(targets).reshape(-1)]
return res.reshape(list(targets.shape)+[nb_classes])
Package
You can use mpu.ml.indices2one_hot. It’s tested and simple to use:
import mpu.ml
one_hot = mpu.ml.indices2one_hot([1, 3, 0], nb_classes=5)
I’m not sure the performance, but the following code works and it’s neat.
x = np.array([0, 5])
x_onehot = np.identity(6)[x]
You could use List comprehension:
[0 if i !=5 else 1 for i in range(10)]
turns to
[0,0,0,0,0,1,0,0,0,0]
If the input is zero I want to make an array which looks like this:
[1,0,0,0,0,0,0,0,0,0]
and if the input is 5:
[0,0,0,0,0,1,0,0,0,0]
For the above I wrote:
np.put(np.zeros(10),5,1)
but it did not work.
Is there any way in which, this can be implemented in one line?
Something like :
np.array([int(i == 5) for i in range(10)])
Should do the trick.
But I suppose there exist other solutions using numpy.
edit : the reason why your formula does not work : np.put does not return anything, it just modifies the element given in first parameter. The good answer while using np.put() is :
a = np.zeros(10)
np.put(a,5,1)
The problem is that it can’t be done in one line, as you need to define the array before passing it to np.put()
The problem here is that you save your array nowhere. The put function works in place on the array and returns nothing. Since you never give your array a name you can not address it later. So this
one_pos = 5
x = np.zeros(10)
np.put(x, one_pos, 1)
would work, but then you could just use indexing:
one_pos = 5
x = np.zeros(10)
x[one_pos] = 1
In my opinion that would be the correct way to do this if no special reason exists to do this as a one liner. This might also be easier to read and readable code is good code.
Taking a quick look at the manual, you will see that np.put does not return a value. While your technique is fine, you are accessing None instead of your result array.
For a 1-D array it is better to just use direct indexing, especially for such a simple case.
Here is how to rewrite your code with minimal modification:
arr = np.zeros(10)
np.put(arr, 5, 1)
Here is how to do the second line with indexing instead of put:
arr[5] = 1
The np.put mutates its array arg in-place. It’s conventional in Python for functions / methods that perform in-place mutation to return None; np.put adheres to that convention. So if a is a 1D array and you do
a = np.put(a, 5, 1)
then a will get replaced by None.
Your code is similar to that, but it passes an un-named array to np.put.
A compact & efficient way to do what you want is with a simple function, eg:
import numpy as np
def one_hot(i):
a = np.zeros(10, 'uint8')
a[i] = 1
return a
a = one_hot(5)
print(a)
output
[0 0 0 0 0 1 0 0 0 0]
import time
start_time = time.time()
z=[]
for l in [1,2,3,4,5,6,1,2,3,4,4,6,]:
a= np.repeat(0,10)
np.put(a,l,1)
z.append(a)
print("--- %s seconds ---" % (time.time() - start_time))
#--- 0.00174784660339 seconds ---
import time
start_time = time.time()
z=[]
for l in [1,2,3,4,5,6,1,2,3,4,4,6,]:
z.append(np.array([int(i == l) for i in range(10)]))
print("--- %s seconds ---" % (time.time() - start_time))
#--- 0.000400066375732 seconds ---
Use np.identity or np.eye. You can try something like this with your input i, and the array size s:
np.identity(s)[i:i+1]
For example, print(np.identity(5)[0:1]) will result:
[[ 1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]
If you are using TensorFlow, you can use tf.one_hot: https://www.tensorflow.org/api_docs/python/array_ops/slicing_and_joining#one_hot
Usually, when you want to get a one-hot encoding for classification in machine learning, you have an array of indices.
import numpy as np
nb_classes = 6
targets = np.array([[2, 3, 4, 0]]).reshape(-1)
one_hot_targets = np.eye(nb_classes)[targets]
The one_hot_targets is now
array([[[ 0., 0., 1., 0., 0., 0.],
[ 0., 0., 0., 1., 0., 0.],
[ 0., 0., 0., 0., 1., 0.],
[ 1., 0., 0., 0., 0., 0.]]])
The .reshape(-1) is there to make sure you have the right labels format (you might also have [[2], [3], [4], [0]]). The -1 is a special value which means “put all remaining stuff in this dimension”. As there is only one, it flattens the array.
Copy-Paste solution
def get_one_hot(targets, nb_classes):
res = np.eye(nb_classes)[np.array(targets).reshape(-1)]
return res.reshape(list(targets.shape)+[nb_classes])
Package
You can use mpu.ml.indices2one_hot. It’s tested and simple to use:
import mpu.ml
one_hot = mpu.ml.indices2one_hot([1, 3, 0], nb_classes=5)
I’m not sure the performance, but the following code works and it’s neat.
x = np.array([0, 5])
x_onehot = np.identity(6)[x]
You could use List comprehension:
[0 if i !=5 else 1 for i in range(10)]
turns to
[0,0,0,0,0,1,0,0,0,0]