One Hot Encoding using numpy [duplicate]

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?

Solution 1:

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)

Solution 2:

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()

Solution 3:

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]

Solution 4:

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