One Hot Encoding using numpy

问题:

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?

回答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)

回答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()

回答3:

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

回答4:

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.

回答5:

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]

回答6:

Use np.identify or np.eye. You can try something like this with your input i, and the array size s:

np.identify(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

回答7:

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]

回答8:

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 ---

回答9:

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]