Convert array of indices to one-hot encoded array in NumPy

Question

Given a 1D array of indices:

a = array([1, 0, 3])

I want to one-hot encode this as a 2D array:

b = array([[0,1,0,0], [1,0,0,0], [0,0,0,1]])

Answer 1

Simple example using np.put_along_axis:

x = rng.integers(0, 10, 20)
t = np.zeros([20, 10])
np.put_along_axis(t, indices=np.expand_dims(x, 1), values=1, axis=1)

print(x)
print(t)

[8 6 5 2 3 0 0 0 1 8 6 9 5 6 9 7 6 5 5 9]
[[0. 0. 0. 0. 0. 0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 0. 0. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 0. 0. 0. 0.]
 [0. 0. 1. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
 [1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 1. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 0. 0. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
 [0. 0. 0. 0. 0. 1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
 [0. 0. 0. 0. 0. 0. 0. 1. 0. 0.]
 [0. 0. 0. 0. 0. 0. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 1.]]

Answer 2

For a more general cases where you have a ndarray, you could use numpy broadcasting:

a = array([[[[1, 0, 3]]]]) # (1, 1, 1, 3)
b = (a[..., np.newaxis] == np.arange(np.max(a) + 1)).astype(np.int32)

which would give you:

array([[[[[0, 1, 0, 0],
          [1, 0, 0, 0],
          [0, 0, 0, 1]]]]], dtype=int32)

The result shape is (1, 1, 1, 3, 4).

Answer 3

Create a zeroed array b with enough columns, i.e. a.max() + 1.
Then, for each row i, set the a[i]th column to 1.

>>> a = np.array([1, 0, 3])
>>> b = np.zeros((a.size, a.max() + 1))
>>> b[np.arange(a.size), a] = 1

>>> b
array([[ 0.,  1.,  0.,  0.],
       [ 1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.]])

Answer 4

I find the easiest solution combines np.take and np.eye

def one_hot(x, depth: int):
  return np.take(np.eye(depth), x, axis=0)

works for x of any shape.

Answer 5

def one_hot(n, class_num, col_wise=True):
  a = np.eye(class_num)[n.reshape(-1)]
  return a.T if col_wise else a

# Column for different hot
print(one_hot(np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9, 9, 8, 7]), 10))
# Row for different hot
print(one_hot(np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9, 9, 8, 7]), 10, col_wise=False))

Answer 6

If using tensorflow, there is one_hot():

import tensorflow as tf
import numpy as np

a = np.array([1, 0, 3])
depth = 4
b = tf.one_hot(a, depth)
# <tf.Tensor: shape=(3, 3), dtype=float32, numpy=
# array([[0., 1., 0.],
#        [1., 0., 0.],
#        [0., 0., 0.]], dtype=float32)>

Answer 7

For 1-hot-encoding

   one_hot_encode=pandas.get_dummies(array)

For Example

ENJOY CODING

Answer 8

Using a Neuraxle pipeline step:

1. Set up your example

import numpy as np
a = np.array([1,0,3])
b = np.array([[0,1,0,0], [1,0,0,0], [0,0,0,1]])

2. Do the actual conversion

from neuraxle.steps.numpy import OneHotEncoder
encoder = OneHotEncoder(nb_columns=4)
b_pred = encoder.transform(a)

3. Assert it works

assert b_pred == b

Link to documentation: neuraxle.steps.numpy.OneHotEncoder

Answer 9

You can also use eye function of numpy:

numpy.eye(number of classes)[vector containing the labels]

Answer 10

You can use the following code for converting into a one-hot vector:

let x is the normal class vector having a single column with classes 0 to some number:

import numpy as np
np.eye(x.max()+1)[x]

if 0 is not a class; then remove +1.

Answer 11

>>> values = [1, 0, 3]
>>> n_values = np.max(values) + 1
>>> np.eye(n_values)[values]
array([[ 0.,  1.,  0.,  0.],
       [ 1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.]])

Answer 12

Use the following code. It works best.

def one_hot_encode(x):
"""
    argument
        - x: a list of labels
    return
        - one hot encoding matrix (number of labels, number of class)
"""
encoded = np.zeros((len(x), 10))

for idx, val in enumerate(x):
    encoded[idx][val] = 1

return encoded

Found it here P.S You don't need to go into the link.

Answer 13

• p will be a 2d ndarray.
• We want to know which value is the highest in a row, to put there 1 and everywhere else 0.

clean and easy solution:

max_elements_i = np.expand_dims(np.argmax(p, axis=1), axis=1)
one_hot = np.zeros(p.shape)
np.put_along_axis(one_hot, max_elements_i, 1, axis=1)

Answer 14

Such type of encoding are usually part of numpy array. If you are using a numpy array like this :

a = np.array([1,0,3])

then there is very simple way to convert that to 1-hot encoding

out = (np.arange(4) == a[:,None]).astype(np.float32)

That's it.

Answer 15

Here is what I find useful:

def one_hot(a, num_classes):
  return np.squeeze(np.eye(num_classes)[a.reshape(-1)])

Here num_classes stands for number of classes you have. So if you have a vector with shape of (10000,) this function transforms it to (10000,C). Note that a is zero-indexed, i.e. one_hot(np.array([0, 1]), 2) will give [[1, 0], [0, 1]].

Exactly what you wanted to have I believe.

PS: the source is Sequence models - deeplearning.ai

Answer 16

Here's a dimensionality-independent standalone solution.

This will convert any N-dimensional array arr of nonnegative integers to a one-hot N+1-dimensional array one_hot, where one_hot[i_1,...,i_N,c] = 1 means arr[i_1,...,i_N] = c. You can recover the input via np.argmax(one_hot, -1)

def expand_integer_grid(arr, n_classes):
    """

    :param arr: N dim array of size i_1, ..., i_N
    :param n_classes: C
    :returns: one-hot N+1 dim array of size i_1, ..., i_N, C
    :rtype: ndarray

    """
    one_hot = np.zeros(arr.shape + (n_classes,))
    axes_ranges = [range(arr.shape[i]) for i in range(arr.ndim)]
    flat_grids = [_.ravel() for _ in np.meshgrid(*axes_ranges, indexing='ij')]
    one_hot[flat_grids + [arr.ravel()]] = 1
    assert((one_hot.sum(-1) == 1).all())
    assert(np.allclose(np.argmax(one_hot, -1), arr))
    return one_hot

Answer 17

In case you are using keras, there is a built in utility for that:

from keras.utils.np_utils import to_categorical   

categorical_labels = to_categorical(int_labels, num_classes=3)

And it does pretty much the same as @YXD's answer (see source-code).

Answer 18

I am adding for completion a simple function, using only numpy operators:

   def probs_to_onehot(output_probabilities):
        argmax_indices_array = np.argmax(output_probabilities, axis=1)
        onehot_output_array = np.eye(np.unique(argmax_indices_array).shape[0])[argmax_indices_array.reshape(-1)]
        return onehot_output_array

It takes as input a probability matrix: e.g.:

[[0.03038822 0.65810204 0.16549407 0.3797123 ] ... [0.02771272 0.2760752 0.3280924 0.33458805]]

And it will return

[[0 1 0 0] ... [0 0 0 1]]

Answer 19

I recently ran into a problem of same kind and found said solution which turned out to be only satisfying if you have numbers that go within a certain formation. For example if you want to one-hot encode following list:

all_good_list = [0,1,2,3,4]

go ahead, the posted solutions are already mentioned above. But what if considering this data:

problematic_list = [0,23,12,89,10]

If you do it with methods mentioned above, you will likely end up with 90 one-hot columns. This is because all answers include something like n = np.max(a)+1. I found a more generic solution that worked out for me and wanted to share with you:

import numpy as np
import sklearn
sklb = sklearn.preprocessing.LabelBinarizer()
a = np.asarray([1,2,44,3,2])
n = np.unique(a)
sklb.fit(n)
b = sklb.transform(a)

I hope someone encountered same restrictions on above solutions and this might come in handy

Answer 20

Just to elaborate on the excellent answer from K3---rnc, here is a more generic version:

def onehottify(x, n=None, dtype=float):
    """1-hot encode x with the max value n (computed from data if n is None)."""
    x = np.asarray(x)
    n = np.max(x) + 1 if n is None else n
    return np.eye(n, dtype=dtype)[x]

Also, here is a quick-and-dirty benchmark of this method and a method from the currently accepted answer by YXD (slightly changed, so that they offer the same API except that the latter works only with 1D ndarrays):

def onehottify_only_1d(x, n=None, dtype=float):
    x = np.asarray(x)
    n = np.max(x) + 1 if n is None else n
    b = np.zeros((len(x), n), dtype=dtype)
    b[np.arange(len(x)), x] = 1
    return b

The latter method is ~35% faster (MacBook Pro 13 2015), but the former is more general:

>>> import numpy as np
>>> np.random.seed(42)
>>> a = np.random.randint(0, 9, size=(10_000,))
>>> a
array([6, 3, 7, ..., 5, 8, 6])
>>> %timeit onehottify(a, 10)
188 µs ± 5.03 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> %timeit onehottify_only_1d(a, 10)
139 µs ± 2.78 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Answer 21

Here is an example function that I wrote to do this based upon the answers above and my own use case:

def label_vector_to_one_hot_vector(vector, one_hot_size=10):
    """
    Use to convert a column vector to a 'one-hot' matrix

    Example:
        vector: [[2], [0], [1]]
        one_hot_size: 3
        returns:
            [[ 0.,  0.,  1.],
             [ 1.,  0.,  0.],
             [ 0.,  1.,  0.]]

    Parameters:
        vector (np.array): of size (n, 1) to be converted
        one_hot_size (int) optional: size of 'one-hot' row vector

    Returns:
        np.array size (vector.size, one_hot_size): converted to a 'one-hot' matrix
    """
    squeezed_vector = np.squeeze(vector, axis=-1)

    one_hot = np.zeros((squeezed_vector.size, one_hot_size))

    one_hot[np.arange(squeezed_vector.size), squeezed_vector] = 1

    return one_hot

label_vector_to_one_hot_vector(vector=[[2], [0], [1]], one_hot_size=3)

Answer 22

You can use sklearn.preprocessing.LabelBinarizer:

Example:

import sklearn.preprocessing
a = [1,0,3]
label_binarizer = sklearn.preprocessing.LabelBinarizer()
label_binarizer.fit(range(max(a)+1))
b = label_binarizer.transform(a)
print('{0}'.format(b))

output:

[[0 1 0 0]
 [1 0 0 0]
 [0 0 0 1]]

Amongst other things, you may initialize sklearn.preprocessing.LabelBinarizer() so that the output of transform is sparse.

Answer 23

I think the short answer is no. For a more generic case in n dimensions, I came up with this:

# For 2-dimensional data, 4 values
a = np.array([[0, 1, 2], [3, 2, 1]])
z = np.zeros(list(a.shape) + [4])
z[list(np.indices(z.shape[:-1])) + [a]] = 1

I am wondering if there is a better solution -- I don't like that I have to create those lists in the last two lines. Anyway, I did some measurements with timeit and it seems that the numpy-based (indices/arange) and the iterative versions perform about the same.

Answer 24

Here is a function that converts a 1-D vector to a 2-D one-hot array.

#!/usr/bin/env python
import numpy as np

def convertToOneHot(vector, num_classes=None):
    """
    Converts an input 1-D vector of integers into an output
    2-D array of one-hot vectors, where an i'th input value
    of j will set a '1' in the i'th row, j'th column of the
    output array.

    Example:
        v = np.array((1, 0, 4))
        one_hot_v = convertToOneHot(v)
        print one_hot_v

        [[0 1 0 0 0]
         [1 0 0 0 0]
         [0 0 0 0 1]]
    """

    assert isinstance(vector, np.ndarray)
    assert len(vector) > 0

    if num_classes is None:
        num_classes = np.max(vector)+1
    else:
        assert num_classes > 0
        assert num_classes >= np.max(vector)

    result = np.zeros(shape=(len(vector), num_classes))
    result[np.arange(len(vector)), vector] = 1
    return result.astype(int)

Below is some example usage:

>>> a = np.array([1, 0, 3])

>>> convertToOneHot(a)
array([[0, 1, 0, 0],
       [1, 0, 0, 0],
       [0, 0, 0, 1]])

>>> convertToOneHot(a, num_classes=10)
array([[0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0, 0, 0, 0]])