Let's say I have a 1d numpy array
a = array([1,0,3])I would like to encode this as a 2D one-hot array
b = array([[0,1,0,0], [1,0,0,0], [0,0,0,1]])Is there a quick way to do this? Quicker than just looping over a to set elements of b, that is.
22 Answers
Your array a defines the columns of the nonzero elements in the output array. You need to also define the rows and then use fancy indexing:
>>> 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.]])>>> 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.]])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).
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
1You can also use eye function of numpy:
numpy.eye(number of classes)[vector containing the labels]
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.
For 1-hot-encoding
one_hot_encode=pandas.get_dummies(array)ENJOY CODING
4You 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.
1Here 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]])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]] = 1I 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.
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 bThe 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)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))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
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.
- 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)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.
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)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_arrayIt 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]]
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_hotUse 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 encodedFound it here P.S You don't need to go into the link.
4Using a Neuraxle pipeline step:
- 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]])- Do the actual conversion
from neuraxle.steps.numpy import OneHotEncoder
encoder = OneHotEncoder(nb_columns=4)
b_pred = encoder.transform(a)- Assert it works
assert b_pred == bLink to documentation: neuraxle.steps.numpy.OneHotEncoder
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)>
