인덱스 배열을 1-hot 인코딩 된 numpy 배열로 변환

인덱스 배열을 1-hot 인코딩 된 numpy 배열로 변환

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1d numpy 배열이 있다고 가정 해 봅시다.

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

이것을 2d 1-hot 어레이로 인코딩하고 싶습니다.

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

이 작업을 수행하는 빠른 방법이 있습니까? 의 a요소를 설정하기 위해 반복하는 것보다 빠릅니다 b.

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배열 a은 출력 배열에서 0이 아닌 요소의 열을 정의합니다. 또한 행을 정의한 다음 멋진 색인 작성을 사용해야합니다.

>>> a = np.array([1, 0, 3])
>>> b = np.zeros((3, 4))
>>> b[np.arange(3), a] = 1
>>> b
array([[ 0.,  1.,  0.,  0.],
       [ 1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.]])

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>>> 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.]])

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여기 내가 유용하다고 생각하는 것이 있습니다.

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

여기 num_classes당신이 가진 수업의 수를 나타냅니다. 따라서 (10000)a 모양의 벡터가 있으면 이 함수 는 벡터 를 (10000, C)로 변환합니다 . 주 제로 색인은, 즉 줄 것이다 .aone_hot(np.array([0, 1]), 2)[[1, 0], [0, 1]]

정확히 당신이 믿고 싶은 것.

추신 : 소스는 시퀀스 모델입니다-deeplearning.ai

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당신은 사용할 수 있습니다 sklearn.preprocessing.LabelBinarizer:

예:

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))

산출:

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

무엇보다도 sklearn.preprocessing.LabelBinarizer()출력 transform이 희박 하도록 초기화 할 수 있습니다 .

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keras를 사용하는 경우이를위한 내장 유틸리티가 있습니다.

from keras.utils.np_utils import to_categorical   

categorical_labels = to_categorical(int_labels, num_classes=3)

그리고 @YXD의 답변 과 거의 동일합니다 ( source-code 참조 ).

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numpy의 눈 기능을 사용할 수도 있습니다 .

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

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다음은 1-D 벡터를 2D one-hot array로 변환하는 함수입니다.

#!/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)

다음은 사용법 예입니다.

>>> 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]])

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나는 짧은 대답이 아니오라고 생각합니다. 보다 일반적인 n차원의 경우 , 나는 이것을 생각해 냈습니다.

# 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.

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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)

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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.

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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

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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.

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• 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)

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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)

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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]]

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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

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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.

참고URL : https://stackoverflow.com/questions/29831489/convert-array-of-indices-to-1-hot-encoded-numpy-array