给定一个一维下标数组:

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

我想把它编码成一个2D数组:

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

当前回答

对于1-hot-encoding

   one_hot_encode=pandas.get_dummies(array)

例如

享受编码

其他回答

使用下面的代码。这样效果最好。

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

在这里找到了p.s.你不需要进入链接。

下面是我根据上面的答案和我自己的用例写的一个示例函数:

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)
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))

如果使用tensorflow,则存在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)>

创建一个有足够列的零数组b,即a.max() + 1。 然后,对于每一行i,设置第a[i]列为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.]])