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))

你也可以使用numpy的eye函数:

numpy。眼(类数)[包含标签的向量]

下面是一个将一维向量转换为二维单热数组的函数。

#!/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)

下面是一些用法示例:

>>> 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]])

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))

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)

我认为简短的答案是否定的。对于n维的更一般的情况，我想到了这个:

# 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

我想知道是否有更好的解决方案——我不喜欢我必须在最后两行创建这些列表。不管怎样，我用timeit做了一些测量，似乎基于numpy的(索引/范围)和迭代版本的表现是一样的。