Question

我想将一个句子转换为一个单热矢量数组。这些向量将是字母表的一个热门表示。它看起来如下：

"hello" # h=7, e=4 l=11 o=14

会变成

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

不幸的是，sklearn的OneHotEncoder不接受输入字符串。

Answer 1

这是回归神经网络中的一项常见任务，如果你想使用它，tensorflow就是为了这个目的而有一个特定的功能。

In [24]: alphabets = {'a' : 0, 'b': 1, 'c':2, 'd':3, 'e':4, 'f':5, 'g':6, 'h':7, 'i':8, 'j':9, 'k':10, 'l':11, 'm':12, 'n':13, 'o':14}

In [25]: idxs = [alphabets[ch] for ch in 'hello']

In [26]: idxs
Out[26]: [7, 4, 11, 11, 14]

# @divakar's approach
In [153]: idxs = np.fromstring("hello",dtype=np.uint8)-97

In [13]: one_hot = tf.one_hot(idxs, 26, dtype=tf.uint8)

In [14]: sess = tf.InteractiveSession()

In [15]: one_hot.eval()
Out[15]: 
array([[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0],
       [0, 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]], dtype=uint8)

Answer 2

只需将传递的字符串中的字母与给定的字母表进行比较：

def string_vectorizer(strng, alphabet=string.ascii_lowercase):
    vector = [[0 if char != letter else 1 for char in alphabet] 
                  for letter in strng]
    return vector

请注意，使用自定义字母表（例如＆＃34; defbcazk＆＃34;时，将按照每个元素出现在原始列表中的顺序排列列。）

string_vectorizer('hello')的输出：

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

Answer 3

你问过＆＃34;句子＆＃34;但是你的例子只提供了一个单词，所以我不确定你想对空格做些什么。但就单个单词而言，您的示例可以通过以下方式实现：

def onehot(ltr):
 return [1 if i==ord(ltr) else 0 for i in range(97,123)]

def onehotvec(s):
 return [onehot(c) for c in list(s.lower())]

onehotvec("hello")
[[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0],
 [0, 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

Answer 4

使用pandas，您可以通过传递分类系列来使用pd.get_dummies：

<payloadFactory media-type="xml">
    <format>
        <book category="cooking">
            <title lang="en">$1</title>
            <author>$2</author> 
            <year>$3</year>
        </book>
    </format>
    <args>
        <arg evaluator="xml" expression="//book/title/text()"/>
        <arg evaluator="xml" expression="//book/author/text()"/>
        <arg evaluator="xml" expression="//book/year/text()"/>
    </args>
</payloadFactory>

Answer 5

这是一种使用NumPy broadcasting为我们提供(N,26)形数组的矢量化方法 -

ints = np.fromstring("hello",dtype=np.uint8)-97
out = (ints[:,None] == np.arange(26)).astype(int)

如果您正在寻找性能，我建议使用初始化数组然后分配 -

out = np.zeros((len(ints),26),dtype=int)
out[np.arange(len(ints)), ints] = 1

示例运行 -

In [153]: ints = np.fromstring("hello",dtype=np.uint8)-97

In [154]: ints
Out[154]: array([ 7,  4, 11, 11, 14], dtype=uint8)

In [155]: out = (ints[:,None] == np.arange(26)).astype(int)

In [156]: print out
[[0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 0]
 [0 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 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 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]]

如何在角色级别对单个热门编码？

5 个答案: