blocks|key|5440|text|数组a定义了输出数组中非零元素的列。您还需要定义行，然后使用花哨的索引：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5441|>>>+a+=+np.array([1,+0,+3])
>>>+b+=+np.zeros((a.size,+a.max()%2B1))
>>>+b[np.arange(a.size),a]+=+1
>>>+b
array([[+0.,++1.,++0.,++0.],
+++++++[+1.,++0.,++0.,++0.],
+++++++[+0.,++0.,++0.,++1.]])|code-block|syntax|javascript|5442|entityMap^0|2|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Your array <code>a</code> defines the columns of the nonzero elements in the output array. You need to also define the rows and then use fancy indexing:

<pre><code>&gt;&gt;&gt; a = np.array([1, 0, 3])
&gt;&gt;&gt; b = np.zeros((a.size, a.max()+1))
&gt;&gt;&gt; b[np.arange(a.size),a] = 1
&gt;&gt;&gt; b
array([[ 0., 1., 0., 0.],
 [ 1., 0., 0., 0.],
 [ 0., 0., 0., 1.]])
</code></pre>

blocks|key|5452|text|>>>+values+=+[1,+0,+3]
>>>+n_values+=+np.max(values)+%2B+1
>>>+np.eye(n_values)[values]
array([[+0.,++1.,++0.,++0.],
+++++++[+1.,++0.,++0.,++0.],
+++++++[+0.,++0.,++0.,++1.]])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5453|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>&gt;&gt;&gt; values = [1, 0, 3]
&gt;&gt;&gt; n_values = np.max(values) + 1
&gt;&gt;&gt; np.eye(n_values)[values]
array([[ 0., 1., 0., 0.],
 [ 1., 0., 0., 0.],
 [ 0., 0., 0., 1.]])
</code></pre>

blocks|key|5508|text|下面是一个将一维向量转换为二维一热数组的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5509|#!/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)%2B1
++++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)|code-block|syntax|javascript|5510|下面是一些示例用法：|5511|>>>+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]])|5512|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here is a function that converts a 1-D vector to a 2-D one-hot array.

<pre><code>#!/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) &gt; 0

 if num_classes is None:
 num_classes = np.max(vector)+1
 else:
 assert num_classes &gt; 0
 assert num_classes &gt;= np.max(vector)

 result = np.zeros(shape=(len(vector), num_classes))
 result[np.arange(len(vector)), vector] = 1
 return result.astype(int)
</code></pre>

Below is some example usage:

<pre><code>&gt;&gt;&gt; a = np.array([1, 0, 3])

&gt;&gt;&gt; convertToOneHot(a)
array([[0, 1, 0, 0],
 [1, 0, 0, 0],
 [0, 0, 0, 1]])

&gt;&gt;&gt; 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]])
</code></pre>

blocks|key|4513092|text|我认为简短的答案是否定的。对于n维度中更一般的情况，我提出了以下内容：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4513093|#+For+2-dimensional+data,+4+values
a+=+np.array([[0,+1,+2],+[3,+2,+1]])
z+=+np.zeros(list(a.shape)+%2B+[4])
z[list(np.indices(z.shape[:-1]))+%2B+[a]]+=+1|code-block|syntax|javascript|4513094|我想知道是否有更好的解决方案--我不喜欢在最后两行中创建这些列表。无论如何，我用timeit做了一些测量，看起来numpy-based+(indices/arange)和迭代版本的性能大致相同。|4513095|entityMap^0|F|1|0|0|14|6|1K|5|1X|7|25|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

I think the short answer is no. For a more generic case in <code>n</code> dimensions, I came up with this:

<pre><code># 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
</code></pre>

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 <code>timeit</code> and it seems that the <code>numpy</code>-based (<code>indices</code>/<code>arange</code>) and the iterative versions perform about the same.

blocks|key|4513118|text|您可以使用sklearn.preprocessing.LabelBinarizer|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4513119|示例：|4513120|import+sklearn.preprocessing
a+=+[1,0,3]
label_binarizer+=+sklearn.preprocessing.LabelBinarizer()
label_binarizer.fit(range(max(a)%2B1))
b+=+label_binarizer.transform(a)
print('{0}'.format(b))|code-block|syntax|javascript|4513121|输出：|4513122|[[0+1+0+0]
+[1+0+0+0]
+[0+0+0+1]]|4513123|其中，您可以初始化sklearn.preprocessing.LabelBinarizer()，以便transform的输出是稀疏的。|4513124|entityMap|0|LINK|mutability|MUTABLE|url|http://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.LabelBinarizer.html#sklearn.preprocessing.LabelBinarizer^0|5|10|5|10|0|0|0|0|0|0|9|12|1E|9|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|17|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|18|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

You can use <a href="http://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.LabelBinarizer.html#sklearn.preprocessing.LabelBinarizer" rel="noreferrer"><code>sklearn.preprocessing.LabelBinarizer</code></a>:

Example:

<pre><code>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))
</code></pre>

output:

<pre><code>[[0 1 0 0]
 [1 0 0 0]
 [0 0 0 1]]
</code></pre>

Amongst other things, you may initialize <code>sklearn.preprocessing.LabelBinarizer()</code> so that the output of <code>transform</code> is sparse.

blocks|key|5666|text|如果你正在使用keras，有一个内置的实用程序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5667|from+keras.utils.np_utils+import+to_categorical+++

categorical_labels+=+to_categorical(int_labels,+num_classes=3)|code-block|syntax|javascript|5668|它的功能与@YXD's+answer+(请参阅source-code)非常相似。|offset|length|5669|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/29831596/4228275|1|https://github.com/keras-team/keras/blob/2d183db0372e5ac2a686608cb9da0a9bd4319764/keras/utils/np_utils.py#L9^0|0|0|5|D|0|N|B|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]|$I|10|J|11|1|12]]|A|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]|S|$5|N|O|P|A|$Q|T]]]]

In case you are using keras, there is a built in utility for that:

<pre><code>from keras.utils.np_utils import to_categorical 

categorical_labels = to_categorical(int_labels, num_classes=3)
</code></pre>

And it does pretty much the same as <a href="https://stackoverflow.com/a/29831596/4228275">@YXD's answer</a> (see <a href="https://github.com/keras-team/keras/blob/2d183db0372e5ac2a686608cb9da0a9bd4319764/keras/utils/np_utils.py#L9" rel="noreferrer">source-code</a>).

blocks|key|5702|text|以下是我根据上面的答案和我自己的用例编写的一个示例函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5703|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)|code-block|syntax|javascript|5704|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is an example function that I wrote to do this based upon the answers above and my own use case:

<pre><code>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)
</code></pre>

blocks|key|4513233|text|为了详细说明来自K3---rnc的excellent+answer，下面是一个更通用的版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4513234|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)+%2B+1+if+n+is+None+else+n
++++return+np.eye(n,+dtype=dtype)[x]|code-block|syntax|javascript|4513235|此外，下面是该方法的快速基准测试和YXD的currently+accepted+answer方法(略有更改，因此它们提供了相同的API，但后者仅适用于1Dndarray)：|4513236|def+onehottify_only_1d(x,+n=None,+dtype=float):
++++x+=+np.asarray(x)
++++n+=+np.max(x)+%2B+1+if+n+is+None+else+n
++++b+=+np.zeros((len(x),+n),+dtype=dtype)
++++b[np.arange(len(x)),+x]+=+1
++++return+b|4513237|后一种方法快约35%25+(MacBook专业版13+2015)，但前者更通用：|4513238|>>>+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])
>>>+%25timeit+onehottify(a,+10)
188+µs+±+5.03+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+10000+loops+each)
>>>+%25timeit+onehottify_only_1d(a,+10)
139+µs+±+2.78+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+10000+loops+each)|4513239|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/1090455/k3-rnc|1|https://stackoverflow.com/a/39480638/2054455|2|https://stackoverflow.com/users/553404/yxd|3|https://stackoverflow.com/a/29831596/2054455^0|8|8|0|H|G|1|0|0|H|3|2|L|P|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]|$A|18|B|19|1|1A]]|C|$]]|$1|D|3|E|5|F|7|1B|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]|$A|1G|B|1H|1|1I]]|C|$]]|$1|K|3|L|5|F|7|1J|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1K|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1L|8|@]|9|@]|C|$G|H]]|$1|Q|3|-4|5|6|7|1M|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]|10|$5|T|U|V|C|$W|11]]|12|$5|T|U|V|C|$W|13]]]]

Just to elaborate on the <a href="https://stackoverflow.com/a/39480638/2054455">excellent answer</a> from <a href="https://stackoverflow.com/users/1090455/k3-rnc">K3---rnc</a>, here is a more generic version:

<pre><code>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]
</code></pre>

Also, here is a quick-and-dirty benchmark of this method and a method from the <a href="https://stackoverflow.com/a/29831596/2054455">currently accepted answer</a> by <a href="https://stackoverflow.com/users/553404/yxd">YXD</a> (slightly changed, so that they offer the same API except that the latter works only with 1D ndarrays):

<pre><code>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
</code></pre>

The latter method is ~35% faster (MacBook Pro 13 2015), but the former is more general:

<pre><code>&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; np.random.seed(42)
&gt;&gt;&gt; a = np.random.randint(0, 9, size=(10_000,))
&gt;&gt;&gt; a
array([6, 3, 7, ..., 5, 8, 6])
&gt;&gt;&gt; %timeit onehottify(a, 10)
188 µs ± 5.03 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
&gt;&gt;&gt; %timeit onehottify_only_1d(a, 10)
139 µs ± 2.78 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
</code></pre>

blocks|key|5788|text|我最近遇到了一个类似的问题，并找到了上述解决方案，结果证明，只有当你的数字在一定的形式内时，它才能令人满意。例如，如果您想要对以下列表进行单热编码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5789|all_good_list+=+[0,1,2,3,4]|code-block|syntax|javascript|5790|请继续，上面已经提到了发布的解决方案。但是，如果考虑到这些数据：|5791|problematic_list+=+[0,23,12,89,10]|5792|如果你使用上面提到的方法，你最终可能会得到90个单热点列。这是因为所有答案都包含类似n+=+np.max(a)%2B1的内容。我找到了一个更通用的解决方案，它适合我，并想与你分享：|offset|length|style|CODE|5793|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)|5794|我希望有人在上面的解决方案上遇到了同样的限制，这可能会派上用场。|5795|entityMap^0|0|0|0|0|16|F|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

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:

<pre><code>all_good_list = [0,1,2,3,4]
</code></pre>

go ahead, the posted solutions are already mentioned above. But what if considering this data:

<pre><code>problematic_list = [0,23,12,89,10]
</code></pre>

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 <code>n = np.max(a)+1</code>. I found a more generic solution that worked out for me and wanted to share with you:

<pre><code>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)
</code></pre>

I hope someone encountered same restrictions on above solutions and this might come in handy

blocks|key|4513322|text|您也可以使用numpy的eye函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4513323|numpy.eye(number+of+classes)[vector+containing+the+labels]|style|CODE|4513324|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.eye.html^0|C|3|0|0|0|1M|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@$A|U|B|V|F|G]]|9|@]|C|$]]|$1|H|3|-4|5|6|7|W|8|@]|9|@]|C|$]]]|I|$J|$5|K|L|M|C|$N|O]]]]

You can also use <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.eye.html" rel="noreferrer">eye</a> function of numpy:

<code>numpy.eye(number of classes)[vector containing the labels]</code>

blocks|key|4513343|text|这是一个独立于维度的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4513344|这会将任何包含非负整数的N维数组arr转换为一个单精度的N%2B1维数组one_hot，其中one_hot[i_1,...,i_N,c]+=+1表示arr[i_1,...,i_N]+=+c。您可以通过np.argmax(one_hot,+-1)恢复输入|offset|length|style|CODE|4513345|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%2B1+dim+array+of+size+i_1,+...,+i_N,+C
++++:rtype:+ndarray

++++"""
++++one_hot+=+np.zeros(arr.shape+%2B+(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+%2B+[arr.ravel()]]+=+1
++++assert((one_hot.sum(-1)+==+1).all())
++++assert(np.allclose(np.argmax(one_hot,+-1),+arr))
++++return+one_hot|code-block|syntax|javascript|4513346|entityMap^0|0|G|3|Y|7|18|Q|20|K|2Q|M|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]|$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$]]

Here's a dimensionality-independent standalone solution.

This will convert any N-dimensional array <code>arr</code> of nonnegative integers to a one-hot N+1-dimensional array <code>one_hot</code>, where <code>one_hot[i_1,...,i_N,c] = 1</code> means <code>arr[i_1,...,i_N] = c</code>. You can recover the input via <code>np.argmax(one_hot, -1)</code>

<pre><code>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
</code></pre>

blocks|key|4513355|text|这种类型编码通常是numpy数组的一部分。如果您使用的是这样的numpy数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4513356|a+=+np.array([1,0,3])|code-block|syntax|javascript|4513357|然后有一种非常简单的方法可以将其转换为1-hot编码|4513358|out+=+(np.arange(4)+==+a[:,None]).astype(np.float32)|4513359|就这样。|4513360|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Such type of encoding are usually part of numpy array. If you are using a numpy array like this :

<pre><code>a = np.array([1,0,3])
</code></pre>

then there is very simple way to convert that to 1-hot encoding

<pre><code>out = (np.arange(4) == a[:,None]).astype(np.float32)
</code></pre>

That's it.

blocks|key|5935|text|使用以下代码。它的效果最好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5936|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|code-block|syntax|javascript|5937|Found+it+here附言:你不需要进入链接。|offset|length|5938|entityMap|0|LINK|mutability|MUTABLE|url|https://towardsdatascience.com/cifar-10-image-classification-in-tensorflow-5b501f7dc77c^0|0|0|0|D|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Use the following code. It works best.

<pre><code>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
</code></pre>

<a href="https://towardsdatascience.com/cifar-10-image-classification-in-tensorflow-5b501f7dc77c" rel="nofollow noreferrer">Found it here</a> P.S You don't need to go into the link.

blocks|key|5979|text|您可以使用以下代码将其转换为一个热向量：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5980|设x是正态类向量，只有一列，类0到某个数字：|5981|import+numpy+as+np
np.eye(x.max()%2B1)[x]|code-block|syntax|javascript|5982|如果0不是类，则删除%2B1。|5983|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

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:

<pre><code>import numpy as np
np.eye(x.max()+1)[x]
</code></pre>

if 0 is not a class; then remove +1.

blocks|key|5997|text|对于1-热编码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5998|+++one_hot_encode=pandas.get_dummies(array)|code-block|syntax|javascript|5999|For+Example|offset|length|6000|享受编码|6001|entityMap|0|LINK|mutability|MUTABLE|url|https://i.stack.imgur.com/Fkyvz.png^0|0|0|0|B|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]]|A|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

For 1-hot-encoding

<pre><code> one_hot_encode=pandas.get_dummies(array)
</code></pre>

<a href="https://i.stack.imgur.com/Fkyvz.png" rel="noreferrer">For Example</a>

ENJOY CODING

blocks|key|6015|text|如果使用tensorflow，则有one_hot()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|6016|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)>|code-block|syntax|javascript|6017|entityMap|0|LINK|mutability|MUTABLE|url|https://www.tensorflow.org/api_docs/python/tf/one_hot^0|4|A|H|9|H|9|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

If using <code>tensorflow</code>, there is <a href="https://www.tensorflow.org/api_docs/python/tf/one_hot" rel="nofollow noreferrer"><code>one_hot()</code></a>:
<pre class="lang-py prettyprint-override"><code>import tensorflow as tf
import numpy as np

a = np.array([1, 0, 3])
depth = 4
b = tf.one_hot(a, depth)
# &lt;tf.Tensor: shape=(3, 3), dtype=float32, numpy=
# array([[0., 1., 0.],
# [1., 0., 0.],
# [0., 0., 0.]], dtype=float32)&gt;
</code></pre>

blocks|key|6025|text|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))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|6026|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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))
</code></pre>

blocks|key|6053|text|我发现将np.take和np.eye结合起来是最简单的解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|6054|def+one_hot(x,+depth:+int):
++return+np.take(np.eye(depth),+x,+axis=0)|code-block|syntax|javascript|6055|适用于任何形状的x。|6056|entityMap^0|4|7|C|6|0|0|8|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

I find the easiest solution combines <code>np.take</code> and <code>np.eye</code>
<pre class="lang-py prettyprint-override"><code>def one_hot(x, depth: int):
 return np.take(np.eye(depth), x, axis=0)
</code></pre>
works for <code>x</code> of any shape.

Let's say I have a 1d numpy array
<pre><code>a = array([1,0,3])
</code></pre>
I would like to encode this as a 2D one-hot array
<pre><code>b = array([[0,1,0,0], [1,0,0,0], [0,0,0,1]])
</code></pre>
Is there a quick way to do this? Quicker than just looping over <code>a</code> to set elements of <code>b</code>, that is.

Convert array of indices to 1-hot encoded numpy array

假设我有一个一维numpy数组a = array([1,0,3])我想将其编码为2D one-hot数组b = array([[0,1,0,0], [1,0,0,0], [0,0,0,1]])有没有一种快速的方法可以做到这一点？这比在a上循环设置b的元素更快。

问将索引数组转换为1-hot编码的numpy数组
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