[英]Convert array of indices to 1-hot encoded numpy array
Let's say I have a 1d numpy array假设我有一个 1d numpy 数组
a = array([1,0,3])
Your array a
defines the columns of the nonzero elements in the output array.您的数组
a
定义了输出数组中非零元素的列。 You need to also define the rows and then use fancy indexing:您还需要定义行,然后使用花式索引:
>>> 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.]])
>>> 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.]])
In case you are using keras, there is a built in utility for that:如果您使用 keras,有一个内置的实用程序:
from keras.utils.np_utils import to_categorical
categorical_labels = to_categorical(int_labels, num_classes=3)
And it does pretty much the same as @YXD's answer (see source-code ).它与@YXD 的答案几乎相同(请参阅源代码)。
Here is what I find useful:以下是我认为有用的内容:
def one_hot(a, num_classes):
return np.squeeze(np.eye(num_classes)[a.reshape(-1)])
Here num_classes
stands for number of classes you have.这里
num_classes
代表您拥有的类数。 So if you have a
vector with shape of (10000,) this function transforms it to (10000,C) .因此,如果您有
a
形状为(10000,) a
向量,则此函数会将其转换为(10000,C) 。 Note that a
is zero-indexed, ie one_hot(np.array([0, 1]), 2)
will give [[1, 0], [0, 1]]
.请注意,
a
是零索引的,即one_hot(np.array([0, 1]), 2)
将给出[[1, 0], [0, 1]]
。
Exactly what you wanted to have I believe.正是你想要的,我相信。
PS: the source is Sequence models - deeplearning.ai PS:来源是序列模型-deeplearning.ai
您还可以使用 numpy 的眼睛功能:
numpy.eye(number of classes)[vector containing the labels]
You can use sklearn.preprocessing.LabelBinarizer
:您可以使用
sklearn.preprocessing.LabelBinarizer
:
Example:例子:
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))
output:输出:
[[0 1 0 0]
[1 0 0 0]
[0 0 0 1]]
Amongst other things, you may initialize sklearn.preprocessing.LabelBinarizer()
so that the output of transform
is sparse.除其他外,您可以初始化
sklearn.preprocessing.LabelBinarizer()
以便transform
的输出是稀疏的。
You can use the following code for converting into a one-hot vector:您可以使用以下代码转换为 one-hot 向量:
let x is the normal class vector having a single column with classes 0 to some number:让 x 是具有单个列的普通类向量,其中类为 0 到某个数字:
import numpy as np
np.eye(x.max()+1)[x]
if 0 is not a class;如果 0 不是一个类; then remove +1.
然后删除+1。
For 1-hot-encoding对于 1-hot-encoding
one_hot_encode=pandas.get_dummies(array)
ENJOY CODING享受编码
Here is a function that converts a 1-D vector to a 2-D one-hot array.这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)
Below is some example usage:下面是一些示例用法:
>>> 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]])
I think the short answer is no.我认为简短的回答是否定的。 For a more generic case in
n
dimensions, I came up with this:对于
n
维更通用的情况,我想出了这个:
# 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
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.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中创建这些列表。 Anyway, I did some measurements with
timeit
and it seems that the numpy
-based ( indices
/ arange
) and the iterative versions perform about the same.无论如何,我用
timeit
做了一些测量,似乎基于numpy
的( indices
/ arange
)和迭代版本的表现大致相同。
Just to elaborate on the excellent answer from K3---rnc , here is a more generic version:只是为了详细说明K3---rnc的优秀答案,这里有一个更通用的版本:
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):此外,这里是此方法的快速基准测试和YXD当前接受的答案中的一种方法(略有更改,因此它们提供相同的 API,只是后者仅适用于 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:后一种方法快约 35%(MacBook Pro 13 2015),但前者更通用:
>>> 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)
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.我最近遇到了同样的问题,并找到了上述解决方案,结果证明只有当你的数字符合某种形式时才会令人满意。 For example if you want to one-hot encode following list:
例如,如果您想对以下列表进行单热编码:
all_good_list = [0,1,2,3,4]
go ahead, the posted solutions are already mentioned above.继续,上面已经提到了已发布的解决方案。 But what if considering this data:
但是如果考虑这些数据呢:
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.如果您使用上述方法进行操作,您可能会得到 90 个单热列。 This is because all answers include something like
n = np.max(a)+1
.这是因为所有答案都包含类似
n = np.max(a)+1
。 I found a more generic solution that worked out for me and wanted to share with you:我找到了一个更通用的解决方案,它对我有用,并想与您分享:
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我希望有人在上述解决方案上遇到相同的限制,这可能会派上用场
Such type of encoding are usually part of numpy array.这种类型的编码通常是 numpy 数组的一部分。 If you are using a numpy array like this :
如果您使用的是这样的 numpy 数组:
a = np.array([1,0,3])
then there is very simple way to convert that to 1-hot encoding那么有一种非常简单的方法可以将其转换为 1-hot 编码
out = (np.arange(4) == a[:,None]).astype(np.float32)
That's it.就是这样。
clean and easy solution:干净简单的解决方案:
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)
Here is an example function that I wrote to do this based upon the answers above and my own use case:这是我根据上述答案和我自己的用例编写的示例函数:
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)
I am adding for completion a simple function, using only numpy operators:我正在添加一个简单的函数来完成,仅使用 numpy 运算符:
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: eg:它需要一个概率矩阵作为输入:例如:
[[0.03038822 0.65810204 0.16549407 0.3797123 ] ... [0.02771272 0.2760752 0.3280924 0.33458805]]
[[0.03038822 0.65810204 0.16549407 0.3797123] ... [0.02771272 0.2760752 0.3280924 0.33458805]]
And it will return它会回来
[[0 1 0 0] ... [0 0 0 1]]
[[0 1 0 0] ... [0 0 0 1]]
Here's a dimensionality-independent standalone solution.这是一个与维度无关的独立解决方案。
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
.这会将任何非负整数的 N 维数组
arr
转换为单热 N+1 维数组one_hot
,其中one_hot[i_1,...,i_N,c] = 1
表示arr[i_1,...,i_N] = c
。 You can recover the input via np.argmax(one_hot, -1)
您可以通过
np.argmax(one_hot, -1)
恢复输入
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
Use the following code.使用以下代码。 It works best.
它效果最好。
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
Found it here PS You don't need to go into the link. 在这里找到它PS 你不需要进入链接。
import numpy as np
a = np.array([1,0,3])
b = np.array([[0,1,0,0], [1,0,0,0], [0,0,0,1]])
from neuraxle.steps.numpy import OneHotEncoder
encoder = OneHotEncoder(nb_columns=4)
b_pred = encoder.transform(a)
assert b_pred == b
Link to documentation: neuraxle.steps.numpy.OneHotEncoder文档链接: neuraxle.steps.numpy.OneHotEncoder
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))
I find the easiest solution combines np.take<\/code> and
np.eye<\/code>
我发现最简单的解决方案结合了
np.take<\/code>和
np.eye<\/code>
def one_hot(x, num_classes: int):
return np.take(np.eye(num_classes), x, axis=0)
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