问题:连接两个一维NumPy数组

我在NumPy中有两个简单的一维数组。我应该能够使用numpy.concatenate将它们连接起来。但是我收到以下代码的错误:

TypeError:只有length-1数组可以转换为Python标量

import numpy
a = numpy.array([1, 2, 3])
b = numpy.array([5, 6])
numpy.concatenate(a, b)

为什么?

I have two simple one-dimensional arrays in NumPy. I should be able to concatenate them using numpy.concatenate. But I get this error for the code below:

TypeError: only length-1 arrays can be converted to Python scalars

Code

import numpy
a = numpy.array([1, 2, 3])
b = numpy.array([5, 6])
numpy.concatenate(a, b)

Why?


回答 0

该行应为:

numpy.concatenate([a,b])

要连接的数组需要作为一个序列而不是作为单独的参数传递。

NumPy文档中

numpy.concatenate((a1, a2, ...), axis=0)

将一系列数组连接在一起。

它试图将您解释b为axis参数,这就是为什么它抱怨无法将其转换为标量。

The line should be:

numpy.concatenate([a,b])

The arrays you want to concatenate need to be passed in as a sequence, not as separate arguments.

From the NumPy documentation:

numpy.concatenate((a1, a2, ...), axis=0)

Join a sequence of arrays together.

It was trying to interpret your b as the axis parameter, which is why it complained it couldn’t convert it into a scalar.


回答 1

连接一维数组有多种可能性,例如,

numpy.r_[a, a],
numpy.stack([a, a]).reshape(-1),
numpy.hstack([a, a]),
numpy.concatenate([a, a])

对于大型阵列,所有这些选项都同样快。对于小型的,concatenate有一点优势:

在此处输入图片说明

该图是使用perfplot创建的:

import numpy
import perfplot

perfplot.show(
    setup=lambda n: numpy.random.rand(n),
    kernels=[
        lambda a: numpy.r_[a, a],
        lambda a: numpy.stack([a, a]).reshape(-1),
        lambda a: numpy.hstack([a, a]),
        lambda a: numpy.concatenate([a, a]),
    ],
    labels=["r_", "stack+reshape", "hstack", "concatenate"],
    n_range=[2 ** k for k in range(19)],
    xlabel="len(a)",
)

There are several possibilities for concatenating 1D arrays, e.g.,

numpy.r_[a, a],
numpy.stack([a, a]).reshape(-1),
numpy.hstack([a, a]),
numpy.concatenate([a, a])

All those options are equally fast for large arrays; for small ones, concatenate has a slight edge:

enter image description here

The plot was created with perfplot:

import numpy
import perfplot

perfplot.show(
    setup=lambda n: numpy.random.rand(n),
    kernels=[
        lambda a: numpy.r_[a, a],
        lambda a: numpy.stack([a, a]).reshape(-1),
        lambda a: numpy.hstack([a, a]),
        lambda a: numpy.concatenate([a, a]),
    ],
    labels=["r_", "stack+reshape", "hstack", "concatenate"],
    n_range=[2 ** k for k in range(19)],
    xlabel="len(a)",
)

回答 2

的第一个参数concatenate本身应该是要串联的数组序列

numpy.concatenate((a,b)) # Note the extra parentheses.

The first parameter to concatenate should itself be a sequence of arrays to concatenate:

numpy.concatenate((a,b)) # Note the extra parentheses.

回答 3

另一种方法是使用“ concatenate”的缩写形式,即“ r _ […]”或“ c _ […]”,如下面的示例代码所示(请参见http://wiki.scipy.org / NumPy_for_Matlab_Users以获取更多信息):

%pylab
vector_a = r_[0.:10.] #short form of "arange"
vector_b = array([1,1,1,1])
vector_c = r_[vector_a,vector_b]
print vector_a
print vector_b
print vector_c, '\n\n'

a = ones((3,4))*4
print a, '\n'
c = array([1,1,1])
b = c_[a,c]
print b, '\n\n'

a = ones((4,3))*4
print a, '\n'
c = array([[1,1,1]])
b = r_[a,c]
print b

print type(vector_b)

结果是:

[ 0.  1.  2.  3.  4.  5.  6.  7.  8.  9.]
[1 1 1 1]
[ 0.  1.  2.  3.  4.  5.  6.  7.  8.  9.  1.  1.  1.  1.] 


[[ 4.  4.  4.  4.]
 [ 4.  4.  4.  4.]
 [ 4.  4.  4.  4.]] 

[[ 4.  4.  4.  4.  1.]
 [ 4.  4.  4.  4.  1.]
 [ 4.  4.  4.  4.  1.]] 


[[ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 4.  4.  4.]] 

[[ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 1.  1.  1.]]

An alternative ist to use the short form of “concatenate” which is either “r_[…]” or “c_[…]” as shown in the example code beneath (see http://wiki.scipy.org/NumPy_for_Matlab_Users for additional information):

%pylab
vector_a = r_[0.:10.] #short form of "arange"
vector_b = array([1,1,1,1])
vector_c = r_[vector_a,vector_b]
print vector_a
print vector_b
print vector_c, '\n\n'

a = ones((3,4))*4
print a, '\n'
c = array([1,1,1])
b = c_[a,c]
print b, '\n\n'

a = ones((4,3))*4
print a, '\n'
c = array([[1,1,1]])
b = r_[a,c]
print b

print type(vector_b)

Which results in:

[ 0.  1.  2.  3.  4.  5.  6.  7.  8.  9.]
[1 1 1 1]
[ 0.  1.  2.  3.  4.  5.  6.  7.  8.  9.  1.  1.  1.  1.] 


[[ 4.  4.  4.  4.]
 [ 4.  4.  4.  4.]
 [ 4.  4.  4.  4.]] 

[[ 4.  4.  4.  4.  1.]
 [ 4.  4.  4.  4.  1.]
 [ 4.  4.  4.  4.  1.]] 


[[ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 4.  4.  4.]] 

[[ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 4.  4.  4.]
 [ 1.  1.  1.]]

回答 4

以下是使用numpy.ravel(),的更多方法:numpy.array()利用一维数组可以解包为普通元素的事实:

# we'll utilize the concept of unpacking
In [15]: (*a, *b)
Out[15]: (1, 2, 3, 5, 6)

# using `numpy.ravel()`
In [14]: np.ravel((*a, *b))
Out[14]: array([1, 2, 3, 5, 6])

# wrap the unpacked elements in `numpy.array()`
In [16]: np.array((*a, *b))
Out[16]: array([1, 2, 3, 5, 6])

Here are more approaches for doing this by using numpy.ravel(), numpy.array(), utilizing the fact that 1D arrays can be unpacked into plain elements:

# we'll utilize the concept of unpacking
In [15]: (*a, *b)
Out[15]: (1, 2, 3, 5, 6)

# using `numpy.ravel()`
In [14]: np.ravel((*a, *b))
Out[14]: array([1, 2, 3, 5, 6])

# wrap the unpacked elements in `numpy.array()`
In [16]: np.array((*a, *b))
Out[16]: array([1, 2, 3, 5, 6])

回答 5

来自numpy 文档的更多事实:

语法为 numpy.concatenate((a1, a2, ...), axis=0, out=None)

轴= 0用于行连接轴= 1用于列连接

>>> a = np.array([[1, 2], [3, 4]])
>>> b = np.array([[5, 6]])

# Appending below last row
>>> np.concatenate((a, b), axis=0)
array([[1, 2],
       [3, 4],
       [5, 6]])

# Appending after last column
>>> np.concatenate((a, b.T), axis=1)    # Notice the transpose
array([[1, 2, 5],
       [3, 4, 6]])

# Flattening the final array
>>> np.concatenate((a, b), axis=None)
array([1, 2, 3, 4, 5, 6])

希望对您有所帮助!

Some more facts from the numpy docs :

With syntax as numpy.concatenate((a1, a2, ...), axis=0, out=None)

axis = 0 for row-wise concatenation axis = 1 for column-wise concatenation

>>> a = np.array([[1, 2], [3, 4]])
>>> b = np.array([[5, 6]])

# Appending below last row
>>> np.concatenate((a, b), axis=0)
array([[1, 2],
       [3, 4],
       [5, 6]])

# Appending after last column
>>> np.concatenate((a, b.T), axis=1)    # Notice the transpose
array([[1, 2, 5],
       [3, 4, 6]])

# Flattening the final array
>>> np.concatenate((a, b), axis=None)
array([1, 2, 3, 4, 5, 6])

I hope it helps !


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