numpy里*与dot与multiply

一、*  , dot()   multiply()

1, 对于array来说,(* 和 dot()运算不同, * 和 multiply()运算相同)

*和multiply() 是每个元素对应相乘

dot() 是矩阵乘法

2, 对于matrix来说,(* 和 multiply()运算不同,* 和 dot()运算相同)

* 和dot() 是矩阵乘法

multiply()  是每个元素对应相乘

3, 混合的时候(与矩阵同)

multiply 为对应乘

dot为矩阵乘法(矩阵在前数组在后时,均为一维时数组可适应,即能做矩阵乘法)

*为 矩阵乘法(但无上述适应性)

总结:dot为矩阵乘法multiply是对应乘* 看元素,元素为矩阵(包括含矩阵)时为矩阵乘法,元素为数组时为对应乘法

Python 3.6.4 (default, Jan 7 2018, 03:52:16)
[GCC 4.2.1 Compatible Android Clang 5.0.300080 ] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import numpy as np
>>> a1=np.array([1,2,3])
>>> a1*a1
array([1, 4, 9])
>>> np.dot(a1,a1)
14
>>> np.multiply(a1,a1)
array([1, 4, 9])
>>> m1 = np.mat(a1)
>>> m1
matrix([[1, 2, 3]])
>>> m1*m1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/data/data/com.termux/files/usr/lib/python3.6/site-packages/numpy-1.13.3-py3.6-linux-aarch64.egg/numpy/matrixlib/defmatrix.py", line 309, in __mul__
return N.dot(self, asmatrix(other))
ValueError: shapes (1,3) and (1,3) not aligned: 3 (dim 1) != 1 (dim 0)
>>> np.dot(m1,m1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: shapes (1,3) and (1,3) not aligned: 3 (dim 1) != 1 (dim 0)
>>> np.multiply(m1,m1)
matrix([[1, 4, 9]])
>>> np.multiply(a1,m1)
matrix([[1, 4, 9]])
>>> np.multiply(m1,a1)
matrix([[1, 4, 9]])
>>> np.dot(a1,m1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: shapes (3,) and (1,3) not aligned: 3 (dim 0) != 1 (dim 0)
>>> np.dot(m1,a1)
matrix([[14]])
>>> a1*m1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/data/data/com.termux/files/usr/lib/python3.6/site-packages/numpy-1.13.3-py3.6-linux-aarch64.egg/numpy/matrixlib/defmatrix.py", line 315, in __rmul__
return N.dot(other, self)
ValueError: shapes (3,) and (1,3) not aligned: 3 (dim 0) != 1 (dim 0)
>>> m1*a1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/data/data/com.termux/files/usr/lib/python3.6/site-packages/numpy-1.13.3-py3.6-linux-aarch64.egg/numpy/matrixlib/defmatrix.py", line 309, in __mul__
return N.dot(self, asmatrix(other))
ValueError: shapes (1,3) and (1,3) not aligned: 3 (dim 1) != 1 (dim 0)
>>> a1,m1
(array([1, 2, 3]), matrix([[1, 2, 3]]))
>>>

 -------------------------------------------------------------------

当array是二维时,我们可以把它看做是矩阵(但是乘法的法则还是按照array来算)

In [45]: a_x
Out[45]:
array([[1, 2, 3],
       [2, 4, 0],
       [1, 2, 1]])

In [46]: a_y
Out[46]: array([[-1,  1, -1]])

先看dot

# 这时用dot做矩阵乘法要注意:满足左边的列数等于右边的行数

In [47]: np.dot(a_x,a_y)
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-47-3b884e90b0ef> in <module>()
----> 1 np.dot(a_x,a_y)

ValueError: shapes (3,3) and (1,3) not aligned: 3 (dim 1) != 1 (dim 0)

In [48]: np.dot(a_y,a_x)
Out[48]: array([[ 0,  0, -4]])

# 当左边的变为一维的数组时,结果还是一个二维的数组(矩阵形式)

In [49]: a_y_ = a_y.flatten()

In [50]: a_y_
Out[50]: array([-1,  1, -1])

然后还有一点很重要 np.dot(a_y_,a_y_) 可以将两个一维的数组(这时没有行列向量之说,从转置后还为本身也可以看出)直接做内积
In [109]: np.dot(a_y_,a_y_)
Out[109]: 3


In [51]: np.dot(a_y,a_x)
Out[51]: array([[ 0,  0, -4]])

再看multiply(与*的效果一致):

In [52]: a_y
Out[52]: array([[-1,  1, -1]])

In [53]: a_y_
Out[53]: array([-1,  1, -1])

In [54]: np.multiply(a_y,a_y_)
Out[54]: array([[1, 1, 1]])

In [55]: np.multiply(a_y_,a_y)
Out[55]: array([[1, 1, 1]])

得到结论:都是以高维的那个元素决定结果的维度。(重要)

In [56]: np.multiply(a_y_,a_y_.T)   # 在a_y_是一维的行向量,a_y_ = a_y_.T,所以结果一样
Out[56]: array([1, 1, 1])

In [57]: np.multiply(a_y_,a_y.T)    # a_y 是二维的行向量,转置一下就变成二维的列向量
Out[57]:
array([[ 1, -1,  1],
       [-1,  1, -1],
       [ 1, -1,  1]])
In [58]: np.multiply(a_y,a_y.T)     # 与上述效果相同
Out[58]:
array([[ 1, -1,  1],
       [-1,  1, -1],
       [ 1, -1,  1]])

上面两个就是下面这俩对应元素相乘!

In [59]: a_y
Out[59]: array([[-1,  1, -1]])

In [60]: a_y.T
Out[60]:
array([[-1],
       [ 1],
       [-1]])

与下面这种张量积效果相同

In [61]: np.outer(a_y,a_y)
Out[61]:
array([[ 1, -1,  1],
       [-1,  1, -1],
       [ 1, -1,  1]])

In [62]: np.outer(a_y,a_y.T)     # 这时用不用转置的效果都一样
Out[62]:
array([[ 1, -1,  1],
       [-1,  1, -1],
       [ 1, -1,  1]])

In [63]: np.outer(a_y_,a_y_)    # 两个一维的行向量外积的效果也一样
Out[63]:
array([[ 1, -1,  1],
       [-1,  1, -1],
       [ 1, -1,  1]])

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                                                                                    下面看一下 .T 和 .transpose()

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

在小于等于二维的情况下,两者效果是相同的!

In [64]: x
Out[64]:
matrix([[1, 2, 3],
        [2, 4, 0],
        [1, 2, 1]])

In [65]: a_x
Out[65]:
array([[1, 2, 3],
       [2, 4, 0],
       [1, 2, 1]])

In [68]: (x.T == x.transpose()).all()     # 为矩阵时相同
Out[68]: True

In [70]: (a_x.T == a_x.transpose()).all()    # 为数组时相同
Out[70]: True

In [71]: (a_y.T == a_y.transpose()).all()   # a_y为一维数组(行向量)相同
Out[71]: True

In [72]: (a_y_.T == a_y_.transpose()).all() # a_y_为二维数组(行向量)也相同
Out[72]: True

下面才是两种区别的重点:

当维数大于等于三时(此时只有数组了),

T 其实就是把顺序全部颠倒过来,

而transpose(), 里面可以指定顺序,输入为一个元组(x0,x1,x2...x(n-1) )为(0,1,2,...,n-1)的一个排列,其中n为数组维数,

效果与np.transpose(arr, axes=(x0,x1,x2...x(n-1)) ) 相同,

    def transpose(self, *axes): # real signature unknown; restored from __doc__
        """
        a.transpose(*axes)
        
            Returns a view of the array with axes transposed.
        
            For a 1-D array, this has no effect. (To change between column and
            row vectors, first cast the 1-D array into a matrix object.)
            For a 2-D array, this is the usual matrix transpose.
            For an n-D array, if axes are given, their order indicates how the
            axes are permuted (see Examples). If axes are not provided and
            ``a.shape = (i[0], i[1], ... i[n-2], i[n-1])``, then
            ``a.transpose().shape = (i[n-1], i[n-2], ... i[1], i[0])``.
        
            Parameters
            ----------
            axes : None, tuple of ints, or `n` ints
        
             * None or no argument: reverses the order of the axes.
        
             * tuple of ints: `i` in the `j`-th place in the tuple means `a`'s
               `i`-th axis becomes `a.transpose()`'s `j`-th axis.
        
             * `n` ints: same as an n-tuple of the same ints (this form is
               intended simply as a "convenience" alternative to the tuple form)
        
            Returns
            -------
            out : ndarray
                View of `a`, with axes suitably permuted.
        
            See Also
            --------
            ndarray.T : Array property returning the array transposed.
        
            Examples
            --------
            >>> a = np.array([[1, 2], [3, 4]])
            >>> a
            array([[1, 2],
                   [3, 4]])
            >>> a.transpose()
            array([[1, 3],
                   [2, 4]])
            >>> a.transpose((1, 0))
            array([[1, 3],
                   [2, 4]])
            >>> a.transpose(1, 0)
            array([[1, 3],
                   [2, 4]])
        """
        pass

这两者的源代码还是又挺大区别的,虽然意思差不多

def transpose(a, axes=None):
    """
    Permute the dimensions of an array.
    Parameters
    ----------
    a : array_like
        Input array.
    axes : list of ints, optional
        By default, reverse the dimensions, otherwise permute the axes
        according to the values given.
    Returns
    -------
    p : ndarray
        `a` with its axes permuted.  A view is returned whenever
        possible.
    See Also
    --------
    moveaxis
    argsort
    Notes
    -----
    Use `transpose(a, argsort(axes))` to invert the transposition of tensors
    when using the `axes` keyword argument.
    Transposing a 1-D array returns an unchanged view of the original array.
    Examples
    --------
    >>> x = np.arange(4).reshape((2,2))
    >>> x
    array([[0, 1],
           [2, 3]])
    >>> np.transpose(x)
    array([[0, 2],
           [1, 3]])
    >>> x = np.ones((1, 2, 3))
    >>> np.transpose(x, (1, 0, 2)).shape
    (2, 1, 3)
    """
return _wrapfunc(a, 'transpose', axes)

还是看实验吧:

In [73]: arr=np.arange(16).reshape(2,2,4)

In [74]: arr
Out[74]:
array([[[ 0,  1,  2,  3],
        [ 4,  5,  6,  7]],

       [[ 8,  9, 10, 11],
        [12, 13, 14, 15]]])

In [75]: arr.transpose((1,0,2))   # 注意这个参数不带key,不能写axes=(1,0,2),若写成这个,则报错
Out[75]:
array([[[ 0,  1,  2,  3],
        [ 8,  9, 10, 11]],

       [[ 4,  5,  6,  7],
        [12, 13, 14, 15]]])

In [78]: np.transpose(arr, (1, 0, 2))  #这个参数可带key,刻写成axes=(1,0,2),若写成这个,则通过
Out[78]:
array([[[ 0,  1,  2,  3],
        [ 8,  9, 10, 11]],

       [[ 4,  5,  6,  7],
        [12, 13, 14, 15]]])

 

三个维度的编号对应为(0,1,2),比如这样,我们需要拿到7这个数字,怎么办,肯定需要些三个维度的值,7的第一个维度为0,第二个维度为1,第三个3,

即坐标轴(0,1,2)的对应的坐标为(0,1,3)所以arr[0,1,3]则拿到了7

所以相当于把原来坐标轴[0,1,2]的转置到[1,0,2],即把之前第三个维度转为第一个维度,之前的第二个维度不变,之前的第一个维度变为第三个维度

理解了上面,再来理解swapaxes()就很简单了,swapaxes接受一对轴编号,其实这里我们叫一对维度编号更好吧

arr.swapaxes(2,1)  就是将第三个维度和第二个维度交换

注意这个函数必须接受两个参数,(两个轴),不能写成元组(2,1)的形式再传入。

In [81]: arr.swapaxes(2,1)
Out[81]:
array([[[ 0,  4],
        [ 1,  5],
        [ 2,  6],
        [ 3,  7]],

       [[ 8, 12],
        [ 9, 13],
        [10, 14],
        [11, 15]]])

In [82]: arr.swapaxes((2,1))
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-82-24459d1e2980> in <module>()
----> 1 arr.swapaxes((2,1))

TypeError: swapaxes() takes exactly 2 arguments (1 given)

还是那我们的数字7来说,之前的索引是(0,1,3),那么交换之后,就应该是(0,3,1)

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原文地址:https://www.cnblogs.com/cymwill/p/8364428.html