波场和dot的区别,dot认证与3c认证的区别

1.dot

首先看下dot源码中的注释部分

def dot(a, b, out=None): “”” dot(a, b, out=None) Dot product of two arrays. Specifically, – If both `a` and `b` are 1-D arrays, it is inner product of vectors (without complex conjugation). – If both `a` and `b` are 2-D arrays, it is matrix multiplication, but using :func:`matmul` or “a @ b“ is preferred. – If either `a` or `b` is 0-D (scalar), it is equivalent to :func:`multiply` and using “numpy.multiply(a, b)“ or “a * b“ is preferred. – If `a` is an N-D array and `b` is a 1-D array, it is a sum product over the last axis of `a` and `b`. – If `a` is an N-D array and `b` is an M-D array (where “M>=2“), it is a sum product over the last axis of `a` and the second-to-last axis of `b`:: dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m]) …..

关注一下最常用的两种情况:

If bothaandbare 1-D arrays, it is inner product of vectors
这就是两个向量dot,最后得到的两个向量的内积。

If bothaandbare 2-D arrays, it is matrix multiplication, but using :func:matmulor “a @ b“ is preferred.
2-D arrays指的就是矩阵了。根据上面的解释不难看出,如果是两个矩阵dot,执行的就是矩阵相乘运算。

写段代码测试下

def demo2(): a1 = np.arange(1, 5) a2 = a1[::-1] print(a1) print(a2) # 两个向量dot为内积 print(a1.dot(a2)) print(np.dot(a1, a2)) print(“\n\n”) b1 = np.arange(1, 5).reshape(2, 2) b2 = np.arange(5, 9).reshape(2, 2) b3 = np.arange(9, 15).reshape(3, 2) print(b1) print(b2) print(b3) print(np.dot(b1, b2)) # 会报错, 不满足矩阵相乘条件 # print(np.dot(b1, b3))

代码执行的结果

[1 2 3 4][4 3 2 1]2020[[1 2] [3 4]][[5 6] [7 8]][[ 9 10] [11 12] [13 14]][[19 22] [43 50]] 2.multiply

同样的看一下multiply对应源码的注释部分。

def multiply(x1, x2, *args, **kwargs): # real signature unknown; NOTE: unreliably restored from __doc__ “”” multiply(x1, x2, /, out=None, *, where=True, casting=’same_kind’, order=’K’, dtype=None, subok=True[, signature, extobj]) Multiply arguments element-wise. Parameters ———- x1, x2 : array_like Input arrays to be multiplied. If “x1.shape != x2.shape“, they must be broadcastable to a common shape (which becomes the shape of the output). out : ndarray, None, or tuple of ndarray and None, optional …..

明白multiply方法的关键就是上面的一句注释:

Multiply arguments element-wise.

说人话就是:按对应的元素相乘。

def demo3(): a1 = np.arange(1, 5) a2 = a1[::-1] print(a1) print(a2) print(np.multiply(a1, a2)) print(“\n\n”) b1 = np.arange(1, 5).reshape(2, 2) b2 = np.arange(5, 9).reshape(2, 2) print(b1) print(b2) print(np.multiply(b1, b2))

运行得到结果

[1 2 3 4][4 3 2 1][4 6 6 4][[1 2] [3 4]][[5 6] [7 8]][[ 5 12] [21 32]]

参考对应的代码,应该就很容易理解了。

3. *运算符

乘法运算符,最后得到的结果,跟multiply方法得到的结果是一样的。

def demo4(): a1 = np.arange(1, 5) a2 = a1[::-1] print(a1) print(a2) print(a1 * a2) print(“\n\n”) b1 = np.arange(1, 5).reshape(2, 2) b2 = np.arange(5, 9).reshape(2, 2) print(b1) print(b2) print(b1 * b2)

最终结果

[1 2 3 4][4 3 2 1][4 6 6 4][[1 2] [3 4]][[5 6] [7 8]][[ 5 12] [21 32]]

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