x = np.array([[1,2], [3,4]])
print x # Prints "[[1 2]
# [3 4]]"
print x.T # Prints "[[1 3]
# [2 4]]"
# Note that taking the transpose of a rank 1 array does nothing:
v = np.array([1,2,3])
print v # Prints "[1 2 3]"
print v.T # Prints "[1 2 3]"
点乘和矩阵乘法
和MATLAB不同，* 是元素逐个相乘，而不是矩阵乘法。在Numpy中使用dot来进行矩阵乘法：
x = np.array([[1,2],[3,4]])
y = np.array([[5,6],[7,8]])
v = np.array([9,10])
w = np.array([11, 12])
# Inner product of vectors; both produce 219
print v.dot(w)
print np.dot(v, w)
# Matrix / vector product; both produce the rank 1 array [29 67]
print x.dot(v)
print np.dot(x, v)
# Matrix / matrix product; both produce the rank 2 array
# [[19 22]
# [43 50]]
print x.dot(y)
print np.dot(x, y)
x = np.array([[1,2,3], [4,5,6], [7,8,9], [10, 11, 12]])
v = np.array([1, 0, 1])
y = np.empty_like(x) # Create an empty matrix with the same shape as x
# Add the vector v to each row of the matrix x with an explicit loop
for i in range(4):
y[i, :] = x[i, :] + v
# Now y is the following
# [[ 2 2 4]
# [ 5 5 7]
# [ 8 8 10]
# [11 11 13]]
print y
这样是行得通的，但是当x矩阵非常大，利用循环来计算就会变得很慢很慢。我们可以换一种思路：
x = np.array([[1,2,3], [4,5,6], [7,8,9], [10, 11, 12]])
v = np.array([1, 0, 1])
vv = np.tile(v, (4, 1)) # Stack 4 copies of v on top of each other
print vv # Prints "[[1 0 1]
# [1 0 1]
# [1 0 1]
# [1 0 1]]"
y = x + vv # Add x and vv elementwise
print y # Prints "[[ 2 2 4
# [ 5 5 7]
# [ 8 8 10]
# [11 11 13]]"
Numpy广播机制可以让我们不用创建vv，就能直接运算，看看下面例子：
x = np.array([[1,2,3], [4,5,6], [7,8,9], [10, 11, 12]])
v = np.array([1, 0, 1])
y = x + v # Add v to each row of x using broadcasting
print y # Prints "[[ 2 2 4]
# [ 5 5 7]
# [ 8 8 10]
# [11 11 13]]"
# Compute outer product of vectors
v = np.array([1,2,3]) # v has shape (3,)
w = np.array([4,5]) # w has shape (2,)
# To compute an outer product, we first reshape v to be a column
# vector of shape (3, 1); we can then broadcast it against w to yield
# an output of shape (3, 2), which is the outer product of v and w:
# [[ 4 5]
# [ 8 10]
# [12 15]]
print np.reshape(v, (3, 1)) * w
# Add a vector to each row of a matrix
x = np.array([[1,2,3], [4,5,6]])
# x has shape (2, 3) and v has shape (3,) so they broadcast to (2, 3),
# giving the following matrix:
# [[2 4 6]
# [5 7 9]]
print x + v
# Add a vector to each column of a matrix
# x has shape (2, 3) and w has shape (2,).
# If we transpose x then it has shape (3, 2) and can be broadcast
# against w to yield a result of shape (3, 2); transposing this result
# yields the final result of shape (2, 3) which is the matrix x with
# the vector w added to each column. Gives the following matrix:
# [[ 5 6 7]
# [ 9 10 11]]
print (x.T + w).T
# Another solution is to reshape w to be a row vector of shape (2, 1);
# we can then broadcast it directly against x to produce the same
# output.
print x + np.reshape(w, (2, 1))
# Multiply a matrix by a constant:
# x has shape (2, 3). Numpy treats scalars as arrays of shape ();
# these can be broadcast together to shape (2, 3), producing the
# following array:
# [[ 2 4 6]
# [ 8 10 12]]
print x * 2