# Create the following rank 2 array with shape (3, 4) # [[ 1 2 3 4] # [ 5 6 7 8] # [ 9 10 11 12]] a = np.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])
# Use slicing to pull out the subarray consisting of the first 2 rows # and columns 1 and 2; b is the following array of shape (2, 2): # [[2 3] # [6 7]] b = a[:2, 1:3]
# A slice of an array is a view into the same data, so modifying it # will modify the original array. print(a[0, 1]) # Prints "2" b[0, 0] = 77 # b[0, 0] is the same piece of data as a[0, 1] print(a[0, 1]) # Prints "77"
# Create the following rank 2 array with shape (3, 4) # [[ 1 2 3 4] # [ 5 6 7 8] # [ 9 10 11 12]] a = np.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])
# Two ways of accessing the data in the middle row of the array. # Mixing integer indexing with slices yields an array of lower rank, # while using only slices yields an array of the same rank as the # original array: row_r1 = a[1, :] # Rank 1 view of the second row of a row_r2 = a[1:2, :] # Rank 2 view of the second row of a print(row_r1, row_r1.shape) # Prints "[5 6 7 8] (4,)" print(row_r2, row_r2.shape) # Prints "[[5 6 7 8]] (1, 4)"
# We can make the same distinction when accessing columns of an array: col_r1 = a[:, 1] col_r2 = a[:, 1:2] print(col_r1, col_r1.shape) # Prints "[ 2 6 10] (3,)" print(col_r2, col_r2.shape) # Prints "[[ 2] # [ 6] # [10]] (3, 1)"
整数数组 索引允许你使用另一个数组的数据构造任意数组 大批。 这是一个例子：
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import numpy as np
a = np.array([[1,2], [3, 4], [5, 6]])
# An example of integer array indexing. # The returned array will have shape (3,) and print(a[[0, 1, 2], [0, 1, 0]]) # Prints "[1 4 5]"
# The above example of integer array indexing is equivalent to this: print(np.array([a[0, 0], a[1, 1], a[2, 0]])) # Prints "[1 4 5]"
# When using integer array indexing, you can reuse the same # element from the source array: print(a[[0, 0], [1, 1]]) # Prints "[2 2]"
# Equivalent to the previous integer array indexing example print(np.array([a[0, 1], a[0, 1]])) # Prints "[2 2]"
print(np.sum(x)) # Compute sum of all elements; prints "10" print(np.sum(x, axis=0)) # Compute sum of each column; prints "[4 6]" print(np.sum(x, axis=1)) # Compute sum of each row; prints "[3 7]"
转置矩阵， 只需使用 T数组对象的属性：
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import numpy as np
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]"
# We will add the vector v to each row of the matrix x, # storing the result in the matrix 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)
Numpy 广播允许我们执行此计算，而无需实际执行 创建多个副本 v。 考虑这个版本，使用广播：
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import numpy as np
# We will add the vector v to each row of the matrix x, # storing the result in the matrix y 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]]"
线路 y = x + v尽管有效 x有形状 (4, 3)和 v有形状 (3,)由于广播； 这条线的工作原理就像 v实际上有形状 (4, 3), 其中每一行都是一个副本 v，并且按元素求和。
# 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 column 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)