数据处理之Numpy(二)¶
1.Numpy基础运算¶
import numpy as np
A = np.arange(2,14).reshape((3,4))
print(A)
# 最小元素索引
print(np.argmin(A)) # 0
# 最大元素索引
print(np.argmax(A)) # 11
# 求整个矩阵的均值
print(np.mean(A)) # 7.5
print(np.average(A)) # 7.5
print(A.mean()) # 7.5
# 中位数
print(np.median(A)) # 7.5
# 累加
print(np.cumsum(A))
# [ 2 5 9 14 20 27 35 44 54 65 77 90]
# 累差运算
B = np.array([[3,5,9],
[4,8,10]])
print(np.diff(B))
'''
[[2 4]
[4 2]]
'''
C = np.array([[0,5,9],
[4,0,10]])
print(np.nonzero(B))
print(np.nonzero(C))
'''
# 将所有非零元素的行与列坐标分割开，重构成两个分别关于行和列的矩阵
(array([0, 0, 0, 1, 1, 1], dtype=int64), array([0, 1, 2, 0, 1, 2], dtype=int64))
(array([0, 0, 1, 1], dtype=int64), array([1, 2, 0, 2], dtype=int64))
'''
# 仿照列表排序
A = np.arange(14,2,-1).reshape((3,4)) # -1表示反向递减一个步长
print(A)
'''
[[14 13 12 11]
[10 9 8 7]
[ 6 5 4 3]]
'''
print(np.sort(A))
'''
# 只是对每行进行递增排序
[[11 12 13 14]
[ 7 8 9 10]
[ 3 4 5 6]]
'''
# 矩阵转置
print(np.transpose(A))
'''
[[14 10 6]
[13 9 5]
[12 8 4]
[11 7 3]]
'''
print(A.T)
'''
[[14 10 6]
[13 9 5]
[12 8 4]
[11 7 3]]
'''
print(A)
print(np.clip(A,5,9))
'''
clip(Array,Array_min,Array_max)
将Array_min<X<Array_max X表示矩阵A中的数，如果满足上述关系，则原数不变。
否则，如果X<Array_min，则将矩阵中X变为Array_min;
如果X>Array_max，则将矩阵中X变为Array_max.
[[9 9 9 9]
[9 9 8 7]
[6 5 5 5]]
'''
2.Numpy索引¶
import numpy as np
A = np.arange(3,15)
print(A)
# [ 3 4 5 6 7 8 9 10 11 12 13 14]
print(A[3])
# 6
B = A.reshape(3,4)
print(B)
'''
[[ 3 4 5 6]
[ 7 8 9 10]
[11 12 13 14]]
'''
print(B[2])
# [11 12 13 14]
print(B[0][2])
# 5
print(B[0,2])
# 5
# list切片操作
print(B[1,1:3]) # [8 9] 1:3表示1-2不包含3
for row in B:
print(row)
'''
[3 4 5 6]
[ 7 8 9 10]
[11 12 13 14]
'''
# 如果要打印列，则进行转置即可
for column in B.T:
print(column)
'''
[ 3 7 11]
[ 4 8 12]
[ 5 9 13]
[ 6 10 14]
'''
# 多维转一维
A = np.arange(3,15).reshape((3,4))
# print(A)
print(A.flatten())
# flat是一个迭代器，本身是一个object属性
for item in A.flat:
print(item)
3.Numpy array合并¶
3.1 数组合并¶
import numpy as np
A = np.array([1,1,1])
B = np.array([2,2,2])
print(np.vstack((A,B)))
# vertical stack 上下合并,对括号的两个整体操作。
'''
[[1 1 1]
[2 2 2]]
'''
C = np.vstack((A,B))
print(C)
print(A.shape,B.shape,C.shape)
# (3,) (3,) (2, 3)
# 从shape中看出A,B均为拥有3项的数组(数列)
# horizontal stack左右合并
D = np.hstack((A,B))
print(D)
# [1 1 1 2 2 2]
print(A.shape,B.shape,D.shape)
# (3,) (3,) (6,)
# 对于A,B这种，为数组或数列，无法进行转置，需要借助其他函数进行转置
3.2 数组转置为矩阵¶
print(A[np.newaxis,:]) # [1 1 1]变为[[1 1 1]]
print(A[np.newaxis,:].shape) # (3,)变为(1, 3)
print(A[:,np.newaxis])
'''
[[1]
[1]
[1]]
'''
3.3 多个矩阵合并¶
# concatenate的第一个例子
print("------------")
print(A[:,np.newaxis].shape) # (3,1)
A = A[:,np.newaxis] # 数组转为矩阵
B = B[:,np.newaxis] # 数组转为矩阵
# axis=0纵向合并
C = np.concatenate((A,B,B,A),axis=0)
print(C)
'''
[[1]
[1]
[1]
[2]
[2]
[2]
[2]
[2]
[2]
[1]
[1]
[1]]
'''
# axis=1横向合并
C = np.concatenate((A,B),axis=1)
print(C)
'''
[[1 2]
[1 2]
[1 2]]
'''
# concatenate的第二个例子
print("-------------")
a = np.arange(8).reshape(2,4)
b = np.arange(8).reshape(2,4)
print(a)
print(b)
print("-------------")
# axis=0多个矩阵纵向合并
c = np.concatenate((a,b),axis=0)
print(c)
# axis=1多个矩阵横向合并
c = np.concatenate((a,b),axis=1)
print(c)
'''
[[0 1 2 3]
[4 5 6 7]
[0 1 2 3]
[4 5 6 7]]
[[0 1 2 3 0 1 2 3]
[4 5 6 7 4 5 6 7]]
'''
4.Numpy array分割¶
4.1 构造3行4列矩阵¶
import numpy as np
A = np.arange(12).reshape((3,4))
print(A)
'''
[[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]]
'''
4.2 等量分割¶
# 等量分割
# 纵向分割同横向合并的axis
print(np.split(A, 2, axis=1))
'''
[array([[0, 1],
[4, 5],
[8, 9]]), array([[ 2, 3],
[ 6, 7],
[10, 11]])]
'''
# 横向分割同纵向合并的axis
print(np.split(A,3,axis=0))
# [array([[0, 1, 2, 3]]), array([[4, 5, 6, 7]]), array([[ 8, 9, 10, 11]])]
4.3 不等量分割¶
print(np.array_split(A,3,axis=1))
'''
[array([[0, 1],
[4, 5],
[8, 9]]), array([[ 2],
[ 6],
[10]]), array([[ 3],
[ 7],
[11]])]
'''
4.4 其他的分割方式¶
# 横向分割
print(np.vsplit(A,3)) # 等价于print(np.split(A,3,axis=0))
# [array([[0, 1, 2, 3]]), array([[4, 5, 6, 7]]), array([[ 8, 9, 10, 11]])]
# 纵向分割
print(np.hsplit(A,2)) # 等价于print(np.split(A,2,axis=1))
'''
[array([[0, 1],
[4, 5],
[8, 9]]), array([[ 2, 3],
[ 6, 7],
[10, 11]])]
'''
5.Numpy copy与 =¶
5.1 =赋值方式会带有关联性¶
import numpy as np
# `=`赋值方式会带有关联性
a = np.arange(4)
print(a) # [0 1 2 3]
b = a
c = a
d = b
a[0] = 11
print(a) # [11 1 2 3]
print(b) # [11 1 2 3]
print(c) # [11 1 2 3]
print(d) # [11 1 2 3]
print(b is a) # True
print(c is a) # True
print(d is a) # True
d[1:3] = [22,33]
print(a) # [11 22 33 3]
print(b) # [11 22 33 3]
print(c) # [11 22 33 3]
5.2 copy()赋值方式没有关联性¶
a = np.arange(4)
print(a) # [0 1 2 3]
b =a.copy() # deep copy
print(b) # [0 1 2 3]
a[3] = 44
print(a) # [ 0 1 2 44]
print(b) # [0 1 2 3]
# 此时a与b已经没有关联