numpyPractice100-two

import numpy as np

# 21  tile函数使用：相当于是复制作用
np.tile(np.array([[0,1], [1,0]]), (4,4))

a = np.array([0, 1, 2])
np.tile(a, 2)

# 22 : 标准化一个方阵
arr = np.random.random((5,5))
(arr - np.mean(arr)) / (np.std(arr))  # 标准化公式

# 23
color = np.dtype([("r", np.ubyte, 1),
                  ("g", np.ubyte, 1),
                  ("b", np.ubyte, 1),
                  ("a", np.ubyte, 1)])
color

# 24  向量之间的相乘（5*3）*（3*2）= 5*2
# dot() 方法
np.dot(np.ones((5,3)), np.ones((3,2)))

# python 3.5 写法：@
np.ones((5,3)) @ np.ones((3,2))

# 25 将某个区间的数字变成相反数：乘以-1
arr = np.arange(11)
arr[(arr > 3)&(arr < 8)] *= -1
arr

# 26 表达式的结果
# 将 sum()函数中所有数相加：1+2+3+4+（-1）=9
sum(range(5), -1)

# 27 判断表达式的合法性
# Z**Z
# 2 << Z >> 2
# Z <- Z
# 1j*Z
# Z/1/1
# Z<Z>Z   错误

Z = np.array([1,2,3])
Z

Z**Z   # 对应的元素求指数即可

2 << Z >> 2

array([1, 2, 4], dtype=int32)

Z <- Z   # Z和它的相反数进行比较

array([False, False, False])

1j*Z

array([0.+1.j, 0.+2.j, 0.+3.j])

Z/1/1

array([1., 2., 3.])

# 28  表达式的结果
print(np.array(0) / np.array(0))   # nan
print(np.array(0) // np.array(0))  # 0
print(np.array([np.nan]).astype(int).astype(float))  # [-2.14748365e+09]

nan
0
[-2.14748365e+09]

# 29
Z = np.random.uniform(-10,+10,10)
print (np.copysign(np.ceil(np.abs(Z)), Z))

[ -2.   5.   5. -10.  -9. -10.   6.   1.  -3.  -7.]

# 30 
# np.intersectld()：返回的是两个数组中经过排序后的相同元素
# Return the sorted, unique values that are in both of the input arrays.
z1 = np.random.randint(0,10,10)
z2 = np.random.randint(0,10,10)
print(np.intersect1d(z1,z2))

[0 2 3 6 7]

# 31
# How to ignore all numpy warnings 
defaults = np.seterr(all="ignore")
Z = np.ones(1) / 0

# Back to sanity
_ = np.seterr(**defaults)

with np.errstate(divide='ignore'):
    Z = np.ones(1) / 0

# 32 
np.sqrt(-1) == np.emath.sqrt(-1)

False

# 33 获得时间日期：昨天、今天，明天
yesterday = np.datetime64('today', 'D') - np.timedelta64(1, 'D')
today = np.datetime64('today', 'D')
tomorrow = np.datetime64('today', 'D') + np.timedelta64(1, 'D')
print(yesterday)
print(today)
print(tomorrow)

2019-10-13
2019-10-14
2019-10-15

# 34  获取某个时间段内的所有日期：含头不含尾
np.arange('2016-07', '2016-07-27', dtype='datetime64[D]')

array(['2016-07-01', '2016-07-02', '2016-07-03', '2016-07-04',
       '2016-07-05', '2016-07-06', '2016-07-07', '2016-07-08',
       '2016-07-09', '2016-07-10', '2016-07-11', '2016-07-12',
       '2016-07-13', '2016-07-14', '2016-07-15', '2016-07-16',
       '2016-07-17', '2016-07-18', '2016-07-19', '2016-07-20',
       '2016-07-21', '2016-07-22', '2016-07-23', '2016-07-24',
       '2016-07-25', '2016-07-26'], dtype='datetime64[D]')

# 35  计算两个数组之间的数量关系
a = np.ones(3) * 1
b = np.ones(3) * 2
c = np.ones(3) * 3

np.add(a,b,out=a)

array([3., 3., 3.])

np.divide(a,2,out=a)

array([1.5, 1.5, 1.5])

np.negative(a,out=a)

array([-1.5, -1.5, -1.5])

# 36
Z = np.random.uniform(0,10,10)  # 随机生成10个数字

print(Z)
print (Z - Z%1)  # 取整数
print (np.floor(Z))  
print (np.ceil(Z)-1)
print (Z.astype(int))  # 取整后为int类型
print (np.trunc(Z))

[3.95803958 1.32403183 8.64610951 0.83095194 4.94494921 9.23832181
 0.37511147 8.1509876  9.82676114 8.29316378]
[3. 1. 8. 0. 4. 9. 0. 8. 9. 8.]
[3. 1. 8. 0. 4. 9. 0. 8. 9. 8.]
[3. 1. 8. 0. 4. 9. 0. 8. 9. 8.]
[3 1 8 0 4 9 0 8 9 8]
[3. 1. 8. 0. 4. 9. 0. 8. 9. 8.]

# 37 
z = np.zeros((5,5))   # 5*5的全0方阵
z += np.arange(5)    # np.arange(5) 生成一个向量0-4
print(z)

[[0. 1. 2. 3. 4.]
 [0. 1. 2. 3. 4.]
 [0. 1. 2. 3. 4.]
 [0. 1. 2. 3. 4.]
 [0. 1. 2. 3. 4.]]

np.arange(5)

array([0, 1, 2, 3, 4])

# 38
def generate():
    for x in range(10):
        yield x
#         return x

z = np.fromiter(generate(), dtype=float, count=-1)
print(z)

[0. 1. 2. 3. 4. 5. 6. 7. 8. 9.]

# 39
# linspace()的用法
Z = np.linspace(0,1,11,endpoint=False)[1:]   # 生成10个0-1的数字，不包含尾部
print(Z)

[0.09090909 0.18181818 0.27272727 0.36363636 0.45454545 0.54545455
 0.63636364 0.72727273 0.81818182 0.90909091]

# 40 
z = np.random.random(10)
z.sort()  # 排序之后输出
print(z)

[0.09537726 0.11914916 0.12808735 0.28657504 0.32288421 0.39927615
 0.64877625 0.66318512 0.75063225 0.87013168]

# 41
z = np.arange(10)
print(sum(z))
np.add.reduce(z)   # 速度比sum()函数快很多

45
45

# 42 检验两个数组是否相等equal
a = np.random.randint(0,2,5)
b = np.random.randint(0,2,5)

equal = np.allclose(a, b)
print(equal)

equal = np.array_equal(a, b)
print(equal)

False
False

# 43 创建一个只读矩阵
Z = np.zeros(10)
Z.flags.writeable = False
print(Z)
# Z[0]=1   无法修改数据，只是可读状态

[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]

# 44
z = np.random.random((10,2))  # 生成 10*2 的矩阵
x,y = z[:, 0],z[:, 1]   # 取出第一列和第二列的数据
r = np.sqrt(x**2+y**2)
t = np.arctan2(y,x)
print(z)
print(r)
print(t)

[[0.54870349 0.71706842]
 [0.2816144  0.06322106]
 [0.85545704 0.29143293]
 [0.92798256 0.09136097]
 [0.61973184 0.38362868]
 [0.0454154  0.37239676]
 [0.25334226 0.25988723]
 [0.74275296 0.92348859]
 [0.35936247 0.84002203]
 [0.23819685 0.94926246]]
 
[0.90291895 0.28862358 0.90373663 0.93246901 0.72886111 0.37515584
 0.36293756 1.18512157 0.91366208 0.97869145]
 
[0.91763557 0.22083385 0.32834355 0.09813492 0.55429018 1.44944125
 0.79814999 0.89344479 1.1665554  1.32494413]

# 45 将数组中最大数用0代替
z = np.random.random(10)
z[z.argmax()]= 0
print(z)

[0.30695133 0.14808481 0.78878546 0.15906783 0.78569492 0.66986127
 0.5942439  0.14547699 0.         0.5825813 ]

z.argmax()  # 找出最大数

2