# -*- coding: utf-8 -*-

"""

Created on Sun Sep 20 13:55:58 2020

@author: angus

"""

import datetime

print(datetime.date.today())

from datetime import date

print(date.today())

import numpy as np

a = np.arange(0,60,5.0).reshape(3,4)

b = np.array([1,2,3,4],dtype=int)

for x,y in np.nditer([a,b]):

print(x,y)

# 一 、修改维度的函数

# broadcast_to()将数组广播到新形状

a = np.arange(4).reshape(1,4)

a #维数相同，有一个轴长为1

np.broadcast_to(a,(4,4))

# expand_dims()指定位置插入新轴扩展数组形状

x = np.array([[1,2],[3,4]])

x.shape

np.expand_dims(x,axis=0).shape

y = np.expand_dims(x,axis=1)

y.shape

# squeeze()从给定数组形状中，删除长度为1的那个维度。

np.squeeze(y).shape

a = np.arange(3).reshape(1,3,1)

np.squeeze(a)

a.size

# 二、数组形状修改的函数

# 1. reshape()参数必须是分解因子

# 2. flat属性

a = np.arange(8).reshape(2,4)

a

list(a.flat)#数组上一个迭代器

# [0, 1, 2, 3, 4, 5, 6, 7]

# 3. ndarray.flatten()

a.flatten(order='F')#按列取

help(a.flatten)

# 三、数组连接操作

# 1.stack会增加新的维度

a = np.array([[1,2],[3,4]])

b = np.array([[5,6],[7,8]])

np.stack((a,b),axis=0).shape

np.stack((a,b),axis=1).shape

# 水平堆叠

np.hstack((a,b))

np.c_[a,b]

np.r_['1',a,b]

# 垂直堆叠

np.vstack((a,b))

# 等同于 r_是一个实例

np.r_[a,b]

np.c_['0',a,b]

#拼接

np.r_[np.array([1,10]),5,np.arange(10,100,10)]

## array([ 1, 10, 5, 10, 20, 30, 40, 50, 60, 70, 80, 90])

# 查看路径 np.vstack?

np.concatenate((a,b),1)

np.concatenate((a,b),0)

# 四、数组元素操作

## 1. np.resize()

a = np.arange(4)

a.size## 4

np.resize(a,(2,2))

np.resize(a,(2,4))

np.resize(a,(3,3))

a = np.arange(4).reshape(2,2)

np.resize(a,(3,3))

# ndarray内存中存储一样，包装呈现不一样

## 2. np.append()

a

np.append(a,[[4,5]],axis=0)

np.append(a,[[4,5],[6,7]],axis=1)

## 3. np.insert()

a = np.arange(6).reshape(3,2)

np.insert(a,4,[6,7])

np.insert(a,1,[6,7],axis=0)

x = np.insert(a,2,[6],axis=1)

np.append(a,[[0,1]],axis=1)

## 4. np。delete

np.delete(x,1,1)

## 5.np.unique()去重

a = np.array([[8,3,3,5,1],[4,3,4,8,5]])

np.unique(a,return_counts=True)

np.unique(a,return_inverse=True)

##The indices to reconstruct the original array from the unique array. Only

##provided if return_inverse is True.

np.unique(a,return_index = True)

##The indices of the first occurrences of the unique values in the original array.

##Only provided if return_index is True.

#与算术有关的函数

a = np.array([1.0, 3, 1.15, 8.7, 20.53])

np.around(a,decimals=1)

np.floor(a)

np.ceil(a)

a = np.array([10, 20, 30])

b = np.array([3, 5, 7])

np.mod(a,b) # 取余

a % b

np.power(a,2) # 求幂

a ** 2

a = np.arange(9).reshape(3,3)

np.min(a)

np.min(a,1)

np.max(a,1)

np.ptp(a) # 差值

##Range of values (maximum - minimum) along an axis.

np.percentile(a,50)# 百分位数

np.percentile(a,25)

np.median(a)

np.mean(a,1)

np.mean(a,0)

np.std(a,0)##标准差

np.var(a)##方差

#排序相关功能

#获取非零元素

np.nonzero(a)#非零元素的下标

# =============================================================================

# (array([0, 0, 1, 1, 1, 2, 2, 2], dtype=int64),

# array([1, 2, 0, 1, 2, 0, 1, 2], dtype=int64))

# (0,1),(0,2,)...

# =============================================================================

a = np.ones((3,3))

a[[0, 2, 1, 1, 0],[0, 2, 0, 1, 2]] = 0

np.nonzero(a)

#等同于

a[np.where(a != 0)]

np.where(a == 0)

# 使用循环与向量化两种不同方法来计算100以内质数和

def checkprime(x):

return True

def sumprimebyiter(n=100):

return primesum

sumprimebyiter()

import numpy as np

def sumprimebyarr(n=100):

return np.sum(a[check_prime_vec(a)])

sumprimebyarr()

#随机生成10个坐标，计算两两之间距离

Z = np.random.randint(10,size=(10,2))

X,Y = np.atleast_2d(Z[:,0]),np.atleast_2d(Z[:,1])

D = np.sqrt((X-X.T)**2 + (Y-Y.T)**2)

print(D)