/Demo_1/Demo_1.py

import math
print("Hello, wordk!")
classmates = ['Michael', 'Bob', 'Tracy']
print(len(classmates))
for name in classmates:
    print(name)

sum = 0
for x in [1,2,3,4,5,6]:
    sum += x
print(sum)

sum = 0
for x in range(101):
    sum += x
print(sum)

dic = {'Michael':93, 'Bob': 75, 'Tracy': 85}
print(dic['Michael'])
print(dic)
print(dic.get('SomeName', 'Do not have'))
dic.pop('Bob')
print(dic)


s1 = set([5,2,3])
s = set([12, 1 , 6])
print(s1 | s)

arr = [23, 23, 1, 4, 100]
arr.sort()
print(arr)

def area_of_circle(x):
    return (3.141592654*x*x)
print('%.3f' %(area_of_circle(1)))

def move(x, y, stpe, angle = 0):
    return x *10 , y*10, angle*10
tupleInstance = move(1, 2, 3 , 5)
print(tupleInstance)

def move(p):
    return p[0]+11
L = [1,2]
print(move(L))

'''
练习
请定义一个函数quadratic(a, b, c)，接收3个参数，返回一元二次方程：
ax2 + bx + c = 0
的两个解。
提示：计算平方根可以调用math.sqrt()函数：
'''
#必选参数
def quadratic(a, b, c):
    if not isinstance(a, (int, float)) or not isinstance(b, (int, float)) or not isinstance(c, (int, float)):
        raise TypeError('bad operand type')
    if a == 0:
        return -c / b
    else:
        return (math.sqrt(b * b - 4 * a * c) - b)/ 2.0 / a, ( - math.sqrt(b * b - 4 * a * c) - b)/ 2.0 / a
print(quadratic(2, 3, 1))
print(quadratic(1, 3, -4))

def pow(x, n = 2):
    s = 1
    while(n):
        n -= 1
        s *= x
    return s
print(pow(2,5))

def add_end(L = []):
    L.append('End')
    return L
print(add_end())
print(add_end())

def add_end(L = None):
    if L is None:
        L = []
    L.append('End')
    return L
print(add_end())
print(add_end())

#可变参数
def calc(*nums):
    sum = 0
    for i in nums:
        sum += math.pow(i, 2)
    return sum
num = [1,2]
print(calc(1,2))
print(calc(*num))

#关键字参数
"""关键字参数有什么用？它可以扩展函数的功能。"""
"""比如，在person函数里，我们保证能接收到name和age这两个参数，"""
"""但是，如果调用者愿意提供更多的参数，我们也能收到。"""
"""试想你正在做一个用户注册的功能，除了用户名和年龄是必填项外，其他都是可选项，"""
"""利用关键字参数来定义这个函数就能满足注册的需求。"""
def person(name, age, **kw):
    print('name:', name, 'age:', age, 'others:', kw)
    if 'School'in kw:
        print("有School 信息")
person("李金旭", 11, School = ['Neu','Swun'], Contry = 'China')

#命名参数
'''如果要限制关键字参数的名字，就可以用命名关键字参数，
    例如，只接收city和job作为关键字参数。这种方式定义的函数如下：'''
def person(name, age, *, city, job):
    print(name, age, city, job)
person("李金旭", 11, city = ['Neu','Swun'], job = 'China')

def person(name, age, *argv):
    print(name, age, argv)
person("李金旭",11,'Neu')

def person(name, age, *, city, job):
    print(name,"is", age," years old in ", city, "as a ",job)
person("李金旭",11,city = "Bejing", job = "IT front to end Engineer")

#参数组合
'''
    在Python中定义函数，可以用必选参数、默认参数、可变参数、关键字参数和命名关键字参数，这5种参数都可以组合使用。
    但是请注意，参数定义的顺序必须是：必选参数、默认参数、可变参数、命名关键字参数和关键字参数。
'''
'''
    *args是可变参数，args接收的是一个tuple；
    **kw是关键字参数，kw接收的是一个dict。
'''
def func1(a, b, c = 0, *argv, **kw):
    print('a =', a, 'b =', b, 'c =', c, 'args =', args, 'kw =', kw)
def func2(*argv, **kw):
    print(argv, kw)
args = (1,2,3,4)
kw = {'Name':'李金旭', 'Age': 11}
func2(args, kw)

def print_step(x, y):
    print("# ", x," --> ",y)
def move(n, a, b, c):
    if(n == 1):
         print_step(a, c)
    else:
        move(n-1, a, c, b) #a通过c将n-1个移动到b
        print_step(a, c)
        move(n-1,b,a,c )#将n-1个从b通过a移动到c

move(4,'A','B','C')

#练习Leetcode 260. Single Number III
class Solution(object):
    def singleNumber(self, nums):
        xor = 0
        for num in nums:
            xor = xor^num
        diff_pos = 0
        for i in range(31):
            if(xor & (1 << i)):
                diff_pos = i
                break
        rec = [0,0]
        for num in nums:
            if(num & (1 << diff_pos)):
                rec[1] ^= num
            else:
                rec[0] ^= num
        print(rec)
        return rec

t = Solution()
ans = t.singleNumber((1,1,2,2,4,6))
print(ans)

#切片
list_num = list(range(100))
print(list_num)
print(list_num[:3])#取出前三个
print(list_num[::3])#每隔3个取一个
print(list_num[0:10])
print(list_num[0:10:2])
tuple = tuple(range(100))


#列表生成公式
print(list(range(1,11)))
L = []
for i in range(1,11):
    L.append(i * i)
print(L)

L = [ x * x for x in range(2, 10)]
print(L)

L = [x*x for x in range(1,11) if x%2 == 0]
print(L)

'''使用两层循环生成全排列'''
L = [m + n for m in 'ABC' for n in 'XYZ']
print(L)

import  os
print([d for d in os.listdir('../')])

L = ['Hello', 'World', 'IBM', 'Apple']
print([s.lower() for s in L])

print(isinstance(L[0], str))
print(isinstance(L[1], int))

#生成器
g = (x * x for x in range(10))
print(next(g))
print(next(g))
print(next(g))
for n in g:
    print(n)
def fibonacci(max):
    n,a,b = 0, 0, 1
    while n < max:
        print(b)
        a, b = b , a+b
        n = n + 1
    return 'done'
print(fibonacci(10))

def fibonacciGenerator(max):
    n,a,b = 0,0,1
    while n < max:
        yield b
        a,b = b, a+b
        n = n + 1
    return 'done'

print(fibonacciGenerator(6))
for n in fibonacciGenerator(11):
    print(n)

'''generator和函数的执行流程不一样。函数是顺序执行，遇到return语句或者最后一行函数语句就返回。
而变成generator的函数，在每次调用next()的时候执行，遇到yield语句返回，再次执行时从上次返回的
yield语句处继续执行。'''
'''yield 相当于在程序中插入了一个断点，这个断点也是通过return返回的'''
def odd():
    print('step 1')
    yield 1
    print('step 2')
    yield 3
    print('step 3')
    yield 5
    yield 'done'
g = odd()

while True:
    try:
        x = next(g)
        print('g:',x)
    except StopIteration as e:
        print('Generator return value:',e.value)
        break

#练习：生成杨辉三角
L = []
def YanghuiTriangle(max):
    n = 0
    L = [[] for i in range(max)]
    while n < max:
        i = 0
        while i <= n:
            if i == 0 or i == n:
                L[n].append(1)
            else:
                L[n].append(L[n - 1][i] + L[n-1][i - 1])
            i+=1
        yield L[n]
        n += 1
    yield 'done'
gY = YanghuiTriangle(10)
while True:
    try:
        print('#',next(gY))
    except StopIteration as e:
        print('Generator return value : ', e.value)
        break

#迭代器
for x in [1,2,4,5,6,7]:
    print(x)
'''等价于下面iter'''
it = iter([1,2,4,5,6,7])
while True:
    try:
        x = next(it)
        print(x)
    except StopIteration:
        break

####函数式编程####
'''Python 支持函数式编程，因为可以使用变量作为参数，也支持用函数作为参数，所以Python不是纯函数式编程语言'''
print(abs)
f = abs
'''函数名也是变量名'''
print(f(-11))

def add(x,y,f):
    return f(x) + f(y)
print(add(-4,-5,f))

#map/reduce


from  functools import reduce
print(list(map(str,[-1,-2,-3,-4,-5])))
def fn(x, y):
    print(x, y)
    return x*10 + y
r = reduce(fn,[1,3,5,7,9])
print(r)

def func_sum(x, y):
    return x + y
def square(x):
    return x*x
list_r = map(square,[-1,-2,-3,-4,-5])
ll_r = list(list_r)
print('r = ',ll_r)
ans = reduce(func_sum, ll_r)
print('sum is ',ans)

def add_100(a, b, c):
    print(a,b,c)
    return a * 10000+ b *100 + c
list1 = [11,22,33]
list2 = [44,55,66]
list3 = [77,88,99]
rec = map(add_100,list1, list2, list3)
print(list(rec))

from functools import reduce
def fn(x, y):
    print(x, y)
    return x * 10 + y
def char2num(s):
    return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
    '''{} is a dictionary, [] is index of dict, so [key] return  value'''
print(reduce(fn, map(char2num, '13579')))

def str2int(s):
    def fn(x, y):
        return x * 10 + y
    def char2num(s):
        return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
    return reduce(fn, map(char2num, s))
print(str2int('13578'))
print(int('13573'))
print(str(13572))

def str2int_(s):
    def char2num(s):
        return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
    return reduce(lambda x,y: x * 10 + y, map(char2num, s))
    '''lambda 匿名函数，有些简单函数只需要用1次，所以不给起名字'''
print(str2int_('2012'))

#联系
#1 规范化英文名
def normalize(name):
    return name.capitalize()
L1 = ['adam', 'LISA', 'barT']
L2 = list(map(normalize, L1))
print(L2)
#2 请编写一个prod()函数，可以接受一个list并利用reduce()求积：
from functools import  reduce
def prod(L):
    return reduce(lambda x,y : x * y, L)
print('3 * 5 * 7 * 9 =', prod([3, 5, 7, 9]))
#3 利用map和reduce编写一个str2float函数，把字符串'123.456'转换成浮点数123.456：
from functools import  reduce
def str2float(s):
    def char2num(s):
        return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
    pos = s.find('.')
    s_num = s.split('.')[0] + s.split('.')[1]
    print(pos, s_num)
    L = reduce(lambda x,y : x * 10 + y, map(char2num, s_num))
    return  L/ math.pow(10, pos)
print('str2float(\'123.456\') =', str2float('123.456'))
#Another solution
def str2float_(s):
    def char2num(s):
        return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
    a, b = s.split('.')
    L = reduce(lambda  x,y: x * 10 + y, map(char2num, a + b))
    return L/10**len(b)
print('str2float(\'123.456\') =', str2float('123.456'))