20191009

# -*- coding: utf-8 -*-
"""
Spyder Editor

This is a temporary script file.
"""
# =============================================================================
#练习
# 汉诺塔的移动可以用递归函数非常简单地实现。
# 
# 请编写move(n, a, b, c)函数，它接收参数n，表示3个柱子A、B、C中第1个柱子A的盘子数量，然后打印出把所有盘子从A借助B移动到C的方法，例如：
# 
# =============================================================================
# -*- coding: utf-8 -*-
def move(n, a, b, c):
    if n == 1:
        print(a, '-->', c)
    else:
        move(n-1,a,c,b)
        print(a, '-->', c)
        move(n-1,b,a,c)
move(3, 'A', 'B', 'C')
move(4, 'A', 'B', 'C')

#高级特性
L=[]
n=1
while n<=99:
    L.append(n)
    n=n+1
L1=range(99)
L1=list(range(99))
L1.append(99)
L1.pop(0)

# =============================================================================
# 切片
# =============================================================================
L=['Michael','Sarah','Tracy','Bob','Jack']
[L[0],L[1],L[2]]
r=[]
n=3
for x in range(n):
    r.append(L[x])
#切片[n:m]表示，从索引n开始取，直到索引m-1为止，但不包括索引m。
L[0:3]#从0开始直到索引到L[2]
L[1:4]
#从头索引0可以省略
L[:4]
#倒数切片
L[-2:]
L[-2:-1]
L=list(range(100))
L[:10]#前10个数
L[-10:]#后10个数
L[10:20]#11-20数
L[:10:2]#前十个数字每两个取一个
L[::5]#所有数字5个取一个
L[:]#原list
#tuple也可看做list
(1,2,3,4,5)[:3]
#字符串也可以
'ABCDEFGH'[1:3]
'ABCDEFGH'[::2]

# =============================================================================
# 利用切片操作，实现一个trim()函数，去除字符串首尾的空格，注意不要调用str的strip()方法：
# =============================================================================
def trim(s):
    if len(s) ==0:
        return ''
    while s[0]==' ':
        if len(s)==1:
            return ''
        s=s[1:]
    while s[-1]==' ':
        s=s[:-1]#截取到倒数第一个之前的数字不包含倒数第一个
    return s

# =============================================================================
# 迭代
# =============================================================================
#dictionary 迭代
d={'a':1,'b':2,'c':3}
for key in d:
    print(key)
for value in d.values():
    print(value)
for key,value in d.items():
    print(key,value)
#字符串迭代
for character in 'string':
    print(character)
#判断是否可迭代对象：通过collections模块的Iterable类型判断
from collections import Iterable
isinstance('abc', Iterable) 
isinstance([1,2,3],Iterable)
isinstance(123, Iterable)
#enumerate函数 迭代引用索引
i=0
for i , value in enumerate(['A','B','C']):
    print(i,value)
for value in ['A','B','C']:
    print (i , value)
    i=i+1
#双变量迭代
for x,y in[(1,1),(2,2),(3,3)]:
    print(x,y)

# =============================================================================
# 请使用迭代查找一个list中最小和最大值，并返回一个tuple：
# =============================================================================
def findMinAndMax(L):
    if len(L)==0:
        return None,None
    else:
        MIN=L[0]
        MAX=L[0]
        for x in L:
            if MIN>=x:
                MIN=x
            if MAX<=x:
                MAX=x
        return MIN,MAX

# =============================================================================
# 列表生成式List Comprehensions
# =============================================================================
#要生成list [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]可以用list(range(1, 11))：
L=list(range(1,11))
#要生成[1x1, 2x2, 3x3, ..., 10x10]怎么做？
L=[]
for x in list(range(1,11)):
    L.append(x*x)
#超快的方式[函数式，迭代语句，条件语句]
L=[x*x for x in range(1,11)]
L=[x*x for x in range(1,11) if x%2==0]
#两次迭代
#ABC的全排列
L=[m+n for m in 'ABC' for n in 'ABC' if m!=n]
L=[j+k+l for j in 'ABC' for k in 'ABC' for l in 'ABC' if j is not l and j is not k and k is not l]
#目录下所有文件以及目录名字
import os
[d for d in os.listdir('c:/')]
#dictionary的列表生成
d={'x':'A','y':'B','z':'C'}
for key,value in d.items():
    print(key,'=',value)
[key+'='+value for key,value in d.items()]
#把list变成小写
L=[j+k+l for j in 'ABC' for k in 'ABC' for l in 'ABC' if j is not l and j is not k and k is not l]
[s.lower() for s in L]

# =============================================================================
# 如果list中既包含字符串，又包含整数，由于非字符串类型没有lower()方法，所以列表生成式会报错：
# 使用内建的isinstance函数可以判断一个变量是不是字符串：
# >>> x = 'abc'
# >>> y = 123
# >>> isinstance(x, str)
# True
# >>> isinstance(y, str)
# False
# 请修改列表生成式，通过添加if语句保证列表生成式能正确地执行：
# =============================================================================
L1 = ['Hello', 'World', 18, 'Apple', None]
L2 = [s.lower() for s in L1 if isinstance(s, str)==True]
# 测试:
print(L2)
if L2 == ['hello', 'world', 'apple']:
    print('测试通过!')
else:
    print('测试失败!')

# =============================================================================
# 生成器generator
# 在Python中，这种一边循环一边计算的机制，称为生成器：generator。
# =============================================================================
#要创建一个generator，有很多种方法。第一种方法很简单，只要把一个列表生成式的[]改成()，就创建了一个generator：
L=[x*x for x in range(10)]
L
g=(x*x for x in range(10))
g
#通过next()函数获得generator的下一个返回值：
next(g)
next(g)
next(g)
next(g)
next(g)
next(g)
next(g)
next(g)
next(g)
next(g)
next(g)
g=(x*x for x in range(10))
    for n in g:
        print(n)
#斐波拉契数列（Fibonacci）
def Fibonacci(n,a=1,b=1):
    i=1
    print(a)
    while i<n:
        print(b)
        i,a,b=i+1,b,a+b
    return 'Done'
#斐波拉契数列（Fibonacci）生成器generator
def Fibonacci2(n,a=1,b=1):
    i=1
    print(a)
    while i<n:
        yield(b)
        i,a,b=i+1,b,a+b
    return 'Done'
f=Fibonacci2(6)
for x in f:
    print(x)
#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。
def odd():
    print('step 1')
    yield 1
    print('step 2')
    yield 2
    print('step 3')
    yield 3
o=odd()    
next(o)
next(o)
next(o)
next(o)    

#但是用for循环调用generator时，发现拿不到generator的return语句的返回值。如果想要拿到返回值，必须捕获StopIteration错误，返回值包含在StopIteration的value中：
g=Fibonacci2(6)
while True:
    try:
        x=next(g)
        print('g:',x)
    except StopIteration as e:
        print('Generator return value:',e.value)
        break
    
# =============================================================================
#   练习
# 杨辉三角定义如下：
# 
#           1
#          / \
#         1   1
#        / \ / \
#       1   2   1
#      / \ / \ / \
#     1   3   3   1
#    / \ / \ / \ / \
#   1   4   6   4   1
#  / \ / \ / \ / \ / \
# 1   5   10  10  5   1
# 把每一行看做一个list，试写一个generator，不断输出下一行的list：
# =============================================================================
def triangles(t=20):
    L=[1]
    while len(L)<20:
        yield L
        i=1
        L1=[1,1]
        while i < len(L):
            L1.insert(i,L[i-1]+L[i])
            i=i+1
        L=L1
#简洁写法：
def triangles1():
    list = [1]
    while True:
        yield (list)
        list = [1]+[ list[i]+list[i+1] for i in range(len(list)-1)] +[1]


# 期待输出:
# [1]
# [1, 1]
# [1, 2, 1]
# [1, 3, 3, 1]
# [1, 4, 6, 4, 1]
# [1, 5, 10, 10, 5, 1]
# [1, 6, 15, 20, 15, 6, 1]
# [1, 7, 21, 35, 35, 21, 7, 1]
# [1, 8, 28, 56, 70, 56, 28, 8, 1]
# [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
n = 0
results = []
for t in triangles():
    results.append(t)
    n = n + 1
    if n == 10:
        break

for t in results:
    print(t)

if results == [
    [1],
    [1, 1],
    [1, 2, 1],
    [1, 3, 3, 1],
    [1, 4, 6, 4, 1],
    [1, 5, 10, 10, 5, 1],
    [1, 6, 15, 20, 15, 6, 1],
    [1, 7, 21, 35, 35, 21, 7, 1],
    [1, 8, 28, 56, 70, 56, 28, 8, 1],
    [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
]:
    print('测试通过!')
else:
    print('测试失败!')