《python笔记》廖雪峰Python3习题

• 1.利用map()函数，把用户输入的不规范的英文名字，变为首字母大写，其他小写的规范名字。输入：['adam', 'LISA', 'barT']，输出：['Adam', 'Lisa', 'Bart']

def f(s):
    s=s[0].upper()+s[1:].lower()
    return s
L1=['adam', 'LISA', 'barT']
L2=list(map(f,L1))
print(L2)

• 2.Python提供的sum()函数可以接受一个list并求和，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]))
if prod([3, 5, 7, 9]) == 945:
    print('测试成功!')
else:
    print('测试失败!')

• 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：

#coding：utf-8
from functools import reduce

def str2float(s):
    DIGITS = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
    s = s.split('.')
    s1 = s[0]
    s2 = s[1]
    s = s1 + s2
    def char2num(s):
        return DIGITS[s]
    def f(a, b):
        return a*10 + b
    x=reduce(f,map(char2num,s))/10**len(s2)
    return x

print('str2float(\'123.456\') =', str2float('123.456'))
if abs(str2float('123.456') - 123.456) < 0.00001:
    print('测试成功!')
else:
    print('测试失败!')

#方法二
from functools import reduce
def str2float(s):
    def fn(x,y):
        return x*10+y
    n=s.index('.')
    s1=list(map(int,[x for x in s[:n]]))
    s2=list(map(int,[x for x in s[n+1:]]))
    return reduce(fn,s1) + reduce(fn,s2)/10**len(s2)#m**n 这个表达的就是 m 的 n 次方
print('\'123.4567\'=',str2float('123.4567'))  

• 4.整理成一个str2int的函数就是：

from functools import reduce

DIGITS = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}

def str2int(s):
    def fn(x, y):
        return x * 10 + y
    def char2num(s):
        return DIGITS[s]
    return reduce(fn, map(char2num, s))

print(str2int('2'))

• 5.用filter求素数

计算素数的一个方法是埃氏筛法，它的算法理解起来非常简单：

首先，列出从2开始的所有自然数，构造一个序列：

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, ...

取序列的第一个数2，它一定是素数，然后用2把序列的2的倍数筛掉：

3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, ...

取新序列的第一个数3，它一定是素数，然后用3把序列的3的倍数筛掉：

5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, ...

取新序列的第一个数5，然后用5把序列的5的倍数筛掉：

7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, ...

不断筛下去，就可以得到所有的素数。

def _odd_iter():
    n = 1
    while True:
        n = n + 2
        yield n

def _not_divisible(n):
    return lambda x: x % n >0

def primes():
    yield 2
    it = _odd_iter()
    while True:
      n = next(it)
      yield n
      it = filter(_not_divisible,it)

for n in primes():
    if n < 10:
        print(n)
    else:
        break

• 6.回数是指从左向右读和从右向左读都是一样的数，例如12321，909。请利用filter()筛选出回数：

#方法一
def f(x):
    n=str(x)
    return n==n[::-1]
print(list(filter(f,[m for m in range(1000) if m> 100])))

[101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 505, 515, 525, 535, 545, 555, 565, 575, 585, 595, 606, 616, 626, 636, 646, 656, 666, 676, 686, 696, 707, 717, 727, 737, 747, 757, 767, 777, 787, 797, 808, 818, 828, 838, 848, 858, 868, 878, 888, 898, 909, 919, 929, 939, 949, 959, 969, 979, 989, 999]

#方法二
def is_palindrome(n):
    a=str(n)
    for i in range(len(a)):
        if a[i]==a[-i-1]:
            return a
        else:
            return False
output = filter(is_palindrome, range(1, 1000))
print('1~1000:', list(output))
if list(filter(is_palindrome, range(1, 200))) == [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191]:
    print('测试成功!')
else:
    print('测试失败!')

• 7.假设我们用一组tuple表示学生名字和成绩：L = [('Bob', 75), ('Adam', 92), ('Bart', 66), ('Lisa', 88)]。请用sorted()对上述列表分别按名字和分数排序：

L = [('Bob', 75), ('Adam', 92), ('Bart', 66), ('Lisa', 88)]
def by_name(t):
    return t[0]
def by_score(t):
    return t[-1]

L1=sorted(L,key=by_name)
L2=sorted(L,key=by_score,reverse=True)
print('L1=',L1)
print('L2=',L2)

• 8.利用闭包返回一个计数器函数，每次调用它返回递增整数：

方法一:

def createCounter():
    s = [0]
    def counter():
        s[0] = s[0]+1
        return s[0]
    return counter

counterA = createCounter()
print(counterA(), counterA(), counterA(), counterA(), counterA()) # 1 2 3 4 5
counterB = createCounter()
if [counterB(), counterB(), counterB(), counterB()] == [1, 2, 3, 4]:
    print('测试通过!')
else:
    print('测试失败!')

方法二：

s = 0
def createCounter():
    def counter():
        global s
        s=s+1
        return s
    return counter
counterA = createCounter()
print(counterA(),counterA())

方法三：

def createCounter():
    m=0
    def counter():
        nonlocal m
        m = m + 1
        return m
    return counter

counterX=createCounter()
print(counterX(),counterX())

• 9.请设计一个decorator，它可作用于任何函数上，并打印该函数的执行时间：

import time, functools

def printcalltime(fun):
    @functools.wraps(fun)
    def wrapper(*args,**kw):
        start = time.time()
        res = fun(*args,**kw)
        end = time.time()
        print('%s executed in %s ms '%(fun.__name__,(end-start)*1000))
        return res
    return wrapper
@printcalltime
def fast(x,y):
    time.sleep(0.0045)
    return x+y
@printcalltime
def slow(x,y):
    time.sleep(0.1345)
    return x*y

f = fast(11,22)
print(f)
s= slow(11,33)
print(s)

if f != 33:
    print('测试失败')
elif s != 363:
    print('测试失败')

• 10.为了统计学生人数，可以给Student类增加一个类属性，每创建一个实例，该属性自动增加：

class Student(object):
    count = 0

    def __init__(self, name):
        self.name = name
        Student.count += 1 #每创建一个实例，count就加1


if Student.count != 0:
    print('测试失败!')
else:
    bart = Student('Bart')
    if Student.count != 1:
        print('测试失败!')
    else:
        lisa = Student('Bart')
        if Student.count != 2:
            print('测试失败!')
        else:
            print('Students:', Student.count)
            print('测试通过!')