题型分析

1、

简单让A、B的值交换

a,b = 1,2
print(a,b)
a = 1
b = 2
a,b = b,a
print(a,b)

a,b=[1,2],[5,9]
print(a,b)

 2、元组         如果元祖里面只有一个元素且不加,那此元素是什么类型，就是什么类型。

tu1 = (1)
tu2 = (1,)
print(tu1,type(tu1))
print(tu2,type(tu2))
tu1 = ([1])
tu2 = ([1],)
print(tu1,type(tu1))
print(tu2,type(tu2))

 3、

l1 = [1,[1],2,3,4]
l2 = l1[:]
l1[1].append('a')
#l2 的结果是什么?
#l1 和l2同时都是[1, [1, 'a'], 2, 3, 4]    浅copy
print(l1,id(l1))
print(l2,id(l2))
print(l1[1] is l2[1])

 生成器相关面试题：

一、

def demo():
    for i in range(4):
        yield i

g=demo()

g1=(i for i in g)
g2=(i for i in g1)

print(list(g1))
print(list(g2))

面试题1

二、

def add(n,i):
    return n+i

def test():
    for i in range(4):
        yield i

g=test()
for n in [1,10]:
    g=(add(n,i) for i in g)

print(list(g))

面试题2

三、

import os

def init(func):
    def wrapper(*args,**kwargs):
        g=func(*args,**kwargs)
        next(g)
        return g
    return wrapper

@init
def list_files(target):
    while 1:
        dir_to_search=yield
        for top_dir,dir,files in os.walk(dir_to_search):
            for file in files:
                target.send(os.path.join(top_dir,file))
@init
def opener(target):
    while 1:
        file=yield
        fn=open(file)
        target.send((file,fn))
@init
def cat(target):
    while 1:
        file,fn=yield
        for line in fn:
            target.send((file,line))

@init
def grep(pattern,target):
    while 1:
        file,line=yield
        if pattern in line:
            target.send(file)
@init
def printer():
    while 1:
        file=yield
        if file:
            print(file)

g=list_files(opener(cat(grep('python',printer()))))

g.send('/test1')

协程应用：grep -rl /dir

tail&grep

 内置函数相关的面试题：

现有两个元组(('a'),('b')),(('c'),('d'))，请使用python中匿名函数生成列表[{'a':'c'},{'b':'d'}]

#答案一
test = lambda t1,t2 :[{i:j} for i,j in zip(t1,t2)]
print(test(t1,t2))
#答案二
print(list(map(lambda t:{t[0]:t[1]},zip(t1,t2))))
#还可以这样写
print([{i:j} for i,j in zip(t1,t2)])

coding

1.下面程序的输出结果是：
d = lambda p:p*2
t = lambda p:p*3
x = 2
x = d(x)
x = t(x)
x = d(x)
print x

2.现有两元组(('a'),('b')),(('c'),('d')),请使用python中匿名函数生成列表[{'a':'c'},{'b':'d'}]

3.以下代码的输出是什么？请给出答案并解释。
def multipliers():
    return [lambda x:i*x for i in range(4)]
print([m(2) for m in multipliers()])
请修改multipliers的定义来产生期望的结果。

练习

第三题答案：

 使用递归函数实现斐波那契数列：

def func(n,a=1,b=1):
    if n == 1 :
        return a
    return func(n-1,b,a+b)
print(func(100))

使用递归实现阶乘算法：

def fac(n):
    if n == 1 :
        return 1
    return n * fac(n-1)
print(fac(10))

 纸牌游戏（《流畅的Python》）

l

import collections
from random import choice
import random
Card = collections.namedtuple('Card',['rank','suit'])
class FrenchDeck:
    ranks = [str(n) for n in range(2,11)] + list('JQKA')
    suits = 'spades diamonds clubs hearts'.split()    #黑桃  方片  梅花  红桃
    def __init__(self):
        self.cards = [Card(rank,suit) for suit in self.suits for rank in self.ranks]
    def __len__(self):
        return len(self.cards)
    def __getitem__(self, position):
        return self.cards[position]
beer_card = Card('7','diamonds')  #方片 7
print(beer_card)
deck = FrenchDeck()
print(deck.__dict__)
print(len(deck))
print(deck[0])
print(deck[-1])
print(choice(deck))          #对应  from random import choice
n = random.choice(deck)
print(n)
print(deck[:3])
print(deck[12::13])
for card in deck:
    print(card)
for card in reversed(deck):
    print(card)
print(Card('Q','hearts') in deck)
print(Card('7','beasts') in deck)
from random import shuffle   #随机洗牌
shuffle(deck)

去重题型

class Person:
    def __init__(self,name,age,sex):
        self.name = name
        self.age = age
        self.sex = sex

    def __hash__(self):
        return hash(self.name+self.sex)

    def __eq__(self, other):
        if self.name == other.name and self.sex == other.sex:return True


p_lst = []
for i in range(84):
    p_lst.append(Person('egon',i,'male'))

print(p_lst)
print(set(p_lst))

一道面试题