python练手习题

不断记录python常见习题,不断寻求更多更好的解决办法。持续更新中.....

练习:

1. list两两元素交换位置,如[1,2,3,4,5,6] 执行后为 -> [2,1,4,3,6,5]

第一种实现方式,比较原始:

    l = [1,2,3,4]
    l1 = l[::2]
    l2 = l[1::2]
    print l1
    print l2
    c = [ ]
    for i in range( max ( len(l1), len(l2) ) ):
        print i
        if l2:
            c.append(l2[i])
        if l1:
            c.append(l1[i])
    print c

 

2. 打乱有序序列l

#!/usr/bin/python
# -*- coding: UTF-8 -*-

from random import shuffle
l = list(range(10))
shuffle(l)
print l

 

3. 一个支持各种操作符的类

 

#!/usr/bin/python
# -*- coding: UTF-8 -*-
import math
import itertools


class Vector(object):
    typecode = 'd'

    def __init__(self, components):
        self._components = [float(i) for i in components]

    def __repr__(self):
        return 'Vector({})'.format(list(self._components))

    def __str__(self):
        return str(tuple(self))

    def __len__(self):
        return len(self._components)

    def __getitem__(self, item):
        return self._components[item]

    def __getattr__(self, item):
        cls = type(self)

    def __iter__(self):
        return iter(self._components)

    def __eq__(self, other):
        return tuple(self) == tuple(other)

    def __abs__(self):
        return math.sqrt(sum(x * x for x in self))

    def __neg__(self):
        return Vector(-x for x in self)

    def __pos__(self):
        return Vector(x for x in self)

    def __add__(self, other):
        pairs = itertools.izip_longest(self, other, fillvalue=0.0)
        return Vector(a + b for a, b in pairs)


if __name__ == '__main__':
    vector2 = Vector(range(5))
    print vector2 == range(5)
    print abs(vector2)
    print str(vector2)
    print repr(vector2)
    print -vector2
    print +vector2
    print vector2+Vector(range(2))

4. 列表操作,属于列表内部的重新组合

"""
给定一个字符串列表`strlist`和整数`k`。
请编写函数`func`,它返回字符串列表中`k`个相邻字符串中最长的第一个。

示例:
func(["this", "is", "an", "example"], 1) --> "example"
func(["this", "is", "another", "example"], 1) --> "another"
func(["this", "is", "another", "example"], 2) --> "anotherexample"

假设字符串列表的长度为`n`,如果`n = 0`或`k > n`或`k <= 0`则返回空字符串。

示例:
func([], 1) --> ""
func(["this", "is", "an", "example"], 5) --> ""
func(["this", "is", "an", "example"], 0) --> ""
"""
#!/usr/bin/env python
# -*- coding: utf-8 -*-


def func(strlist, k):
    if strlist == [] or len(strlist) < k:
return ''
new_list = [strlist[num:num + k] for num in range(len(strlist) - k + 1)]
l = [''.join(n) for n in new_list]
l_num = [len(nu) for nu in l]
return l[l_num.index(max(l_num))]
 

 

 5. 列表操作,属于从一个list中抽取所有特性的子list。

"""
给定一个仅包含整数,且按照大小顺序排好序的列表,列表内不存在重复的整数。
实现一个函数,将列表格式化为由`,`(逗号)分隔的字符串;如果相邻的整数(至少3个)是
连续的(值相差1为连续),则将这几个相邻的整数格式化为由`-`(减号)分隔、左右分别为
起始和终止位整数的字符串。

示例:
func([-6,-3,-2,-1,0,1,3,4,5,7,8,9,10,11,14,15,17,18,19,20]) --> '-6,-3-1,3-5,7-11,14,15,17-20'
func([-3,-2,-1,2,10,15,16,18,19,20]) --> '-3--1,2,10,15,16,18-20'

特殊情况示例:
func([]) --> ''
"""

 

def func(args):
    if len(args) == 0:
        return ''
    li = []
    flag = 0
    args_index = 0
    while args_index < len(args):
        args_index_value = args[args_index]
        if args_index+1 == len(args):
            li.append(str(args_index_value))
            break
        for args_back_value in args[args_index+1:]:
            if args_index_value-args_back_value+flag == -1:
                flag += 1
            else:
                if flag >= 2:
                    args_index += flag
                    li.append(str(args_index_value)+'-'+str(args[args_index]))
                else:   
                    li.append(str(args_index_value))
                flag = 0
                args_index += 1
                break
    return ','.join(li)

 6. 合并两个有序列表

list_1 = [1,2,3,4]
list_2 = [3,5,7,8]
list_merge = list_1 + list_2
print sorted(list_merge)

7. 获取list的交集、差集、并集等,思路先把list转换成集合,集合操作后转换回list

list_1 = [1, 2, 3, 4]
list_2 = [3, 5, 7, 8]
print list(set(list_1) & set(list_2))
print list(set(list_1)|set(list_2))
print list(set(list_1) - set(list_2))

 8. tuple与list之间的互相转换

tp = ('a', 'b', 'c')
li = ['a', 'b', 'c']
print list(tp)
print tuple(li)

 9. 

"""
你控制着你的游戏角色在地图上行走,每走一步都会消耗1点体力。
编写一个函数为你的角色规划行进路线,在不改变原有行进顺序的情况下,用最节省体力的方式到达目的地。

示例:

func(["NORTH", "SOUTH", "EAST"]) --> ["EAST"]
向"NORTH"行进后立即向"SOUTH"行进,"NORTH"和"SOUTH"是相反的方向,所以["NORTH", "SOUTH"]是可以减省掉的。

func(["EAST", "WEST", "WEST", "WEST"]) --> ["WEST", "WEST"]
向"EAST"行进后立即向"WEST"行进,"EAST"和"WEST"是相反的方向,所以["EAST", "WEST"]是可以减省掉的。

特殊情况说明:

func(["NORTH", "WEST", "EAST", "SOUTH"]) --> func(["NORTH", "SOUTH"]) --> []
["WEST", "EAST"]减省掉后,行进路线变成了["NORTH", "SOUTH"],可以进一步减省掉。

func(["NORTH", "WEST", "SOUTH", "EAST"]) --> ["NORTH", "WEST", "SOUTH", "EAST"]
["NORTH", "WEST"], ["WEST", "SOUTH"], ["SOUTH", "EAST"]任意两组相邻的方向,都不是相反的方向,所以不能减省掉。
"""
def func(directions):
    tmp = ''
    for index, li in enumerate(directions):
        if tmp == '':
            tmp = li
        else:
            if tmp == 'NORTH' and li == 'SOUTH' or tmp == 'SOUTH' and li == 'NORTH' or \
                                    tmp == 'EAST' and li == 'WEST' or tmp == 'WEST' and li == 'EAST':
                tmp = ''
                directions = func(directions[:index - 1] + directions[index + 1:])
            else:
                tmp = li
    return directions

10 正则相关

"""
请实现一个函数来对目标字符串进行校验,使其满足以下全部条件:

* 不少于6个字符
* 包含至少一个小写字母
* 包含至少一个大写字母
* 包含至少一个数字
* 只能包含大小写字母和数字

示例:
func("12ABcd") --> True
func("12ABc") --> False
func("12ABCD") --> False
func("12abcd") --> False
func("ABCdef") --> False
func("12AB cd") --> False
"""
import re

def func(string):
    if len(string)< 6:
        return False
    if not string.isalnum():
        return False
    if string.isdigit():
        return False
    if string.isalpha():
        return False
    if len(re.findall(r'\d+', string)) == 0 :
        return False
    if not re.match(r'(?=.*[a-z])', string):
        return False
    if not re.match(r'(?=.*[A-Z])', string):
        return False
    return True

 11. 有一个长度是101的数组,存在1~100的数字,有一个是重复的,找出重复出来

    import random
    l = list(range(100))
    random.shuffle(l)
    random_num = random.randint(0, 99)
    l.insert(random_num, random_num)
    print([num for num in l if l.count(num) == 2])
    print([(index, num) for index, num in enumerate(l) if l.count(num) == 2])

 

 

posted @ 2018-11-18 10:44  曹毅  阅读(379)  评论(0编辑  收藏  举报