python常用的几个功能/模块

1 求list的子集

import itertools

# 有序
print(list(itertools.permutations([1, 2, 3, 4], 2)))
# [(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)]

# 无序
print(list(itertools.combinations([1, 2, 3, 4], 2)))
# [(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]

2 求两个集合的交集、并集

a=[1,3,5]
b=[1,2,3]
set(a) & set(b)  #交集 
set(a) | set(b)  #并集
set(a) - set(b)  #差集（得到的结果是set([5])）

3 排列组合

1）一个list里面取n个值（无序，组合）：

import copy


def combine(l, n):
    answers = []
    one = [0] * n

    def next_c(li=0, ni=0):
        if ni == n:
            answers.append(copy.copy(one))
            return
        for lj in range(li, len(l)): # python2 和python3一个用range，一个用xrange
            one[ni] = l[lj]
            next_c(lj + 1, ni + 1)
    next_c()
    return answers


list = [1, 2, 3, 4, 5]
sublists = combine(list, 3)

输出的结果为()：

[[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]

2）一个list里面取n个值（有序，排列，此处以n = 3为例）：

list = [1, 2, 3, 4, 5]
sublists = []
for i in range(len(list)):
    for j in range(len(list)):
        if j == i:
            continue
        else:
            for k in range(len(list)):
                if k == j or k == i:
                    continue
                else:
                    sublists.append([list[i], list[j], list[k]])

输出的结果为（）：

[[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 2], [1, 3, 4], [1, 3, 5], [1, 4, 2], [1, 4, 3], [1, 4, 5], [1, 5, 2], [1, 5, 3], [1, 5, 4], [2, 1, 3], [2, 1, 4], [2, 1, 5], [2, 3, 1], [2, 3, 4], [2, 3, 5], [2, 4, 1], [2, 4, 3], [2, 4, 5], [2, 5, 1], [2, 5, 3], [2, 5, 4], [3, 1, 2], [3, 1, 4], [3, 1, 5], [3, 2, 1], [3, 2, 4], [3, 2, 5], [3, 4, 1], [3, 4, 2], [3, 4, 5], [3, 5, 1], [3, 5, 2], [3, 5, 4], [4, 1, 2], [4, 1, 3], [4, 1, 5], [4, 2, 1], [4, 2, 3], [4, 2, 5], [4, 3, 1], [4, 3, 2], [4, 3, 5], [4, 5, 1], [4, 5, 2], [4, 5, 3], [5, 1, 2], [5, 1, 3], [5, 1, 4], [5, 2, 1], [5, 2, 3], [5, 2, 4], [5, 3, 1], [5, 3, 2], [5, 3, 4], [5, 4, 1], [5, 4, 2], [5, 4, 3]]

4 斐波那契数列（变形，跳台阶问题）

一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法。 
dp[n]=dp[n−1]+dp[n−2]

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number):
        # write code here
        '''
        n = 1 : 1 
        n = 2 : 1+1 = 2
        n = 3 : dp[n-2]+dp[n-1]
        '''
        if number == 1 or number == 2:
            return number
        dp = [1,2]
        for i in range(number-2):
            dp.append(dp[-1]+dp[-2])
        return dp[-1]

5 字典

# -*- coding: utf-8 -*-
keys = ['a', 'b', 'c']
values = [1, 2, 3]
dictionary = dict(zip(keys, values))
print dictionary
"""
输出:
{'a': 1, 'c': 3, 'b': 2}
"""
print(dictionary['a'])

6 创建一个1到n的list

list=range(1,n+1)

7 list/string逆序列

>>> a = '1232'
>>> a[::-1]
'2321'

8 list/string每隔n个元素取一个

list1[::2] #每隔一个元素取一个

如list1 = [1,2,3,4,5,6], 则list1[::2] = [1,3,5]