python之基本数据类型及深浅拷贝

一.数据基本类型之set集合

set和dict类似,也是一组key的集合,但不存储value。由于key不能重复,所以,在set中,没有重复的key

set集合,是一个无序且不重复的元素集合

1.创建

s = set()  #创建空集合

s = {'values1','values2'}  #非空集合

2.转换

l = [1,2,5,11]
t = (11,22,12)
#元组转集合
st2 = set(t)
#列表转集合
st = set(l)
print(st)
print(st2)

 

3.常用支持操作

  添加元素-->add(key)

s = set()
s.add(1)
print(s)

  删除元素-->remove(key)

s = set([1,2,3])
s.remove(1)
print(s)

 

   清除元素-->clear()

s = set([1,2,3,4,5])
print(s)
s.clear()
print(s)

  比较元素-->difference()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#set1中有而set2中没有的值
ret = set1.difference(set2)
print(ret)

  删除两集合中相同的元素-->difference_update()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#从set1中删除和set2中相同的元素
set1.difference_update(set2)
print(set1)
print(set2)

  移除元素-->discard(values)

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#移除指定元素,不存在不会报错,remove()不存在会报错,建议discard
set1.discard(44)
print(set1)

  取交集值-->intersection()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#取两个set集合的交集值
ret = set1.intersection(set2)
print(ret)

  取交集并更新-->intersection_update()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#取交集并更新到set1中
set1.intersection_update(set2)
print(set1)

  判断是否交集-->isdisjoint()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#如果两个集合有交集返回false,反之返回true
print(set1.isdisjoint(set2))

  子序列-->issubset()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#判断是否是子集,是返回true,反之返回Flase
print(set1.issubset(set2))

  父序列-->issuperset()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#是否是父序列,是返回True,反之返回Flase
print(set1.issuperset(set2))

  对称交集-->symmetric_difference()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#对称交集,取两个集合中互不存在的元素,生成一个新的集合
ret = set1.symmetric_difference(set2)
print(ret)

  对称交集并更新-->symmetric_difference_update()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#对称交集,并更新元素到set1中
set1.symmetric_difference_update(set2)
print(set1)

  并集-->union()

set1 = {1,44,87,23,55}
set2 = {1,44,88,23,67}
#并集并更新到新的集合中
ret = set1.union(set2)
print(ret)

二.深浅拷贝

1.数字和字符串

  对于 数字 和 字符串 而言,赋值、浅拷贝和深拷贝无意义,因为其永远指向同一个内存地址

 

import copy
# ######### 数字、字符串 #########
n1 = 123
# n1 = "i am alex age 10"
print(id(n1))
# ## 赋值 ##
n2 = n1
print(id(n2))
# ## 浅拷贝 ##
n2 = copy.copy(n1)
print(id(n2))
  
# ## 深拷贝 ##
n3 = copy.deepcopy(n1)
print(id(n3))

 

 

二,其他数据类型

对于list dict,tuple  浅拷贝只拷贝最外一层,深拷贝除了最后一层(因最后一层是字符串)其余的都拷贝

  • 赋值

  赋值,只是创建一个变量,该变量指向原来内存地址,如

n1 = {"k1": "wu", "k2": 123, "k3": ["alex", 456]}
  
n2 = n1

  解析图如下:

  • 浅拷贝

  浅拷贝,在内存中只额外创建第一层数据

  

import copy
  
n1 = {"k1": "wu", "k2": 123, "k3": ["alex", 456]}
  
n3 = copy.copy(n1)

解析图如下:

  • 深拷贝:

  深拷贝,在内存中将所有的数据重新创建一份(排除最后一层,即:python内部对字符串和数字的优化)

import copy
  
n1 = {"k1": "wu", "k2": 123, "k3": ["alex", 456]}
  
n4 = copy.deepcopy(n1)

解析图如下:

 

三.集合作业

1.寻找差异,并将new_dict的值更新到old_dict中

old_dict = {
    "#1":11,
    "#2":22,
    "#3": 100
}

new_dict = {
    "#1":33,
    "#4":22,
    "#7": 100
}
#!/usr/bin/env python
# -*- coding:utf-8 -*-
old_dict = { "#1":11, "#2":22, "#3": 100 } new_dict = { "#1":33, "#4":22, "#7": 100 } old_keys = old_dict.keys() new_keys = new_dict.keys() old_set = set(old_keys) new_set = set(new_keys) #old存在new不存在,并删除new中不存在,old 中存在的元素 del_set = old_set.difference(new_set) for i in del_set: del old_dict[i] #new存在old不存在 add_set = new_set.difference(old_set) for a in add_set: old_dict[a] = new_dict[a] #更新旧数据表 update_set = old_set.intersection(new_set) for u in update_set: old_dict[u] = new_dict[u] print(old_dict)

 

posted @ 2016-05-08 19:08  jl_bai  阅读(398)  评论(0编辑  收藏  举报

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