一 字典的相关函数
1.1 增函数
dictvar = {"a":1,"b":2}
dictvar["c"] = 3
print(dictvar)
执行
[root@node10 python]# python3 test.py test.py
{'a': 1, 'b': 2, 'c': 3}
fromkeys()
使用一组键和默认值创建字典 (不常用 赋初始值)
listvar = ['a','b','c']
res = {}.fromkeys(listvar,None)
print(res)
res = {}.fromkeys(listvar,[1,2])
print(res)
执行
[root@node10 python]# python3 test.py test.py
{'a': None, 'b': None, 'c': None}
{'a': [1, 2], 'b': [1, 2], 'c': [1, 2]}
1.2 删函数
pop()
通过键去删除键值对 (若没有该键可设置默认值,预防报错)
dictvar = {"tiantang":"天堂","renjian":"人间","diyu":"地狱"}
res = dictvar.pop("tiantang")
print(res)
print(dictvar)
# pop可以在第二个参数上指定默认值,预防不存在改键时报错
res = dictvar.pop("ppoiiiiuiuiuiiiuiui","对不起,改键不存在")
print(res)
执行
天堂
{'renjian': '人间', 'diyu': '地狱'}
对不起,改键不存在
popitem() 删除最后一个键值对
dictvar = {"tiantang":"天堂","renjian":"人间","diyu":"地狱"}
res= dictvar.popitem()
print(res)
print(dictvar)
执行
[root@node10 python]# python3 test.py test.py
('diyu', '地狱')
{'tiantang': '天堂', 'renjian': '人间'}
clear() 清空字典
dictvar = {"tiantang":"天堂","renjian":"人间","diyu":"地狱"}
dictvar.clear()
print(dictvar)
执行
[root@node10 python]# python3 test.py test.py
{}
1.3 更改函数
update()
批量更新(有该键就更新,没该键就添加)
dictvar = {"tiantang":"天堂","renjian":"人间","diyu":"地狱"}
dictvar.update({'tiantang':111,"jungle":"盘古"}) # 在括号里面写字典数据 (一个字典即可)
print(dictvar)
# 写法二
dictvar.update(a=1,b=2) #(在括号里面写关键字参数,是多个)
print(dictvar)
执行
[root@node10 python]# python3 test.py test.py
{'tiantang': 111, 'renjian': '人间', 'diyu': '地狱', 'jungle': '盘古'}
{'tiantang': 111, 'renjian': '人间', 'diyu': '地狱', 'jungle': '盘古', 'a': 1, 'b': 2}
1.4 查找函数
get()
通过键获取值(若没有该键可设置默认值,预防报错)
dictvar = {"tiantang":"天堂","renjian":"人间","diyu":"地狱"}
res = dictvar.get("tiantang123")
print(res)
# 如果这个键不存在,可以指定默认值, 如果不写第二个参数值,默认返回None
res = dictvar.get("tiantang123","对不起,该键不存在")
print(res)
res = dictvar['tiantang']
print(res)
执行
[root@node10 python]# python3 test.py test.py None 对不起,该键不存在 天堂
keys()
将字典的键组成新的可迭代对象
dictvar = {"tiantang":"天堂","renjian":"人间","diyu":"地狱"}
res = dictvar.keys()
print(res)
for i in dictvar.keys():
print(i)
print("==============================")
for i in dictvar:
print(i)
执行
tiantang renjian diyu ============================== tiantang renjian diyu
values()
将字典中的值组成新的可迭代对象
dictvar = {"tiantang":"天堂","renjian":"人间","diyu":"地狱"}
res = dictvar.values()
print(res)
for i in res:
print(i)
执行
dict_values(['天堂', '人间', '地狱']) 天堂 人间 地狱
items()
将字典的键值对凑成一个个元组,组成新的可迭代对象
dictvar = {"tiantang":"天堂","renjian":"人间","diyu":"地狱"}
res = dictvar.items()
print(res)
for i in res:
print(i)
print ("================================")
for a,b in res:
print(a,b)
执行
dict_items([('tiantang', '天堂'), ('renjian', '人间'), ('diyu', '地狱')])
('tiantang', '天堂')
('renjian', '人间')
('diyu', '地狱')
================================
tiantang 天堂
renjian 人间
diyu 地狱
二 集合的相关操作
作用:交差并补
2.1 intersection() 交集
set1 = {"曹操","刘禅","孙权","刘备"}
set2 = {"郭嘉","刘禅","张昭","诸葛亮"}
res = set1.intersection(set2)
print(res)
res = set1 & set2
print(res)
2.2 difference() 差集
set1 = {"曹操","刘禅","孙权","刘备"}
set2 = {"郭嘉","刘禅","张昭","诸葛亮"}
res = set1.difference(set2)
print(res)
res = set2.difference(set1)
print(res)
res = set1 - set2
print(res)
执行
{'刘备', '孙权', '曹操'}
{'郭嘉', '张昭', '诸葛亮'}
{'刘备', '孙权', '曹操'}
2.3 union() 并集
set1 = {"曹操","刘禅","孙权","刘备"}
set2 = {"郭嘉","刘禅","张昭","诸葛亮"}
res = set1.union(set2)
print(res)
res = set1 | set2
print(res)
执行
[root@node10 python]# python3 test.py test.py
{'孙权', '曹操', '刘备', '郭嘉', '张昭', '刘禅', '诸葛亮'}
{'孙权', '曹操', '刘备', '郭嘉', '张昭', '刘禅', '诸葛亮'}
2.4 symmetric_difference() 对称差集 (补集情况涵盖在其中)
set1 = {"曹操","刘禅","孙权","刘备"}
set2 = {"郭嘉","刘禅","张昭","诸葛亮"}
res = set1.symmetric_difference(set2)
print(res)
res = set1 ^ set2
print(res)
执行
[root@node10 python]# python3 test.py test.py
{'曹操', '郭嘉', '刘备', '孙权', '张昭', '诸葛亮'}
{'曹操', '郭嘉', '刘备', '孙权', '张昭', '诸葛亮'}
2.5 issubset() 判断是否是子集
真子集:子集元素一定少于父集,完全被包含在其中
set_father = {"曹操","刘禅","孙权","刘备"}
set_son = {"曹操","刘禅"}
res = set_son.issubset(set_father)
print(res)
res = set_son < set_father
print(res)
print ("==========================================")
set_father = {"曹操","刘禅","孙权","刘备"}
set_son = {"曹操","刘禅","孙权","刘备"}
res = set_son <= set_father
print(res)
执行
True True ========================================== True
2.6 判断是否是父集
set_father = {"曹操","刘禅","孙权","刘备"}
set_son = {"曹操","刘禅"}
res = set_father.issuperset(set_son)
print(res)
res = set_father > set_son
print(res)
print ("==========================================")
set_father = {"曹操","刘禅","孙权","刘备"}
set_son = {"曹操","刘禅","孙权","刘备"}
res = set_son >= set_father
print(res)
执行
[root@node10 python]# python3 test.py test.py True True ========================================== True
2.7 isdisjoint() 检测两集合是否不相交
不相交 True 相交False
set_father = {"曹操","刘禅","孙权","刘备"}
set_son = {"曹操","刘禅"}
res = set_father.isdisjoint(set_son)
print(res)
执行
[root@node10 python]# python3 test.py test.py False
三 集合的相关函数
3.1 增函数
add() 向集合中添加数据
setvar = {"曹操","刘禅","孙权","刘备"}
setvar.add("曹丕")
print(setvar)
执行
[root@node10 python]# python3 test.py test.py
{'曹丕', '刘禅', '曹操', '刘备', '孙权'}
update() 迭代着增加
setvar = {"曹操","刘禅","孙权","刘备"}
lst = ['曹植','曹丕']
setvar.update(lst)
print(setvar)
print("==================================")
setvar = {"曹操","刘禅","孙权","刘备"}
setavar = "abcd"
setvar.update(setvar)
print(setvar)
把列表当中的元素一个一个拿出来放进集合中,需要时可迭代性数据
执行
{'孙权', '刘禅', '曹操', '刘备', '曹丕', '曹植'}
==================================
{'刘禅', '孙权', '曹操', '刘备'}
3.2 删除函数
clear() 清空集合
setvar = {"曹操","刘禅","孙权","刘备"}
setvar.clear()
print(setvar)
执行
[root@node10 python]# python3 test.py test.py set()
pop() 随机删除集合中的一个数据
setvar = {"曹操","刘禅","孙权","刘备"}
setvar.clear()
print(setvar)
执行三次
[root@node10 python]# python3 test.py test.py
刘备
{'孙权', '曹操', '刘禅'}
[root@node10 python]# python3 test.py test.py
曹操
{'刘备', '刘禅', '孙权'}
[root@node10 python]# python3 test.py test.py
刘禅
{'曹操', '刘备', '孙权'}
remove() 删除集合中指定的值(不存在则报错)
setvar = {"曹操","刘禅","孙权","刘备"}
setvar.remove("刘禅")
print(setvar)
执行
[root@node10 python]# python3 test.py test.py
{'曹操', '刘备', '孙权'}
discard() 删除集合中指定的值(不存在的不删除 推荐使用)
setvar = {"曹操","刘禅","孙权","刘备"}
setvar.discard("刘禅121211212") # 如果这个值不存在,就不删除,也不报错
print(setvar)
setvar.discard("刘禅") # 如果这个值不存在,就不删除,也不报错
print(setvar)
执行
[root@node10 python]# python3 test.py test.py
{'刘备', '刘禅', '曹操', '孙权'}
{'刘备', '曹操', '孙权'}
3.3 冰冻集合
frozenset 可强转容器类型数据变为冰冻集合 冰冻集合一旦创建,不能在进行任何修改,只能做交叉并补操作
定义一个空冰冻集合
fz = frozenset()
print(fz,type(fz))
fz1 = frozenset([1,"2",3,4])
fz2 = frozenset("7892")
print(fz1,fz2)
执行
frozenset() <class 'frozenset'>
frozenset({1, 3, 4, '2'}) frozenset({'9', '7', '8', '2'})
并冻集合只能做交叉并补
fz1 = frozenset([1,"2",3,4])
fz2 = frozenset("7892")
res = fz1 & fz2
print(res)
执行
[root@node10 python]# python3 test.py test.py
frozenset({'2'})