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元组
#定义:与列表类型比,只不过[]换成() age=(11,22,33,44,55)本质 age=tuple((11,22,33,44,55)) #作用:存多个值,对比列表来说,元组不可变(是可以当做字典的key的),主要是用来读 |
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Index()
print(t.index('b')) #索引出元素第一次出现的位置,还可以指定在某一范围里查找,这里默认在整个元组里查找输出1 |
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Count()
print(t.count('b')) #计算元素出现的次数,这里输出2 |
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Len()
print(len(t)) #输出远组的长度,这里输出4 |
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For循环
for i in t: print(i) #循环打印出元组数据 |
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切片
print(t[1:3]) #切片 输出('b','b') |
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字典
#作用:存多个值,key-value存取,取值速度快 #定义:key必须是不可变类型,value可以是任意类型 info={'name':'egon','age':18,'sex':'male'} #本质info=dict({....}) 或 info=dict(name='egon',age=18,sex='male') 或 info=dict([['name','egon'],('age',18)]) 或 {}.fromkeys(('name','age','sex'),None) |
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Get() 获取values
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如果存在key值,则返回, 不存在可以指定返回内容, 不指定返回内容为None
d = {'Michael': 95, 'Bob': 75, 'Tracy': 85} d.get('Bob') #根据key获取values,如果不存在返回None,这里输出75 |
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Pop() 删除一个key
d.pop('Bob') #根据键删除某一元素 d={'Michael': 95, 'Tracy': 85} |
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添加一个元素
d['Jason']=99 #新增元素 d={'Michael': 95, 'Tracy': 85, 'Jason': 99} |
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Len()
print(len(d)) #输出字典长度,这里输出3 |
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In 判断是否存在
print('Jason' in d) #python3 中移除了 has_key,要判断键是否存在用in |
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For循环
for i in d: #循环默认keys输出 print(i) ========================================= for i in d.values(): #循环按值输出values print(i) ========================================= for k,v in d.items(): #循环按输出key,values print(k,v) ========================================= |
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字典的独有功能
'''1. 获取字典中所有的键''' print(userinfo.keys()) ====================================================== # for item in userinfo.keys(): # print(item) '''2. 获取字典中所有的值''' userinfo.values() ====================================================== # for i in userinfo.values(): # print(i) ====================================================== '''3. 获取字典中所有的键值对''' userinfo.items() # print(userinfo) ====================================================== # for v1,v2 in userinfo.items(): # print(v1,v2) |
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Update
userinfo.update({'zhan','san'}) # 存在则更新, 不存在则添加 |
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公共
len('username') # len u = userinfo['pass'] # 索引 userinfo['pass'] = 124 # 修改 del userinfo['pass'] # 删除键值对 userinfo['pass'] = 156 # 字典里,如果存在则修改,不存在则添加 |
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例题
''' mess = 'k1|v1,k2|v2,k3|v3' 转换成字典info info = {k1:v1,k2:v2,k3:v3} ''' # info = {} # # mess = 'k1|v1,k2|v2,k3|v3' # # for item in mess.split(','): # # v1,v2 = item.split('|') # # info[v1]=v2 # # print(info) ====================================================== # dict # 默认按照键判断, 判断x是否是字典的键 v1 = {'k1':1,'k2':2,'k3':3} if 'x' in v1: # keys pass ====================================================== # 判断k1是否在其中 if 'k1' in v1: pass ====================================================== # 判断v2是否在其中 # 1. fla = '不存在' for cc in v1.values(): # values if cc == 'v2': fla = '存在' print(fla) ====================================================== # 2. if 'v2' in list(v1.values()): pass ====================================================== # 3. if 'v2' in v1.values(): print('1') else: print('2') ====================================================== # 判断k2:v2 是否在其中 vl = v1.get('k2') if v1 == 'v2': pass else: pass |
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集合
#作用:去重,关系运算, #定义: 知识点回顾 可变类型是不可hash类型 不可变类型是可hash类型 #定义集合: 集合:可以包含多个元素,用逗号分割, 集合的元素遵循三个原则: 1:每个元素必须是不可变类型(可hash,可作为字典的key) 2: 没有重复的元素 3:无序 注意集合的目的是将不同的值存放到一起,不同的集合间用来做关系运算,无需纠结于集合中单个值 |
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Add() 添加元素
a = {'nick','jenny','suo'} a.add('The knife girl') print(a) |
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Update更新列表
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} a.update(b) print(a) |
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#清空 a = {'nick','jenny','suo'} a.clear() print(a) |
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Discard 删除指定的元素
#移除指定元素,不存在不报错 a = {'nick','jenny','suo'} a.discard('suo') print(a) |
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remove
#移除指定元素,不存在则报错 a = {'nick','jenny','suo'} a.remove('suo') print(a) a.remove('suo') print(a) |
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pop
#移除随机元素,并赋给新值 a = {'nick','jenny','suo'} set = a.pop() print(set) |
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} set = a.difference(b) print(set) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} a.difference_update(b) print(a) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} set = a.intersection(b) print(set) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} a.intersection_update(b) print(a) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} set = a.symmetric_difference(b) print(set) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} a.symmetric_difference_update(b) print(a) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} set = a.union(b) print(set) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny','The knife girl'} set = a.isdisjoint(b) print(set) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny'} set = a.issubset(b) print(set) =========================================
a = {'nick','jenny','suo'} b = {'nick','jenny'} set = a.issuperset(b) print(set) ========================================= |