Python基础 - 05元祖集合字典
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Python基础 - 05元祖集合字典
三、元祖
3.1 定义元祖
tup = ()
print(type(tup)) # <class 'tuple'>
tup1 = ('aa')
print(type(tup1)) # <class 'str'>
tup2 = ('aa',)
print(type(tup2)) # <class 'tuple'> 只有一个元素时，加上逗号表示元祖，否则为字符串
tup3 = ('aa','bb')
print(type(tup3)) # <class 'tuple'>
aTuple = ('1', 'hello', 'AAA')
print(aTuple) # ('1', 'hello', 'AAA')
print(aTuple[2]) # AAA
print(aTuple[:2]) # ('1', 'hello')
print(aTuple[::-1]) # ('AAA', 'hello', '1')
print(aTuple.count('1')) # 1
print(aTuple.index('hello', 1)) # 1
print(aTuple.index('hello')) # 1
aTuple = ('1', 'hello','AAA', 'hello')
print(aTuple.index('hello', 2)) # 3,从下标2开始
print(aTuple.index('hello')) # 1，返回第一个遇到的元素位置
t3 = ('A', 'a', 'b', 'c', 'd', 'e', 'd')
index = t3.index('d', 1, 7)
print(index) # 4
print(len(t3)) # 7
if 'c' in t3:
print('存在')
else:
print('不存在')
for i in t3:
print(i)
# 元祖转列表
print(list(t3)) # ['A', 'a', 'b', 'c', 'd', 'e', 'd']
list3 = ['1', '2', '3']
print(tuple(list3)) # ('1', '2', '3')
from collections import namedtuple
# 命名元祖
NetworkAddress = namedtuple('NetworkAddress',['hostname','port'])
a = NetworkAddress('www.python.org', 80)
print(a.hostname)
print(a.port)
print(type(a)) # <class '__main__.NetworkAddress'>
print(len(a)) # 2
print(isinstance(a, tuple)) # True
四、集合
4.1 集合定义： 无序不可重复
set1 = {'AAA','AAA'}
print(type(set1)) # <class 'set'>
print(set1) # {'AAA'}
list2 = [1,3,6,5,3,2,4,6,8,6,7,9]
set2 = set(list2)
print(set2) # {1, 2, 3, 4, 5, 6, 7, 8, 9}
set3 = {}
print(type(set3)) # <class 'dict'>
set4 = set()
print(type(set4)) # <class 'set'>
print(len(set4)) # 0
4.2 集合的增删改查
set4.add('西游记')
print(set4) # {'西游记'}
set4.update(set1)
print(set4) # {'西游记', 'AAA'}
set5 = {'AA','CC','DD','FF'}
set5.remove('AA')
print(set5) # {'CC', 'DD', 'FF'}
# KeyError: 'AA'
# set5.remove('AA')
del set1
# NameError: name 'set1' is not defined
# print(set1)
set4.clear()
print(set4) # set()
# 随机移除
set5.pop()
print(set5) # {'CC', 'DD'}
4.3 集合的交集、差集、并集
set2 = {1, 2, 3, 4, 5}
set3 = {1, 2, 6, 7, 8, 9}
result = set2.intersection(set3) # 交集
print(result) # {1, 2}
result = set2.union(set3) # 并集
print(result) # {1, 2, 3, 4, 5, 6, 7, 8, 9}
result = set2.difference(set3) # 差集
print(result) # {3, 4, 5}
print(set2 & set3) # {1, 2}
print(set2 | set3) # {1, 2, 3, 4, 5, 6, 7, 8, 9}
print(set2 - set3) # {3, 4, 5}
五、字典
5.1 定义、增删改查
dict1 = {}
print(type(dict1)) # <class 'dict'>
dict1['name'] = '王路飞'
print(dict1) # {'name': '王路飞'}
dict1['age'] = 30
dict1['sex'] = '男'
print(dict1) # {'name': '王路飞', 'age': 30, 'sex': '男'}
# 修改
dict1['age'] = 33
print(dict1) # {'name': '王路飞', 'age': 33, 'sex': '男'}
dict1['score'] = 90
print(dict1) # {'name': '王路飞', 'age': 33, 'sex': '男', 'score': 90}
dict1['score'] += 5
print(dict1) # {'name': '王路飞', 'age': 33, 'sex': '男', 'score': 95}
# pop(key): 根据key实现删除，删除的是键值对，返回值是key对应的value
book = {'name': '《西游记》', 'author': '吴承恩', 'price': 22.38, 'publish': '中信出版社'}
r = book.pop('price')
print(r) # 22.38
print(book) # {'name': '《西游记》', 'author': '吴承恩', 'publish': '中信出版社'}
pub = book.popitem()
print(pub) # ('publish', '中信出版社') 最后一个键值对
print(book) # {'name': '《西游记》', 'author': '吴承恩'}
del book['author'] # 类似于pop(key)
print(book) # {'name': '《西游记》'}
5.2 字典操作
keys() / values() / items() / get() / pop() / popitem() / clear() / copy() / fromkeys(s,t)
value = book.get('name')
print(value) # 《西游记》
print(len(book)) # 4
value1 = book.get('key1', 'ABC')
print(value1) # None, 没有可以默认返回None; ABC, 指定当key 不存在时的返回值ABC
for i in book:
print(i) # name author price publish
print(book.values()) # dict_values(['《西游记》','吴承恩',22.38,'中信出版社'])
print(list(book.values())) # ['《西游记》', '吴承恩', 22.38, '中信出版社']
print(book.items()) # dict_items([('name', '《西游记》'), ('author', '吴承恩'), ('price', 22.38), ('publish', '中信出版社')]
for i in book.items():
print(i) # ('name', '《西游记》') ('author', '吴承恩') ......
for k, v in book.items():
print(k, v) # name 《西游记》 author 吴承恩 ......
book.setdefault('price', 100)
print(book) # {'name': '《西游记》', 'author': '吴承恩', 'price': 22.38, 'publish': '中信出版社'}
book.setdefault('page', 780)
print(book) # {'name': '《西游记》', 'author': '吴承恩', 'price': 22.38, 'publish': '中信出版社', 'page': 780}
dict1 = {'a': 10, 'b': 20}
book.update(dict1)
print(book) # {'name': '《西游记》', 'author': '吴承恩', 'price': 22.38, 'publish': '中信出版社', 'page': 780, 'a': 10, 'b': 20}
print(dict1)
# TypeError: unsupported operand type(s) for +: 'dict' and 'dict'
# print(book + dict1)
result = book.fromkeys(dict1)
print(result) # {'a': None, 'b': None}
result = book.fromkeys(['c', 'd'], [1, 20])
print(result) # {'c': [1, 20], 'd': [1, 20]}
result = book.fromkeys(['name', 'ss'], [20, 90])
print(result) # {'name': [20, 90], 'ss': [20, 90]}
del book['name']
print(book) # {'author': '吴承恩', 'price': 22.38, 'publish': '中信出版社', 'page': 780}
book.clear()
print(book) # {}
字典转列表、元祖
dict1 = {'name': '张三', 'age': 18}
result = list(dict1)
print(result) # ['name', 'age']
result = tuple(dict1)
print(result) # ('name', 'age')
list1 = [('a', 10), ('b', 20), ('c', 30)]
print(dict(list1)) # {'a': 10, 'b': 20, 'c': 30}
list1 = [('a', [1,2,3]), ('b', 20), ('c', 30)]
print(dict(list1)) # {'a': [1, 2, 3], 'b': 20, 'c': 30}
六、通用方法
- list： 允许重复，有序， 有下标 []
- tuple: 允许重复，元素不能增加删除修改，只能查看 ()
- dict： 键值对存在， 键：唯一， 值：允许重复 {}
- set: 不允许重复，无序 {}
6.1 通用方法
| 运算符 | Python表达式 | 结果 | 描述 | 支持的数据类型 |
| + | [1,2]+[3,4] | [1,2,3,4] | 合并 | 字符串，列表，元祖 |
| * | ['He']*3 | ['He','He','He'] | 复制 | 字符串，列表，元祖 |
| in | 3 in (1,2,3) | True | 元素是否存在 | 字符串，列表，元祖，字典 |
| not in | 4 not in (1,2,3) | True | 元素是否不存在 | 字符串，列表，元祖，字典 |
list1 = [1,2,3,5,6,8,9,0]
result = max(list1)
print(result) # 9
result = min(list1)
print(result) # 0
result = sum(list1)
print(result) # 34
result = abs(-1)
print(result) # 1
t1 = (45,12,78,90,10,88)
result = sorted(t1)
print(result) # [10, 12, 45, 78, 88, 90]
result = sorted(t1, reverse=True)
print(result) # [90, 88, 78, 45, 12, 10]
result = chr(67)
print(result) # C
result = ord('A')
print(result) # 65
6.2 带下标的遍历
enumerate()类用于将一个可遍历的数据对象（如列表、元祖或字符串）组合成一个索引序列，同时列出数据和数据下标，一般用于for循环中。
chars = ['a', 'b', 'c', 'd']
for i, char in enumerate(chars):
print("%d, %s" %(i, char))
# 0, a
# 1, b
# 2, c
# 3, d
6.3 内置方法
print()、 input()、 type()、 id()、 len()、 bin()、 oct()、 hex()、 chr()、 ord()
七、列表推导式： 最终得到一个列表
list1 = []
for i in range(1, 10):
list1.append(i)
print(list1) # [1, 2, 3, 4, 5, 6, 7, 8, 9]
# 格式1：
list2 = [i for i in range(1, 10)]
print(list2) # [1, 2, 3, 4, 5, 6, 7, 8, 9]
list3 = [i for i in range(1, 10, 2)]
print(list3) # [1, 3, 5, 7, 9]
# 格式2： [i for i in 可迭代的 if 条件]
list4 = ['62', 'hello', '100', 'world', 'luck', '88']
list5 = [word for word in list4 if word.isalpha()]
print(list5) # ['hello', 'world', 'luck']
list6 = [word.title() if word.startswith('h') else word.upper() for word in list4]
print(list6) # ['62', 'Hello', '100', 'WORLD', 'LUCK', '88']
a = [(x, y) for x in range(1, 3) for y in range(3)]
print(a) # [(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
b = [(x, y, z) for x in range(1, 3) for y in range(3) for z in range(4, 6)]
print(b)
# [(1, 0, 4), (1, 0, 5), (1, 1, 4), (1, 1, 5), (1, 2, 4), (1, 2, 5), (2, 0, 4), (2, 0, 5), (2, 1, 4), (2, 1, 5), (2, 2, 4), (2, 2, 5)]
aa = [x for x in range(1, 101)]
bb = [aa[i:i + 3] for i in range(0, 101, 3)]
print(bb) # [[1, 2, 3], [4, 5, 6],......, [97, 98, 99], [100]]
li = [[i, i + 1, i + 2] if i <= 98 else [i for i in range(i, 101)] for i in range(1, 101, 3)]
print(li)
标签：dict1,name,05,Python,book,result,print,元祖 来源： https://www.cnblogs.com/kingdomer/p/15316126.html