python基础数据类型补充以及编码的进阶

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基础数据类型补充内容

字符串

s = "everyone‘s the hero in their own story"

# print(s.capitalize())                          # 句子首单词的首字母变大写,其余小写  Everyone‘s the hero in their own story
# print(s.swapcase())                            # 全部变大写                EVERYONE‘S THE HERO IN THEIR OWN STORY
# print(s.title())                               # 每个单词的首字母变大写    Everyone‘S The Hero In Their Own Story
# print(len(s))                                  # 38
# print(s.center(39, "*"))                       # 内同居中,总长度,空白处填充  *everyone‘s the hero in their own story
# print(s.center(45, "*"))                       #                         ****everyone‘s the hero in their own story***
# print(s.center(48, "*"))                       #                         *****everyone‘s the hero in their own story*****

# 寻找字符串中的元素是否存在
# print(s.find("hs"))                              # 通过元素找索引,返回的找到的第一个元素的索引,如果找不到返回-1      -1
# print(s.find("ev"))                              # 0
# print(s.find("ev", 9, 18))                       # -1
# print(s.find("th", 11, 20))                      # 11

# print(s.index("hs"))                             # 通过元素找索引,返回的找到的第一个元素的索引,找不到报错。     报错
# print(s.index("ev"))                             # 0
# print(s.index("ev", 9, 18))                      # 报错
# print(s.index("th", 11, 20))                       # 11

元组

python中元组有一个特性,元组中如果只含有一个元素且没有逗号,则该元组不是元组,与该元素的数据类型一致,如果有逗号,那么它是元组。

# tu1 = (1)
# print(tu1, type(tu1))                    # 1    <class ‘int‘>
# tu2 = ("alex")
# print(tu2, type(tu2))                    # alex     <class ‘str‘>
# tu3 = ([1, 2, 3])
# print(tu3, type(tu3))                   # [1, 2, 3]     <class ‘list‘>

# tu1 = (1,)
# print(tu1, type(tu1))                    # (1,)      <class ‘tuple‘>
# tu2 = ("alex",)
# print(tu2, type(tu2))                    # (‘alex‘,)      <class ‘tuple‘>
# tu3 = ([1, 2, 3],)
# print(tu3, type(tu3))                   # ([1, 2, 3],)     <class ‘tuple‘>

index:通过元素找索引(可切片),找到第一个元素就返回,找不到该元素即报错。

tup = (1, 2, "alex", "你好", True, [1, 2, 3,])
# print(tup.index(1))                          #  0
# print(tup.index("alex"))                     #  2
# print(tup.index(True))                       # 0
# print(tup.index([1, 2, 3,]))                 # 5

count: 获取某元素在列表中出现的次数

# print(tup.count(1))                           #   2

列表

li = ["b", "a", "r", "r", "y"]
# print(li.count("r"))                        #   2
# print(li.index("a"))                          #   1
# print(li.index("r"))                          #   2
# li.sort()                                   # 默认从小到大
li.sort(reverse = True)                       # 从大到小
# print(li)                                   #  [‘a‘, ‘b‘, ‘r‘, ‘r‘, ‘y‘]
# li.reverse()
# print(li)                                     #  [‘y‘, ‘r‘, ‘r‘, ‘a‘, ‘b‘]

列表的相加和相乘整数

l1 = [1, 2, 3]
l2 = [2, 4, "tough", 6]
# print(l1 + l2)                                #   [1, 2, 3, 2, 4, ‘tough‘, 6]
# print(l1 * 3)                               #   [1, 2, 3, 1, 2, 3, 1, 2, 3]

循环列表,改变列表大小的问题

l3 = [11, 22, 33, 44, 55]
需求: 请把索引为奇数对应的元素删除(不能一个一个删除,此l1只是举个例子,里面的元素不定)。

# 正常思路做法
# for index, item in enumerate(l3):
#     print(index, item)
#     print(l3[index])
#     if index % 2 != 0:
#          l3.remove(l3[index])
# print(l3)                             # [11, 33, 44]

# for index in range(len(l3)):
#     if index % 2 == 1:
#         l3.pop(index)
# print(l3)                             #  [11, 33, 44]

# method1   直接删除
# del l3[1::2]
# print(l3)                            # [11, 33, 55]

# method 2     倒序删除
# print(len(l3))                        #    5
# for index in range(len(l3) - 1, -1, -1):
#     # print(index)                               # 4   3  2  1
#     if index % 2 != 0:
#         l3.pop(index)
# print(l3)                                    # [11, 33, 55]


# method 3    思维置换
# new_list = []
# for index, item in enumerate(l3):
#     if index % 2 == 0:
#         new_list.append(l3[index])
# print(new_list)                      #  [11, 33, 55]
# l3 = new_list
# print(l3)

# new_list = []
# for index in range(len(l3)):
#     if index % 2 == 0:
#         new_list.append(l3[index])
# print(new_list)                      #  [11, 33, 55]
# l3 = new_list
# print(l3)

总结:循环一个列表时,最好不要改变列表的大小,因为他会影响你的最终结果

dict

fromkeys:创建一个字典:字典的所有键来自一个可迭代对象,字典的值使用同一个值

# dic = dict.fromkeys("abc", 100)
# print(dic)                             # {‘a‘: 100, ‘b‘: 100, ‘c‘: 100}

# dic = dict.fromkeys([1, 2, 3], "hello")
# print(dic)                                 #  {1: ‘hello‘, 2: ‘hello‘, 3: ‘hello‘}

# 坑: 值共享一个
# dic = dict.fromkeys([1, 2, 3], [])
# print(dic)                                    #    {1: [], 2: [], 3: []}
# dic[1].append(666)
# print(dic)                                    #  {1: [666], 2: [666], 3: [666]}
# print(id(dic[1]),id(dic[2]),id(dic[3]))       #  2905006740296     2905006740296    2905006740296
# popitem 3.5版本之前,popitem为随机删除,  3.6之后为删除最后一个,有返回值
dic = {"name": "李海", "age": 19}
# res = dic.popitem()
# print(dic, res)                                 #  {‘name‘: ‘李海‘}         (‘age‘, 19)

# update
# dic.update(gender = "male", weight = 65)         # {‘name‘: ‘李海‘, ‘age‘: 19, ‘gender‘: ‘male‘, ‘weight‘: 65}
# print(dic)
# dic.update((("height", 180), ("hobby", "football"), ("address", "CDC")))
# print(dic)                                          #  {‘name‘: ‘李海‘, ‘age‘: 19, ‘height‘: 180, ‘hobby‘: ‘football‘, ‘address‘: ‘CDC‘}

# dic.update([("height", 180), ("hobby", "football"), ("address", "CDC")])
# print(dic)                                         #  {‘name‘: ‘李海‘, ‘age‘: 19, ‘height‘: 180, ‘hobby‘: ‘football‘, ‘address‘: ‘CDC‘}

# dic1 = {"name":"jin","age":18,"sex":"male"}
# dic2 = {"name":"alex","weight":75}
#
# dic1.update(dic2)
# print(dic1)                                         #  {‘name‘: ‘alex‘, ‘age‘: 18, ‘sex‘: ‘male‘, ‘weight‘: 75}
# print(dic2)                                         #  {‘name‘: ‘alex‘, ‘weight‘: 75}

循环字典,改变字典大小的问题

dic3 = {‘k1‘: ‘太白‘, ‘k2‘: ‘barry‘, ‘k3‘: ‘白白‘, ‘age‘: 18}
需求: 请将字典中所有键带k元素的键值对删除

# for i in dic3:
#     # print(i)                                     #  k1   k2   k3  age
#     if "k" in i:
#         del dic3[i]
# print(dic3)                             #  报错 RuntimeError: dictionary changed size during iteration

# method 1
# li = []
# # for key in dic3.keys():
# for key in dic3:
#     if "k" in key:
#         li.append(key)
# # print(li)                      #  [‘k1‘, ‘k2‘, ‘k3‘]
# for item in li:
#     # dic3.pop(item)
#     del dic3[item]
# print(dic3)

# method 2
for key in list(dic3.keys()):
    if "k" in key:
        dic3.pop(key)
print(dic3)

总结: 在循环一个字典的过程中,不要改变字典的大小(增,删字典的元素),这样会直接报错。

数据类型间的转换问题

现在学过的数据类型有:intboolstrlisttupledictset

int、bool 、str 三者转换

# int ---> bool
i = 100
print(bool(i))  # True  # 非零即True
i1 = 0
print(bool(i1))  # False 零即False

# bool ---> int
t = True
print(int(t))  # 1  True --> 1
t = False
print(int(t))  # 0  False --> 0

# int ---> str
i1 = 100
print(str(i1))  # ‘100‘

# str ---> int  # 全部由数字组成的字符串才可以转化成数字
s1 = 90
print(int(s1))  # 90

# str ---> bool
s1 = 太白
s2 = ‘‘
print(bool(s1))  # True 非空即True
print(bool(s2))  # False
# bool ---> str
t1 = True
print(str(True))  # ‘True‘

str、list 转换

# str ---> list
s = "alex is gay"
print(s.split())                      # [‘alex‘, ‘is‘, ‘gay‘]

# list ---> str  # 前提 list 里面所有的元素必须是字符串类型才可以
li = [alex, is, gay]
print(" ".join(li))                   #  "alex is gay"

list、set 转换

# list ---> set
s1 = [1, 2, 3]
print(set(s1))                       #  {1, 2, 3}

# set ---> list
set1 = {1, 2, 3, 3,}
print(list(set1))                    # [1, 2, 3]

str、bytes 两者转换

# str ---> bytes
s1 = 太白
print(s1.encode("utf-8"))           #  b‘xe5xa4xaaxe7x99xbd‘

# bytes ---> str
n = bxe5xa4xaaxe7x99xbd
print(n.decode("utf-8"))            # ‘太白‘

所有数据都可以转化成bool值

转化成bool值为False的数据类型有:
‘‘, 0, (), {}, [], set(), None

基础数据类型的总结

按存储空间的占用分(从低到高)

数字
字符串
集合:无序,即无序存索引相关信息
元组:有序,需要存索引相关信息,不可变
列表:有序,需要存索引相关信息,可变,需要处理数据的增删改
字典:有序,需要存key与value映射的相关信息,可变,需要处理数据的增删改(3.6之后有序)

按存值个数区分

标量/原子类型 数字,字符串
容器类型 列表,元组,字典

 

 

 

按可变不可变区分

可变 列表,字典
不可变 数字,字符串,元组,布尔值

 

 

 

按访问顺序区分

直接访问 数字
顺序访问(序列类型) 字符串,列表,元组
key值访问(映射类型) 字典

 

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