• python-数据装换


    list

    list to tuple

    li = ['test1', 'test2', 'test3', 123]
    # tuple
    print(tuple(li))  # ('test1', 'test2', 'test3', 123)
    
    

    list to str

    li = ['test1', 'test2', 'test3']
    # str
    print(str(li), type(str(li)))  # ['test1', 'test2', 'test3'] <class 'str'>
    # join 列表中不能有数字
    print(" ".join(li), type(" ".join(li)))  # test1 test2 test3 <class 'str'>
    

    tuple

    tuple to list

    tuple1 = ("test1", "test2", "test3")
    tuple2 = (1, 2, 3)
    # list
    list_tuple = list(tuple1)
    print(list_tuple)  # ['test1', 'test2', 'test3']
    print(list(tuple2))  # [1, 2, 3]
    
    

    tuple to str

    tuple1 = ("test1", "test2", "test3")
    # str
    str_ = str(tuple1)
    print(str_, type(str_))  # ('test1', 'test2', 'test3') <class 'str'>
    # join
    print(" ".join(tuple1))  # test1 test2 test3
    
    
    tuple2 = ("test1", "test2", "test3", 2)
    # 在运用join方法时,元组里面的元素必须的字符串
    # print(" ".join(tuple2))  # TypeError: sequence item 3: expected str instance, int found
     
    # 数字组成的元组也是不能使用join,必须是字符串
    tuple3 = (1, 2, 3)
    # print(" ".join(tuple3))  # TypeError: sequence item 0: expected str instance, int found
    

    dict

    dict to list

    dic = {"name": "Yang", "age": 123}
    print(list(dic)) # ['name', 'age']  返回键
    

    dict to tuple

    dic = {"name": "Yang", "age": 123}
    print(tuple(dic)) # ('name', 'age')  返回键
    

    dict to str

    dic = {"name": "Yang", "age": 123}
    print(str(dic))  # {'name': 'Yang', 'age': 123}
    print(' '.join(dic))  # name age
    

    str

    str to list

    s = "test"
    print(list(s))  #  ['t', 'e', 's', 't']
    s2 = "123abc"
    print(list(s2))  #  ['1', '2', '3', 'a', 'b', 'c']
    

    str to tuple

    s = "test"
    print(tuple(s))  # ('t', 'e', 's', 't')
    s2 = "123abc"
    print(tuple(s2))  # ('1', '2', '3', 'a', 'b', 'c')
    

    str to dict

    str2 = "{'key1':'valu1','key2':'valu2'}"
    dict2 = eval(str2)
    print(dict2, type(dict2))  # {'key1': 'valu1', 'key2': 'valu2'} <class 'dict'>
    
    
    

    常见的坑

    l1 = [11, 22, 33, 44, 55]
    将此列表索引为奇数位对应的元素全部删除.

    l1 = [11, 22, 33, 44, 55]
    # 将此列表索引为奇数位对应的元素全部删除.
    
    # 方法一:
    # del l1[1::2]
    # print(l1)
    
    # 方法二:错误示例:坑
    
    # 循环一个列表时,不要改变列表的大小.这样会影响你最后的结果.
    # for index in range(len(l1)):
    #     if index % 2 == 1:
    #         # index 奇数
    #         l1.pop(index)
    # print(l1)
    
    # 方法三
    # new_l = []
    # for index in range(len(l1)):
    #     if index % 2 == 0:
    #         new_l.append(l1[index])
    # # print(new_l)
    # l1 = new_l
    # print(l1)
    
    # 方法三:
    # for index in range(len(l1)-1,-1,-1):
    #     if index % 2 == 1:
    #         l1.pop(index)
    # print(l1)
    
    

    dic = {'k1': 'v1', 'k2': 'v2', 'k3': 'v3', 'name': 'alex'}
    将字典中的key中含有k元素的所有键值对删除.

    # 错误代码
    '''坑
    dic = {'k1': 'v1', 'k2': 'v2', 'k3': 'v3', 'name': 'Tom'}
    for key in dic:
        if 'k' in key:
            dic.pop(key)
    print(dic)  # untimeError: dictionary changed size during iteration
    '''
    # 循环一个字典时,不能改变字典的大小,这样会报错.
    
    l1 = []  # 存放满足条件的key
    for key in dic:
        if 'k' in key:
            l1.append(key)
    # print(l1)
    for key in l1:  # 在遍历key进行删除
        dic.pop(key) # 或 del dic[key]
    print(dic)
    
  • 相关阅读:
    2017《面向对象程序设计》课程作业八
    2017《面向对象程序设计》课程作业七
    第六次作业
    2017《面向对象程序设计》课程作业五
    课程作业四
    2017《面向对象程序设计》课程作业三
    2017《面向对象程序设计》课程作业二
    2017《面向对象程序设计》课程作业一
    20200924 次小生成树
    水文章_考拉
  • 原文地址:https://www.cnblogs.com/yangchangjie150330/p/10489446.html
Copyright © 2020-2023  润新知