1)
append:添加的元素默认在列表的最后一位;
insert:可以指定添加元素的位置。
1 list1=["Java","Python","Go"] 2 list1.append("Net") 3 print(list1) 4 list1.insert(0,"C") 5 print(list1)
运行结果如下:
pop:返回删除的对象(默认最后一位)
remove:指定删除的对象
1 list1=["Java","Python","Go"] 2 print(list1) 3 print(list1.pop()) 4 list1.remove("Python") 5 print(list1)
运行结果如下:
1 list1=["Java","Python","Go"] 2 list1.clear() 3 print(list1)
运行结果如下:
1 list1=["Java","Python","Go"] 2 print(len(list1))
运行结果如下:
1 list1=["Java","Python","Go"] 2 print(list1.index("Go"))
运行结果如下:
1 list1=["Java","Python","Go"] 2 print(list.count("Go"))
运行结果如下:
1 list1=["Java","Python","Go"] 2 list2=list1.copy() 3 print(list2)
运行结果如下:
1 list1=["Java","Python","Go"] 2 list3=[1,2,3] 3 list1.extend(list3) 4 print(list1) 5 print(list3)
运行结果如下:
1 list1=["Java","Python","Go"] 2 list1[0]="Java language" 3 list1[1]="Python language" 4 list1[2]="Go language" 5 print(list1)
运行结果如下:
list3=[0,1,2,3,4,5,6,7,8,9] list3.reverse() print(list3)
运行结果如下:
list1=[12,62,23,56,75,36,87,38,23] list1.sort(reverse=True) print(list1)
运行结果如下:
1 list1=["Java","Python","Go","C"] 2 for index,item in enumerate(list1): 3 print(index,item) 4 if index==3 and item=="C": 5 print("继续加油!")
运行结果如下:
1 list1=[1,2,3,4,5,6,7,8,9] 2 print(list1[0:3]) #输出从第一个数到第三个数 3 print(list1[-1]) #输出倒数第一个位数 4 print(list1[-2]) #输出倒数第二个位数 5 print(list1[6]) #输出第索引号为6的数 6 print(list1[-3]) #输出倒数第三个数
运行结果如下:
1 list4=[x for x in range(10) if x >2] 2 print(list4) 3 #列表推导式的理解: 4 list5=[] 5 for item in range(10): 6 if item>2: 7 list5.append(item) 8 print(list5)
运行结果如下:
1 list6=list() 2 for item in zip(list1,list2): 3 list6.append(item) 4 print(list6)
运行结果如下:
list6=[]、list6=list()
怎么找出两张excel表中的不同数据?
1)把Excel表中的数据分别保存在两个列表中;
2)然后对这两个列表使用set集合的对称差集方法就可以找出两张excel表中的不同数据。
1 list1=["go","java","python","c"] 2 list2=["net","a","c","java","go"] 3 print(set(list1)&set(list2)) #取出两个对象中相同的数据(交集) 4 print(set(list1)|set(list2)) #并集(列表1和列表2共同拥有的元素) 5 print(set(list1)-set(list2)) #差集(列表1中有而列表2中没有的元素) 6 print(set(list1)^set(list2)) #对称差集(列表1或列表2 中,但不会同时出现在二者中)
运行结果如下:
a.列表是可变的,可以添加新的元素,也可以任意删除元素;
b.元组是不可变的(结构不可变)
当元组里边只有一个元组的时候,记得加逗号。
1 tuple1=("Java","Python","Go",[1,2,3]) 2 print(tuple1[1])
运行结果如下:
1 tuple1=("Java","Python","Go",[1,2,3]) 2 tuple1[3][0]="First" 3 print(tuple1) 4 print(tuple1.count("Python")) 5 print(tuple1.index("Python")) 6 tuple1="Go," 7 print(type(tuple))
运行结果如下:
1)data=[
("wuya","99"),
("zhangsan","88"),
("lisi","77")
("wangwu","66")]
a.输出所有成绩;
b.获取成绩的最高分、最低分、平均分、并对成绩进行排序
解答1:
1 data=[ 2 ("wuya","99"), 3 ("zhangsan","88"), 4 ("lisi","77"), 5 ("wangwu","66")] 6 7 score=[] 8 for item in data: 9 score.append(int(item[1])) 10 print(score) 11 print("最高分:",max(score)) 12 print("最低分:",min(score)) 13 print("平均分:",sum(score)/len(score)) 14 score.sort() 15 print("分数排序:",score)
运行结果如下:
解答2:
1 data=[ 2 ("wuya","99"), 3 ("zhangsan","88"), 4 ("lisi","77"), 5 ("wangwu","66")] 6 7 score=[int(x[1]) for x in data] 8 print(score) 9 print("最高分:",max(score)) 10 print("最低分:",min(score)) 11 print("平均分:",sum(score)/len(score)) 12 score.sort()#sort()函数只能先排序后打印 13 print("分数排序:",score)
运行结果如下:
2)有列表data1=["Go","阿狸","腾讯","百度","字节跳动"],请获取公司数据。
解答:
1 data1=["Go","阿狸","腾讯","百度","字节跳动"] 2 language,*company=data1 # *表示只获取此字段之后的所有值 3 4 print(language)
运行结果如下:
3)有列表data2=[["Go",["Java"]]],只输出Java。
解答:
data2=[["Go",["Java"]]] print(data2[0][1][0])
运行结果如下:
4)有列表lists=["正常双休",
"带薪年假",
"五险一金"]
以字符形式输出此列表内容。
解答:
lists=["正常双休","带薪年假","五险一金"] print(",".join(lists))
运行结果如下:
1)字典是无序的;
2)字典是通过key-value键值对的数据形式来存储数据的。
1 dict1={"name":"cch","age":"18"} 2 #方式1: 3 print(dict1.get("age")) 4 #方式2: 5 print(dict1["name"])
运行结果如下:
1 dict1={"name":"cch","age":18} 2 dict1["address"]="宝鸡" 3 print(dict1)
运行结果如下:
1 dict2={"name":"cch","age":18,"address":"宝鸡"} 2 dict2["address"]="西安" 3 print(dict2)
运行结果如下:
1 dict3={"name":"cch","age":18,"address":"宝鸡"} 2 del dict3["address"] 3 print(dict3)
运行结果如下:
1 dict4={"name":"cch","age":18,"address":"宝鸡"} 2 for key in dict4.keys(): 3 print(key)
运行结果如下:
1 dict4={"name":"cch","age":18,"address":"宝鸡"} 2 for value in dict4.values(): 3 print(value)
运行结果如下:
1 dict4={"name":"cch","age":18,"address":"陕西宝鸡"} 2 for key,value in dict4.items(): 3 print(key,":",value)
运行结果如下:
1 dict4={"name":"cch","age":18,"address":"陕西宝鸡"} 2 dict5={"language":"Python"} 3 dict4.update(dict5) 4 print(dict4)
运行结果如下:
dict4={“name”:“cch”,"age":"18","adress":"陕西宝鸡","data":[{"english":90,"math":100}]}
输出math的值
1 dict4={"name":"cch","age":18,"address":"陕西宝鸡","data":[{"english":90,"math":100}]} 2 print(dict4["data"][0]["math"])
运行结果如下:
思路
1、字典的排序使用的是关键字sorted();
2、dict.items()针对字典进行循环输出key和value;
3、在sorted内部把Key和value分别存储在不同的列表中;
4、然后根据 key=lambda item:item[0]来进行排序处理:
a.根据索引来进行排序;
b.然后把排序的结果信息进行拼接。
5、强制的转为字典数据类型。
1 dict6={"name":"cch","age":18,"sex":"女","address":"陕西宝鸡"} 2 data=dict(sorted(dict6.items(),key=lambda item:item[0])) 3 print(data)
运行结果如下:
按value 排序
1 dict6={"name":"cch","age":"18","sex":"女","address":"陕西宝鸡"} 2 data1=dict(sorted(dict6.items(),key=lambda item:item[1])) 3 print(data1)
运行结果如下:
1 from collections import OrderedDict 2 data3=OrderedDict() 3 data3["age"]=18 4 data3["address"]="宝鸡" 5 data3["name"]="cch" 6 print(dict(data3))
运行结果如下:
1 str ="Java,Python,Go" 2 str_list=str.split(",") 3 print(str_list) 4 print(type(str_list))
运行结果如下:
1 str ="Java,Python,Go" 2 str_list=str.split(",") 3 list_str=",".join(str_list) 4 print(list_str) 5 print(type(list_str))
运行结果如下:
1 tuple1=("Go",) 2 tuple_list=list(tuple1) 3 print(tuple_list) 4 print(type(tuple_list))
运行结果如下:
1 tuple1=("Go",) 2 tuple_list=list(tuple1) 3 list_tuple=tuple(tuple_list) 4 print(list_tuple) 5 print(type(list_tuple))
运行结果如下:
1 list1=["Java","Python","Go"] 2 list_dict=dict(enumerate(list1)) 3 print(list_dict) 4 print(type(list_dict))
运行结果如下:
1 list1=["Java","Python","Go"] 2 list_dict=dict(enumerate(list1)) 3 dict_list=list(list_dict.items()) 4 print(dict_list) 5 print(type(dict_list))
运行结果如下:
1 str1="Java,Python,Go" 2 str2=list(str1.split()) 3 str_dict=dict(enumerate(str2)) 4 print(str_dict) 5 print(type(str_dict))
运行结果如下:
1 str1="Java,Python,Go" 2 str2=list(str1.split()) 3 str_dict=dict(enumerate(str2)) 4 dict_list=list(str_dict.items()) 5 list_str=",".join(str2) 6 print(list_str) 7 print(type(list_str))
运行结果如下:
1 student_score={ 2 "张":100, 3 "王":48, 4 "李":75, 5 "赵":89, 6 "吴":62} 7 8 minScore=min(zip(student_score.values(),student_score.keys())) 9 print(minScore) 10 maxScore=max(zip(student_score.values(),student_score.keys())) 11 print(maxScore)
运行结果如下:
实战练习题:
1、给一组城市相关的数据,提取相应的城市名。
解答:city=[]
for item in data["ZpData"]["hotCityList"]:
print(data["name"])
city.append(item["name"])
print(city)
2、字符串与字典的转换
解答:
1)字符串转为字典
1 str1="Java,Python,Go" 2 str2=list(str1.split()) 3 str_dict=dict(emnerate(str2)) 4 print(str_dict) 5 print(type(str_dict))
运行结果如下图:
2)字典转为字符串
1 str1="Java,Python,Go" 2 str2=list(str1.split()) 3 str_dict=dict(enumerate(str2)) 4 dict_list=list(str_dict.items()) 5 list_str=",".join(str2) 6 print(list_str) 7 print(type(list_str))
运行结果如下图:
3、输出c
解答:
1 lists=['language',['go','python',['java','net',['c']]]] 2 print(lists[1][2][2][0])
运行解果如下:
4、输出分数的最高,最小,平均dict1={'name':'wuya',"score":[{"english":90},{"math":90},{"wuli":90}]
解答1:
1 dict1={'name':'wuya',"score":[ 2 {"score":90}, 3 {'score':88}, 4 {'score':60}]} 5 Score=[] 6 for item in dict1["score"]: 7 Score.append(item["score"]) 8 print(Score) 9 print("最大:",max(Score)) 10 print("最小:",min(Score)) 11 print("平均:",sum(Score)/len(Score))
运行结果如下:
解答2:
1 dict1={'name':'wuya',"score":[ 2 {"score":90}, 3 {'score':88}, 4 {'score':60}]} 5 lists=[] 6 for item in dict1["score"]: 7 for key,value in item.items(): 8 9 lists.append(item["score"]) 10 print(lists) 11 print("最高分:",max(lists)) 12 print("最低分:",min(lists)) 13 print("平均分:",sum(lists)/len(lists))
运行结果如下:
5、输出唐山
解答:datas=["go",["name":"list","age":"18","adress":["xian","唐山"]]]
1 datas=["go",{"name":'lisi',"age":18,'address':["xian",'唐山']}] 2 print(datas[1]["address"][1])
运行结果如下:
6、按照key和value进行排序
解答:dict2={"name":"cch","sex":"girl","job":"testDev","salary":109.09}
1)按照key排序
1 dict2={"name":"cch","sex":"girl","job":"testDev","salary":109.09} 2 dict3=dict(sorted(dict2.items(),key=lambda item:item[0])) 3 print(dict3)
运行结果如下:
2)按照value排序
1 dict2={"name":"cch","sex":"girl","job":"testDev","salary":"109.09"} 2 dict3=dict(sorted(dict2.items(),key=lambda item:item[1])) 3 print(dict3)
运行结果如下:
7、根据数据,输出成城市名称
解:
1 data4={"code":0,"message":"Success","zpData":[{"name":"全国","code":100010000,"url":"/?city=100010000"},{"name":"北京","code":101010100,"url":"/beijing/"},{"name":"上海","code":101020100,"url":"/shanghai/"},{"name":"广州","code":101280100,"url":"/guangzhou/"},{"name":"深圳","code":101280600,"url":"/shenzhen/"},{"name":"杭州","code":101210100,"url":"/hangzhou/"},{"name":"天津","code":101030100,"url":"/tianjin/"},{"name":"西安","code":101110100,"url":"/xian/"},{"name":"苏州","code":101190400,"url":"/suzhou/"},{"name":"武汉","code":101200100,"url":"/wuhan/"},{"name":"厦门","code":101230200,"url":"/xiamen/"},{"name":"长沙","code":101250100,"url":"/changsha/"},{"name":"成都","code":101270100,"url":"/chengdu/"},{"name":"郑州","code":101180100,"url":"/zhengzhou/"},{"name":"重庆","code":101040100,"url":"/chongqing/"},{"name":"佛山","code":101280800,"url":"/foshan/"},{"name":"合肥","code":101220100,"url":"/hefei/"},{"name":"济南","code":101120100,"url":"/jinan/"},{"name":"青岛","code":101120200,"url":"/qingdao/"},{"name":"南京","code":101190100,"url":"/nanjing/"},{"name":"东莞","code":101281600,"url":"/dongguan/"},{"name":"昆明","code":101290100,"url":"/kunming/"},{"name":"南昌","code":101240100,"url":"/nanchang/"},{"name":"石家庄","code":101090100,"url":"/shijiazhuang/"},{"name":"宁波","code":101210400,"url":"/ningbo/"},{"name":"福州","code":101230100,"url":"/fuzhou/"}]} 2 city=[] 3 for item in data4["zpData"]: 4 city.append(item["name"]) 5 print(city)
运行结果如下: