列表名 = [列表元素1,列表元素2,列表元素3,… ]
说明:app
for循环遍历列表spa
nameList = ["张三","李四","王五","赵六"] for name in nameList: print(name)
运行结果以下:code
张三
李四
王五
赵六
while循环遍历列表blog
nameList = ["张三","李四","王五","赵六"] i= 0 while i<len(nameList): print(nameList[i]) i+=1
运行结果以下:排序
张三
李四
王五
赵六
一、append:经过append能够向列表添加元素rem
nameList = ["张三","李四","王五","赵六"] print(nameList) print("----------------分界线-----------------") nameList.append("侯七") print(nameList)
运行结果以下:字符串
['张三', '李四', '王五', '赵六'] ----------------分界线----------------- ['张三', '李四', '王五', '赵六', '侯七']
二、extend:能够将另外一个集合中的元素逐一添加到列表中for循环
listA = [1,2] listB = [3,4] listA.extend(listB) print(listA)
运行结果为:[1, 2, 3, 4]class
三、insert(index, object) 在指定位置index前插入元素object容器
numList = [1,2,3,4] numList.insert(0,"a") print(numList)
运行结果为:['a', 1, 2, 3, 4]
四、经过下标修改元素
numList = [1,2,3,4] numList[1] = "A"
print(numList)
运行结果为:[1, 'A', 3, 4]
五、查找元素
numList = [1,2,3,4] if 2 in numList: print("2在numList里面") else: print("2不在numList里面")
运行结果为:2在numList里面
六、index和count:用法和字符串中差很少
strList = ['a','b','c','d','a','b','b'] ind = strList.index('b') con = strList.count('b') print(ind) print(con)
打印结果为:0和3。index也是会找到第一个知足状况的后面就不会再找了
七、删除元素
nameList = ["张三","李四","王五","赵六","侯七"] del nameList[2] print(nameList)
运行结果为:['张三', '李四', '赵六', '侯七']
nameList = ["张三","李四","王五","赵六","侯七"] nameList.pop() print(nameList)
运行结果为:['张三', '李四', '王五', '赵六']
nameList = ["张三","李四","王五","赵六","侯七"] nameList.remove("李四") print(nameList)
运行结果为:['张三', '王五', '赵六', '侯七']
八、排序(sort,reverse)
numList = [1,4,3,5,2] numList.reverse() print(numList)
运行结果为:[2, 5, 3, 4, 1],倒序
numList = [1,4,3,5,2] numList.sort() print(numList)
运行结果为:[1, 2, 3, 4, 5],升序
numList = [1,4,3,5,2] numList.sort(reverse = True) print(numList)
运行结果为:[5, 4, 3, 2, 1],降序