列表名 = [列表元素1,列表元素2,列表元素3,… ]
說明:app
for循環遍歷列表spa
nameList = ["張三","李四","王五","趙六"] for name in nameList: print(name)
運行結果以下:code
張三
李四
王五
趙六
while循環遍歷列表blog
nameList = ["張三","李四","王五","趙六"] i= 0 while i<len(nameList): print(nameList[i]) i+=1
運行結果以下:排序
張三
李四
王五
趙六
一、append:經過append能夠向列表添加元素rem
nameList = ["張三","李四","王五","趙六"] print(nameList) print("----------------分界線-----------------") nameList.append("侯七") print(nameList)
運行結果以下:字符串
['張三', '李四', '王五', '趙六'] ----------------分界線----------------- ['張三', '李四', '王五', '趙六', '侯七']
二、extend:能夠將另外一個集合中的元素逐一添加到列表中for循環
listA = [1,2] listB = [3,4] listA.extend(listB) print(listA)
運行結果爲:[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)
運行結果爲:['a', 1, 2, 3, 4]
四、經過下標修改元素
numList = [1,2,3,4] numList[1] = "A"
print(numList)
運行結果爲:[1, 'A', 3, 4]
五、查找元素
numList = [1,2,3,4] if 2 in numList: print("2在numList裏面") else: print("2不在numList裏面")
運行結果爲:2在numList裏面
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)
打印結果爲:0和3。index也是會找到第一個知足狀況的後面就不會再找了
七、刪除元素
nameList = ["張三","李四","王五","趙六","侯七"] del nameList[2] print(nameList)
運行結果爲:['張三', '李四', '趙六', '侯七']
nameList = ["張三","李四","王五","趙六","侯七"] nameList.pop() print(nameList)
運行結果爲:['張三', '李四', '王五', '趙六']
nameList = ["張三","李四","王五","趙六","侯七"] nameList.remove("李四") print(nameList)
運行結果爲:['張三', '王五', '趙六', '侯七']
八、排序(sort,reverse)
numList = [1,4,3,5,2] numList.reverse() print(numList)
運行結果爲:[2, 5, 3, 4, 1],倒序
numList = [1,4,3,5,2] numList.sort() print(numList)
運行結果爲:[1, 2, 3, 4, 5],升序
numList = [1,4,3,5,2] numList.sort(reverse = True) print(numList)
運行結果爲:[5, 4, 3, 2, 1],降序