1 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 2 list.append(‘RainbowSix‘) 3 print(list) 4 5 #[‘csgo‘, 100, ‘pacify‘, ‘cs1.6‘, ‘unturned‘, ‘RainbowSix‘] 6 7 #进程已结束,退出代码0
1 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 2 list.append(‘RainbowSix‘) 3 list.insert(-1,‘hell let loose‘) 4 list.insert(0,‘project winter‘) 5 print(list) 6 7 8 #[‘project winter‘, ‘csgo‘, 100, ‘pacify‘, ‘cs1.6‘, ‘unturned‘, ‘hell let #loose‘, ‘RainbowSix‘] 9 10 #进程已结束,退出代码0
PS append比insert好,因为insert会移动其它元素,没有append快
1 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 2 list.remove(‘csgo‘) 3 print(list) 4 5 6 7 #[100, ‘pacify‘, ‘cs1.6‘, ‘unturned‘] 8 9 #进程已结束,退出代码0
若有多个同名元素,则只能删去第一个
1 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 2 list.pop() 3 print(list) #[‘csgo‘, 100, ‘pacify‘, ‘cs1.6‘] 4 5 6 7 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 8 list.pop(0) 9 print(list) #[100, ‘pacify‘, ‘cs1.6‘, ‘unturned‘] 10 11 12 13 14 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 15 a=list.pop(0) 16 print(a) #csgo
1 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 2 del list[0] 3 print(list) #[100, ‘pacify‘, ‘cs1.6‘, ‘unturned‘
list[位置]。直接索引到一个位置的元素,然后可以对其赋值,查询
1 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 2 for i in range(len(list)): 3 print(i) 4 print(list[i]) 5 6 7 #0 8 #csgo 9 #1 10 #100 11 #2 12 #pacify 13 #3 14 #cs1.6 15 #4 16 #unturned 17 18 #进程已结束,退出代码0
1 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 2 print(list) #[‘csgo‘, 100, ‘pacify‘, ‘cs1.6‘, ‘unturned‘] 3 list.reverse() 4 print(list) #[‘unturned‘, ‘cs1.6‘, ‘pacify‘, 100, ‘csgo‘] 5 6 7 8 9 进程已结束,退出代码0
1 list=[‘csgo‘,100,‘pacify‘,‘cs1.6‘,‘unturned‘] 2 a=list.count(‘csgo‘) 3 print(a) #1
1 list=[999,888,899,123456,1231] 2 list.sort() 3 print(list) #[888, 899, 999, 1231, 123456] 4 list.sort(reverse=True) 5 print(list) #[123456, 1231, 999, 899, 888] 6
原文:https://www.cnblogs.com/heike195929/p/14717451.html