一、List:是Python内置的数据类型列表--List,List是一种有序集合,可以随时删除和添加元素。
创建list直接用 []1 letters = ["a","b","c","d","e"] 2 0 1 2 3 4 3 print(letters[0]) 4 a 5 print(letters[1]) 6 b 7 #list中是通过索引来获取列表中的元素,索引是从0开始
如果说想要直接获取list的最后一个元素可以直接用-1的索引获取
1 >>> letters[-1] 2 ‘e‘ 3 >>> letters[-2] 4 ‘d‘ 5 >>>
通过len()函数可以来获取列表中有多少个元素
>>> len(letters) 5 letters列表中有5个元素
List也是有相应的方法:
append():在List最后一位添加元素
>>> letters.append("F") >>> print(letters) [‘a‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘, ‘F‘]
clear():清空List
>>> letters = ["a","b","c","d","e",] >>> letters.clear() >>> print(letters)[]
count(value):在列表中该元素的个数
>>> letters = ["a","b","c","d","e","e","e"] >>> letters.count("e") 3
extend():扩展List,添加到该List的末尾
>>> letters = ["a","b","c","d","e"] >>> numbers = [1,2,3,4,5] >>> letters.extend(numbers) >>> print(letters) [‘a‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘, 1, 2, 3, 4, 5]
index():查找第一个出现的该元素在列表中的索引
>>> letters = ["a","b","c","d","e","e","e"] >>> letters.index("e") 4
insert(index,p_object):在List中添加元素
>> letters = ["a","b","c","d","e"] >>> letters.insert(1,"A") >>> print(letters) [‘a‘, ‘A‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘]
pop(index = None):通过索引删除病返回List中的元素,如果没有输入参数则默认删除List最后一个元素
>>> letters.pop() ‘e‘ >>> letters.pop(3) ‘d‘ >>> print(letters) [‘a‘, ‘b‘, ‘c‘]
remove(value):删除List中第一个出现的该元素
>>> letters = ["a","b","c","d","e","e","e"] >>> letters.remove("e") >>> print(letters) [‘a‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘, ‘e‘]
reverse():倒转List
>>> letters = ["a","b","c","d","e"] >>> letters.reverse() >>> print(letters) [‘e‘, ‘d‘, ‘c‘, ‘b‘, ‘a‘]
sort():将List中的元素从小到大排序
","f","d","a","e","5","e","e","2","$#aa"] >>> letters.sort() >>> print(letters) [‘$#aa‘, ‘2‘, ‘5‘, ‘6‘, ‘a‘, ‘d‘, ‘e‘, ‘e‘, ‘e‘, ‘f‘] >>>
二、浅拷贝,深拷贝
留坑:填
三、切片
关于切片,最初可能有些迷惑,但总结后思路就特别清晰:
[起始:结束:方向/间隔]
>>> letters = ["6","f","d","a","e","5","e","e","2","$#aa",2,4,2,4,5,66,7,33] >>> letters[1:12:3] #表示从索引为1的元素开始取值,到索引12结束(不包括索引12的元素),间隔为3 [‘f‘, ‘e‘, ‘e‘, 2]
[起始:结束:方向/间隔] 注意:如果想倒序切片,需要把"方向/间隔"改成负数
>>> letters = ["6","f","d","a","e","5","e","e","2","$#aa",2,4,2,4,5,66,7,33] >>> letters[12:1:-3] [2, ‘$#aa‘, ‘e‘, ‘a‘]
时间: 2024-10-23 19:45:01