建立表格并打印

bicycles = ["a","b","c"]
print(bicycles)

要求中括号以及逗号分隔 访问单个元素

bicycles = ["a","b","c"]
print(bicycles[0])

修改列表中的单个元素：

bicycles = ["a","b","c"]
print(bicycles[0])
bicycles[0] = "d"
print(bicycles)

添加：

.append()在末尾添加元素：

bicycles = ["a","b","c"]
print(bicycles)
bicycles.append("d")
print(bicycles)

便可以在末尾加入 .insert(数位，)在第n位数 前面加入（从0计数）：

bicycles = ["a","b","c"]
print(bicycles)
bicycles.insert(1,"d")
print(bicycles)

删除：

del  ……[n]删除第几位：

bicycles = ["a","b","c"]
print(bicycles)
del bicycles[0]
print(bicycles)

.pop()删除（栈原理：去掉最后一个）

bicycles = ["a","b","c"]
print(bicycles)

poped_bicycles = bicycles.pop()
print(bicycles)
print(poped_bicycles)

用途：如果列表元素按时间存储，该方法可以查找最后一个进入的元素 注意： poped_bicycles = bicycles.pop() 是先执行了 poped_bicycles = bicycles. 最后 再执行了 bicycles.pop() 举例：

bicycles = ["a","b","c"]
print(bicycles)

bicycles.pop()
poped_bicycles = bicycles.pop()
print(bicycles)
print(poped_bicycles)

.pop(n)可以弹出列表中的任意的第n个元素（注意的是 pop之后这个元素就不存在了） .remove是删除值

bicycles = ["a","b","c"]
print(bicycles)

bicycles.remove("a")
print(bicycles)

但是只能删这个值第一个出现的元素

bicycles = ["a","a","b","c"]
print(bicycles)

bicycles.remove("a")
print(bicycles)

在应用场景中，我们还可以利用原因解释来说明remove的原因

bicycles = ["a","a","b","c"]
print(bicycles)

too_terrible="a"
bicycles.remove("a")
print(bicycles)
print(f"\nA {too_terrible} is too terrible for me")

记得输入字符变量需要 f和{}

列表的组织：

.sort永久的改变（回不去了），按字母排序

bicycles = ["b","a","c"]
print(bicycles)

bicycles.sort()
print(bicycles)

倒序排

bicycles = ["b","a","c"]
print(bicycles)

bicycles.sort(reverse=True)
print(bicycles)

sorted(列表名)临时排序

bicycles = ["b","a","c"]
print(bicycles)

print(sorted(bicycles))
print(bicycles)

.reverse()倒置

bicycles = ["b","a","c"]
print(bicycles)

bicycles.reverse()
print(bicycles)

len(列表)确定列表长度，非常有用，用于确定还有多少个外星人未射杀（pop），确定需要管理多少项可视化数据，计算网站有多少注册用户

bicycles = ["b","a","c"]
print(bicycles)

print(len(bicycles))

输出列表最后一个元素（列表非空时），可以使用 bicycles[-1]

bicycles = ["b","a","c"]
print(bicycles)

print(bicycles[-1])