列表通用操作

Posted 2020-10-24 炎之朱雀

 

---以下部分截取自雨辰的教学

 

x = [1,2,3]
name = list("天下为公")
print(range(5))
print(list(range(5)))
print(x)
print(name)

names = ["Tom" , "Jerry" , "Mike" , "Peter" , "John"]
scores = [99 , 88 ,76.3 , 90.5 , 77]
print("Mike" in names) #判定对象Mike是在再names序列里面
print("mike" in names)
print("mike" not in names)  #判定对象mike是否不在names序列里面
print(88 in scores)
print(101 in scores)
print(101 not in scores)

x = list(range(4))
y = [98 , 99 , 100]
print(x + y)
print(x)
print([3] * 4)
print(names[0])
print(scores[-1])

print(len(scores))#对象个数
 #类型必须可以比较
print(min(scores))#最小值
print(max(scores))#最大值
print(sum(scores))#总和

cities = ["北京" ,"上海" , "广州" , "深圳" , "天津"]
 #索引位置在两个值之间
print(cities[0:2])
print(cities[0:3])
print(cities[-5:-3])
print(cities[:3])
print(cities[1:])
print(cities[:])
#第二个冒号后面的是步长,默认是1步
print(cities[::2])
print(len(cities))
print(cities.index("上海"))#索引位置
print(scores.index(77))
print(cities.count("天津")) #出现次数
print(scores.count(77)) #出现次数

l = [1,2,3,4,2,3,3,3,2]
print(l.count(3))

x = list(range(1, 11))
print(x)
x[0] = 99  # 列表支持原位改变某一个元素值
print(x)
x[:3] =[98,99,100]
print(x)
x[:3] =[98,99,100,101,102]
print(x)
x[:5] =[99] #所赋的值必须是可迭代的对象
print(x)

l = list(range(1,11))
print(l)
l[::2] = [99,99,99,99,99] #所赋的值不但要可迭代，还有数目相匹配
print(l)
del l[0]
print(l)
del l[:3]
print(l)
del l[::2]
print(l)
l[:2] = []
print(l)

s = [1,2,3,4,5,6,7,8,2,3,2,2]
print(s)
s.remove(2) #.remove只删除第一个匹配的值
print(s)
s.clear()
print(s)
s.append(5)
print(s)
s.append(6)
print(s)
s.append([7,8])
print(s) #追加的对象视为一个值
s.extend([9,10]) #扩展一个序列
print(s)
s.insert(2,7)
print(s)
s[3:3] = [8] #不同手段，达到同样目的
print(s) 

 

 

 

s = [1,2,3,4,5,6,7,8,2,3,2,2]
print(s)
print(s.pop())
result = s.pop()
print(result)
print(s.pop(-1))
print(s.reverse())
print(s)
x = s.reverse()
print(x) #reverse只反转当前对象，不返回值

l = list(range(11))
s = l[:]
x = l.copy() #只复制值，不共享索引位置
x[0] = 98
s[0] = 99
print(s)
print(x)
print(l)