日列表功能、任务

split():

split() 方法根据分隔符将字符串划分为子字符串列表。

加入():

join() 方法使用调用它的字符串作为分隔符，将可迭代的元素连接成单个字符串。

使用循环的示例：

s = "today is thursday"
reverse = ""
i = 0
while i<len(s):
    reverse = reverse + s[-(i+1)]
    i+=1
print(reverse)
i = 0 
word = ""
while i<len(reverse):
    if reverse[i]==' ':
        print(word[::-1],end=' ')
        word = ""
        i+=1
        continue
    word = word + reverse[i]
    i+=1
else:
    print(word[::-1])

yadsruht si yadot
thursday is today

使用列表的示例：

sen = "today is thursday"
l =  sen.split(" ")
print(l)
print(l[::-1])
s = " ".join(l[::-1])
print(s)

['today', 'is', 'thursday']
['thursday', 'is', 'today']
thursday is today

深拷贝：

python 中的深拷贝是指创建一个新对象，该对象是原始对象的完全独立副本，包括嵌套在其中的所有对象。

l1 = [10,20,30]
l2 = l1
print(l1)
print(id(l1))
print(l2)
print(id(l2))
l2[0] = 111
print(l1)
print(l2)

[10, 20, 30]
127285488814912
[10, 20, 30]
127285488814912
[111, 20, 30]
[111, 20, 30]

浅复制：

浅拷贝，其中新对象是原始对象的副本，但它包含对相同嵌套对象的引用。

l1 = [10,20,30]
l2 = l1.copy()
print(l1)
print(id(l1))
print(l2)
print(id(l2))
l2[0] = 111
print(l1)
print(l2)

[10, 20, 30]
140500496468800
[10, 20, 30]
140500496470528
[10, 20, 30]
[111, 20, 30]

字典顺序：

词典顺序是单词在词典中按字母顺序排序的方式的概括。

l1 = ['lakshmi', 'guru', 'kuhan']
l2 = ['lakshmi', 'guru', 'kuhan']
print(l1 == l2)
print(l1 != l2)

l2 = ['guru', 'lakshmi', 'kuhan']
print(l1>l2) 
print(l1<l2)

true
false
true
false

示例：

l1 = [10,20,30]
l2 = [5,6,7]

for no in l1:
    for num in l2:
        print(no, num, end=' ')
    print()

10 5 10 6 10 7 
20 5 20 6 20 7 
30 5 30 6 30 7 

如果输出附近有 20 20，则删除：

l1 = [10,20,30]
l2 = [8,20,7]

for no in l1:
    for num in l2:
        if no==num:
            print("     ", end=" ")
        else:
            print(no, num, end=' ')
    print()

10 8 10 20 10 7 
20 8       20 7 
30 8 30 20 30 7 

任务：
查找给定总和的对：5

l = [1,2,3,4,5,6,7,8,9,0]
for i in range(len(l)):
    for j in range(len(l)):
        sum = l[i] + l[j]
        if sum == 5:
            print(l[i], l[j])

1 4
2 3
3 2
4 1
5 0
0 5

找到丢失的号码：

l = [10,20,30,50,60,70,80,90]
i=0
while i<len(l)-1:
    if l[i+1]-l[i]!=10:
        result=(l[i]+l[i+1])/2
        print(int(result))
    i+=1

40

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