python基础知识第七篇(练习)

# a.     获取内容相同的元素列表

 l1   =  [11,22,33]

 l2   =  [22,33,44]

 for l in l1:

     if l in l2:

         print(l)

# b.    获取 l1  中有， l2 中没有的元素列表

 l1   =  [11,22,33]

 l2   =  [22,33,44]

 for l in l1:

     if l not in l2:

         print(l)

# c.    获取 l2  中有， l1 中没有的元素列表

 l1   =  [11,22,33]

 l2   =  [22,33,44]

 for i in l2:

     if i not in l1:

         print(i)

# d.    获取 l1  和 l2   中内容都不同的元素

 l1   =  [11,22,33]

 l2   =  [22,33,44]

 for i in l1:

     if i not in l2:

         print(i)

 for i in l2:

     if i not in l1:

          print(i)


#有12345678 八个数字，如何组成不重复的两位数，共有多少种？

 count=0

 for i in range(1, 9):

     for i1 in range(1,9):

         if i!=i1:

             count +=1

 print(count)

 count=0

 list1=[1,3]

 for i in range(0,len(list1)):

     for i1 in range(0,len(list1)):

         if i!=i1:

             count +=1

#九九乘法表

 for i in range(1,10):

     for i1 in range(1,i+1):

         print(str(i)+"*"+str(i1)+"="+str(i*i1)+"\t",end="")

     print(end="\n")


    #公鸡5文钱一只，母鸡3文钱一只，小鸡3只一文钱，100文钱只能买100只鸡，有哪些组合？

 for x in range(1, (100 // 5+1) ):

     for y in range(1, (100 // 3+1) ):

         for z in range(1, 101):

             if x + y + z == 100 and 5 * x + 3 * y + z / 3 == 1000:

                 print(x,y,z)

#li=["dsk","ds","pp"],用下划线把列表的值连起来

 li=["dsk","ds","pp",12]

 for i in li:

     print(i,end="_")

 li=["dsk","ds","pp",12]

 li[3]=str(li[3])

 v="_".join(li)

 print(v)


#获取元组tu=("dhik","xjo","xsx")，计算元组的长度并输出，获取元组的第二个元素并输出

 tu=("dhik","xjo","xsx")

 v=len(tu)

 print(v)

tu=("dhik","xjo","xsx")

 v=tu[1]

 print(v)


#nums=[1,3,5,2,8,4]
#找出列表中两个元素相加等于9的结合

nums=[1,3,5,2,8,4]

cound=[]

for i in nums:

    for j in nums:

        if i+j==9:

            cound.append((i,j))

print(cound)

巴特西

python基础知识第七篇(练习)

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