test5

arr1 =[11,22,33] arr2 =[44,22,33] arrSame = [] arrOneHaveTwoHaveNot = [] arrOneHaveTwoHaveNot1 = []

获取内容相同的元素列表: for item in arr1: for iter in arr2: if(item==iter): arrSame.append(iter) #获取arr1 和 arr2 中内容都不同的元素:

for item in arr1: oneHas = 0 for iter in arr2: if(item!=iter): oneHas = 0 continue else: oneHas = 1 break if(oneHas == 0): print(item) arrOneHaveTwoHaveNot.append(item) i =3

for item in arr2: oneHas = 0 for iter in arr1: if(item!=iter): oneHas = 0 continue#中断for循环 else: oneHas = 1 break#跳出for循环

 if(oneHas == 0):#判断是否为0
     print(item)
     arrOneHaveTwoHaveNot1.append(item)

#打印两位数: arr1="123456789" arr2="123456789" for item in arr1: for item2 in arr2: print(item+item2)

#乘法口诀表1~9：

for i in range(1,10): for j in range(1,i+1): n=j*i print(j, "X", i, "=", n, " ",end="")#" "是1个空格 print("\n")#换行符

str1="bfsadaabbjhbfdvginhbbbihffhhhh" #函数作用，在参数一的字符串中，寻找参数二的出现的次数 #第一个参数，字符串类型：被查找的字符串 #第二个参数，需要查找的字符 def strDisplayCount(strn,zifu): baCuen=0 for item in strn: if(item==zifu): baCuen=baCuen+1

return baCuen