python: collections counter

 

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FF=[[]]     for i in range(0,10):        kk = []        for k in range(0,10):            if(k==0):                kk.append(i)            else:                kk.append(k)        #print(kk)        FF.insert(i,kk)     print(FF)     # 打印列表     for row in FF:         for col in row:             print(col, end=" ")         print()   rows, cols = (5, 5) arr = [[0]*cols]*rows   N = 5 arr = [0 for i in range(N)]

　　

 

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entrance_fee = [70, 80, 105, 110, 120, 125]  entrance_fee = entrance_fee + [0]  index =6  k = 0  while k<6 and index==6:  # 因为这里index 是6 ，初始值必须是6      if 115<entrance_fee[k]:          index=k      k=k+1  print("index",index)  for k in range(6,index,-1):      entrance_fee[k]=entrance_fee[k-1]  entrance_fee[index]=115  print(entrance_fee)    result = ['Pass', 'Fail', 'Pass', 'Pass', 'Pass', 'Pass', 'Pass', 'Pass', 'Pass', 'Pass']  p=Counter(result)  # 引用库  from collections import Counter  #print(dict(p)) #分类计数的结果  #print(p['Pass'])  if(p["Pass"]>=5):      print("S/he gets the certificate of completion.")  else:      print("S/he needs to pass more tests.")      while True:      num=0      P=0      result = []       #也可以用while 或for      s=0      while num<10 and P<=5: #当达到5时，就可以确定极格，不用再计数了          s=num+1          test=input(str(s)+":input Pass/Fail")          result.append(test.strip())  #去除两头空格          num+=1      '''      for k in range(0,10):          num=k+1          test = input(str(num)+":input Pass/Fail")          result.append(test.strip())  # 去除两头空格      '''        print("score",result)        for k in range(0, 10):          if result[k] == "Pass":              P = P + 1          if P == 5:  #5极格了，就结束              break      if P >= 5:          print("S/he gets the certificate of completion.")      else:          print("S/he needs to pass more tests.")      yn=input("Whether to continue testing(Y/N)?")      if(yn=='Y' or yn=='y'):          continue      else:          break

　

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''' 移除Paul 这个值 index 1     eca=["Tony","Paul","Kelly","Wendy","Jack"]     for k in range(1,4): # index 1         eca[k]=eca[k+1]     eca[4]="" #len-1     '''     while True:         damin=input("please enter dmain name:")         tld=""         N=len(damin)         dot=-1         for i in range(0,N):             if damin[i]=='.':                 dot=i         if dot!=-1:             for i in range(dot+1,N):                 tld=tld+damin[i]             print(tld)         else:             print("Not a proper domain name")           yn=input("Y/N?")         if(yn=='Y' or yn=='y'):             continue         else:             break             word='Work Hard'     sub_word=word[5:7]     print(sub_word)   #移动位置 要移的index 指定的位置index 1     show=['2A','3B','1C','4E','5A']     temp=show[3]     for k in range(3,1,-1):         show[k]=show[k-1]     show[1]=temp     print(show)       show=['2A','3B','1C','4E','5A']     temp=show[3]     for k in range(3,0,-1):         show[k]=show[k-1]     show[0]=temp     print(show) #移除     wait = [3001, 4001, 5001, 2001, 2002, 3002, 2003, 4002, 2004, 2005]     delindex=2     for k in range(delindex,len(wait)-1):         wait[k]=wait[k+1]     wait[len(wait)-1]=0     print(wait)     #添加一个  添加位置 index 4     wait=[3001,4001,5001,2001,2002,3002,2003,4002,2004,2005]     wait=wait+[0]     for k in range(len(wait)-1,4,-1):         wait[k]=wait[k-1]     wait[4]=9001     print(wait)     print('*********')       eca=["Tony","Paul","Kelly","Wendy","Jack"]     eca =eca+['']     nn=len(eca);     for k in range(nn-1,3,-1):         eca[k]=eca[k-1]     eca[3]="viky"     print(eca)         entrance_fee = [70, 80, 105, 110, 120, 125]     entrance_fee = entrance_fee + [0]     index = 6     k = 0     '''     while k < 6 and index == 6:  # 因为这是6，初始值必须是6         if 115 < entrance_fee[k]:             index = k         k = k + 1     '''     ''''''     for f in range(0,6):         if 115>entrance_fee[f]:             index=f     print("index",index)     for k in range(6,index,-1):         entrance_fee[k]=entrance_fee[k-1]         print(entrance_fee[k])     entrance_fee[index]=115     print(entrance_fee)