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https://code.sololearn.com/c7r247tcwo2Y/?ref=app Correct me for error

23rd Jun 2021, 6:37 PM
Zahed Shaikh
4 ответов
+ 1
Sry for late . May it already solved. But adding my correction : Corrected code, making strings of integers instead of int comparisions, because according your logic ,it is simple way... # pattern1 = ["!","@","#","
quot;,"%","&","*"] pattern2 = list(map(str,range(0,10))) #making list to string of digits.. password=input() def func_count(password ,x): d=0 for i in password : if i in x: # using in here, d+=1 return d #for pattern1 in password : count1=func_count(password ,pattern1 ) #for pattern2 in password : count2=func_count(password ,pattern2 ) if count1>=2 and count2 >=2 and len(password)>=7: print ('Strong') else : print ('Weak') print (count1 ) print (count2 ) print (len(password ))
26th Jun 2021, 11:07 AM
Jayakrishna 🇮🇳
+ 2
How take int as input for pattern 2. Done changes suggested, but getting results always 'weak'
24th Jun 2021, 9:44 AM
Zahed Shaikh
+ 1
pattern1 = ["!","@","#","
quot;,"%","&","*"] pattern2 = range(0,10) #pattern2 now here stores an iterator. not a list of unpacked numbers from 0 to 10 . and its numbers is in form of integer but when you checking in loop, there your having charecter like '1' , not 1. password=input() def func(x): d=0 for i in password : if i==x: #this means char==list, false always.. may i in x works for pattern1 only.. for pattern2, as already told it comparing like string: '1' in integers of range(0,10) so always false. you should need int(i) in x for pattern2 d+=1 return d #ident this out of loop #these 2 statement have no use , you are not storing returned values and using again in if condition hence no need. #func(pattern 1) #func(pattern2 ) if func (pattern1 )>=2 and func (pattern2 )>=2 and len(password)>=7: print ('Strong') else : print ('Weak') so because of wrong check in i==x , tjis cide failing.... hope it helps to correct it...
23rd Jun 2021, 7:11 PM
Jayakrishna 🇮🇳
0
password = list(input()) str = " ".join(password) #a string with space b/w the items import re symb = re.findall("[!@#\\$%&\*]+", str) num = re.findall(r"[0-9]+", str) filt = filter((lambda x: x in symb,str), (password)) print ("Strong" if len(symb) >= 2 and len(num) >= 2 and len(password) >=7 else "Weak")
5th Dec 2021, 8:15 PM
Mas'ud Saleh
Актуальное сегодня
Functions