It\'s always gives me the output,(\' No SSN number found for\', name). Where is

Question

It's always gives me the output,(' No SSN number found for', name). Where is the problem? somehow telephone number does not link to SSN

46 if number in number_name dict.values): 47 48 49 50 51 for n in number_name_dict.keys () 52 53 54 name # Find name to which it belongs # As soon as name is found, print it if number name dict [nlnumber: name n break print (number, ' is associated with', name) 56 57 58 59 60 61 62 63 64 65 # Now check if that name is present in ssn-name dict # If yes, print it's ssn number # Else, print that ssn is not found if name in ssn_name_ dict.keys): ssn - ssn_name_dict [name] print (name + ??'s', 'SSN is', ssn) # If ssn found # Check if it's present in ssn-income-dict # If yes, print corresponding incomes # Else, print income is not found if ssn in ssn_income_dict.keys (): 67 68 69 70 71 72 73 74 else: 75 print('Therel 's no name associated with that number.' print (name's', 'annual income is', '$' ssn_income_dict[ssn]) else: print ('No details about income found for, name) else: print ('No SSN number found for', name)

Explanation / Answer

You missed at line 60: the if condition.

It's only checking tha name value once. You need to check name for all the values within ssn_name_dict.keys() . i.e. you need to add a for loop there. From code, I guess you are pretty good at coding. Also, I don't have the whole code else, I would have written the for loop for you.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.