CODEWITHSUNDEEP
pythonpython-3.x
Sanyam Labroo
Sanyam Labroo

Reputation: 29

Program prints ouput twice

I have made a program which shows contact when contact name is given but the contact not found statement is coming twice

answer=""
while answer!= "yes":

    contact_name = input("Entername: ")
    contacts = {'Mom': 9469211475, 'Dad': 9419198443}
    for name in contacts:
        if name == contact_name:
            print("Contact number is: ", contacts[name])

        else:
            print("Contact not found")


    answer = input("Do you wish to exit?  Please Enter Yes or No: ").lower()

print("Thank you")

Upvotes: 0

Views: 70

Answers (3)

Tomerikoo
Tomerikoo

Reputation: 19414

Your code is currently an anti-pattern: you're using a dict but still looping over all its contents (with an O(n) complexity) to check if a certain key exists.

You don't need to loop the names if you're using a dict... Simply check if the name exists in the dict (with an O(1) complexity) and if it's not, print accordingly. In general, the pattern:

for x in iterable:
    if x == y:
        # do something with y
        break

Is usually equivalent to:

if y in iterable:
    # do something with y

So in your case:

contacts = {'Mom': 9469211475, 'Dad': 9419198443}

while answer != "yes":

    contact_name = input("Entername: ")
    if contact_name in contacts:
        print("Contact number is: ", contacts[contact_name])

    else:
        print("Contact not found")

    answer = input("Do you wish to exit?  Please Enter Yes or No: ").lower()

print("Thank you")

Or, if you're comfortable with exception handling, this saves the double look-up:

contact_name = input("Entername: ")
try:
    print("Contact number is: ", contacts[contact_name])

except KeyError:
    print("Contact not found")

Upvotes: 3

ppwater
ppwater

Reputation: 2277

You can set count of loop and change it every time it loops. then compare it with length of the dict. if it's same as the length of the dict, then that means the word is not found.

Like this:

answer=""
while answer!= "yes":

    contact_name = input("Entername: ")
    contacts = {'Mom': 9469211475, 'Dad': 9419198443}
    loop = 0
    for name in contacts:
        loop += 1
        if loop >= len(contacts):
            print("Contact not found")
            break
        if name == contact_name:
            print("Contact number is: ", contacts[name])
            break
    answer = input("Do you wish to exit?  Please Enter Yes or No: ").lower()

print("Thank you")

Or just use for else

answer=""
while answer!= "yes":

    contact_name = input("Entername: ")
    contacts = {'Mom': 9469211475, 'Dad': 9419198443}
    for name in contacts:
        if name == contact_name:
            print("Contact number is: ", contacts[name])
            break

    else:
        print("Contact not found")


    answer = input("Do you wish to exit?  Please Enter Yes or No: ").lower()

print("Thank you")

Upvotes: 0

costaparas
costaparas

Reputation: 5237

The loop for name in contacts: will iterate over the two elements in the contacts list. Hence, the code in the body of the for loop executes twice.

What you should do instead, is remove the else branch of the if statement and check once after the for loop completes whether the contact has been found or not.

A simple way to do this is by setting a new variable found to False before the loop, updating to True inside the loop if the contact is found and then checking the value of flag after the loop terminates.

Upvotes: 0

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