Tuple problem in Python intermediate

Question

The problem wants me to solve it as a tuple, but I had to rewrite the code as a dictionary to get the proper output.

How would I solve this by leaving it as a tuple?
https://code.sololearn.com/c7ZBmPst4pIC/?ref=app

Answer 1

Rik Wittkopp ,

the output of the code you provided does not respect the task description (which was not given by the op).

Billy ,
sure, you can modify the given input data. the task can be done without problems by using a dictionary.

> but the exercise is meant to use it as it is, using tuples.

contacts = [
('James', 42),
('Amy', 24),
('John', 31),
('Amanda', 63),
('Bob', 18)
]
inp = input()
for contact in contacts:
if inp in contact:
print(f'{contact[0]} is {contact[1]}')
break
else:
print('Not Found')

Answer 2

The challenge has a list containing a group of tuples.

Tuples can be sliced similar to lists, you you can iterate through the list items, & slice your tuples to produce a result.

Example

contacts = [
('James', 42),
('Amy', 24),
('John', 31),
('Amanda', 63),
('Bob', 18)
]

for x in contacts:
print(x[0],"is",x[1],"years old")

Answer 3

Billy Next time you want a dictionary from a tuple like contacts, then there is no need to rewrite it. You can just convert it this way:

contacts = dict(contacts)

Answer 4

Quantum
What a great trick, thanks for that piece of info 😁👍

Answer 5

this is the task description fof this exercise:

You are given a list of contacts, where each contact is represented by a tuple, with the name and age of the contact.

Complete the program to get a string as input, search for the name in the list of contacts and output the age of the contact in the format presented below:

Sample Input:
John

Sample Output:
John is 31

If the contact is not found, the program should output "Not Found".

Answer 6

Thanks but it looks like they also changed it to a dictionary. I thought maybe there was an answer where you leave the given code alone

Answer 7

Lothar
Thank you! I didn't recognize the f'' , but I was able to solve with this:

name = input()
for x in contacts:
if name in x:
print(name, "is", x[1])
break
else:
print("Not Found")

One last question, why doesn't this also work in line 3 of my above for-loop:

if name == x:

Answer 8

https://sololearn.com/discuss/3235074/?ref=app

Answer 9

You can use a for loop to loop through each tuple and do your checking.

Answer 10

Rik Wittkopp
Thanks for your comment, but I couldn't figure out a way to use your method where the list is not printed. dict() worked well

Answer 11

I cant find a way to do the exercise by leaving it as tuples. This is how I solved using the dict() function:

contacts = [
('James', 42),
('Amy', 24),
('John', 31),
('Amanda', 63),
('Bob', 18)
]

contacts = dict(contacts)
name = input()
if name in contacts:
print(name, "is", contacts[name])
else:
print ("Not Found")

Answer 12

Billy, try the code below and see for yourself.
name = input()
for x in contacts:
print(name)
print(x)
print(name == x)

Answer 13

Wong Hei Ming
Ohhh I get what I was doing wrong. It needed to be
if name == x[0]:

Thanks for all the help!!