Python Contact List

Question

Test task, given range of contacts, user should input a name, and if name exists the code should output Name and Age *** So, my code has passed the test, but I think that it is too wordy and feels like I could do it much better and shorter, Can you please state how could I do it better?

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

n = input()

if n in contacts[0]:
print(str(contacts[0][0]) + " is " + str(contacts[0][1]))
elif n in contacts[1]:
print(str(contacts[1][0]) + " is " + str(contacts[1][1]))
elif n in contacts[2]:
print(str(contacts[2][0]) + " is " + str(contacts[2][1]))
elif n in contacts[3]:
print(str(contacts[3][0]) + " is " + str(contacts[3][1]))
elif n in contacts[4]:
print(str(contacts[4][0]) + " is " + str(contacts[4][1]))
else:
print("Not found")

Answer 1

#try this

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

n = input()

for i, j in contacts:
if i == n:
print(i , " is " ,j)
break
else:
print("Not found")

Answer 2

Adilet Kambarov ,

you can check each tuple during an iteration like:

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

n = input()
for contact in contacts:
if n in contact:
print(contact[0], "is", contact[1]) # you don't need to convert data formats when commas as a separater is used!
break
else:
print("Not found")

Answer 3

Please help! I found this answer after thinking about it all day and it accept it as correct But I can’t add the else statement because when I do it either throws an error about indentation or it also prints the else statement anyway and it’s not correct anymore

contacts = [
('James', 42),
('Amy', 24),
('John', 31),
('Amanda', 63),
('Bob', 18)
]
nam = input()
i=0
for name, age in contacts:
name, age = contacts[i]
if nam == name:
print(name + ' is ' + str(age))
i+=1
#with the else statement that throws error is like this:

nam = input()
i=0
for name, age in contacts:
name, age = contacts[i]
if nam == name:
print(name + ' is ' + str(age))
i+=1
else:
print('Not found')