A simple finding something in an array

Question

So I've been trying to make a program "how to find if a value is stored in a variable or not". I did this but it doesn't work, it always says that it is not found even tho its inside the variable

a = [2,5,7,9,11]

cari = input("Input the value : ")
ketemu = False
for i in range(0, len(a)):
    if cari == a[i]:
        ketemu = True
        break

if ketemu:
  print("Value: ", cari, "found it")
else:
  print("Value: ", cari, "not found")

`input()` returns a string value. The string `"5"` is not equal to integer `5`. — John Gordon, Oct 12 '21 at 18:07
No need for that loop. `ketemu = int(input("Input the value : ")) in a` then your print statements — Jab, Oct 12 '21 at 18:07

score 0 · Answer 1 · answered Oct 12 '21 at 18:05

0

You need to convert the result of input from string to int.

cari = int(input("Input the value : "))

answered Oct 12 '21 at 18:05

Unmitigated

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Thank you for your answer, tried it and it worked. Why is it have to be like that if you can please answer me ? Appreciate it a lot :D – Wenky Oct 12 '21 at 18:07
@Wenky `input` returns a string and your list contains ints. An `int` will never be equal to a string. – Unmitigated Oct 12 '21 at 18:08
It is like that because python does not make assumptions for you. You might want to have digits as strings. *Explicit is better than implicit* – mozway Oct 12 '21 at 18:10

score 0 · Answer 2 · answered Oct 12 '21 at 18:08

It is because, you written only input() which is taken as string when you enter the input. For example, if you put 5 as input, cari will be '5' and not 5. So change it from input() to int(input()). Code:

a = [2,5,7,9,11]

cari = int(input("Input the value : "))
ketemu = False
for i in range(0, len(a)):
    if cari == a[i]:
        ketemu = True
        break

if ketemu:
  print("Value: ", cari, "found it")
else:
  print("Value: ", cari, "not found")

score 0 · Answer 3 · answered Oct 12 '21 at 18:10

Here's a slight modification to your code that will result the correct out come.

a = [2,5,7,9,11]

cari = int(input("Input the value : "))
ketemu = False
for i in range(0, len(a)):
    if cari == a[i]:
        ketemu = True
        break

if ketemu:
  print(f"Value: {cari} found it")
else:
  print(f"Value: {cari} not found")

score 0 · Answer 4 · edited Oct 13 '21 at 04:35

0

You missed to add the datatype in which you get your answer. If you're getting integer then put int(input()) and when you get string values then put str(input()) according to your query:

cari = int(input("Input the value : "))

edited Oct 13 '21 at 04:35

marc_s

732,580
175
1,330
1,459

answered Oct 12 '21 at 18:26

Muhammad Ali

1

score -1 · Answer 5 · answered Oct 12 '21 at 18:14

-1

You can simplify the checking loop.

a = [2,5,7,9,11]
cari = int(input("Input: "))
if(cari in a):
    print(f"{cari} found")
else:
    print(f"{cari} NOT found")

answered Oct 12 '21 at 18:14

ninjajosh5

39
5

A simple finding something in an array

5 Answers5