Tukukkk

Change

for i in a:

   if inp == a[i]:

       ...

to

for i in a:

   if inp == i:

       ...

Because for loop iterate over elements in list.(not over index)

a = [10, 20, 30, 40, 50, 60]



inp = int(input("Enter digit: "))



if inp in a:

    print("You found {} at {}".format(inp, a.index(inp)))

else:

    print("Not found")

More pythonic way would be:

a = [10, 20, 30, 40, 50, 60]



inp = int(input("Enter digit :"))

if inp in a:

  print "You found", inp, "at index:", a.index(inp)

else:

  print "Not found"

Output:

Enter digit :10

You found 10 at index: 0

I don't think a for loop is necessary:

l = [i for i in range(0,100,10)]



def found_it():

    print("You found it {}".format(l.index(ans)))

    if ans in l:

        print("You found it")

    else:

        print("Try again")

        found_it()

You can use in to check it in the list

lst = [11, 1, 4, 15, 6]

userInput = int(input('Enter Value')) # IF IT IS INT

if userInput in lst:

    print(userInput, lst.index(userInput))

Firstly,always convert the user input into an integer using the int() function, before finding it in the list of integers.

I guess you only need to find whether one element is there in list or not.For that you can try this:

a = [10, 20, 30, 40, 50, 60]



inp = int(input("Enter digit"))

if inp in a:

    print("You found it at index {}".format(a.index(inp)))

else:

    print("Not found")

There are some logical mistakes in the code :

1. The input function return the string value & you are comparing the
   string value 'inp' with the array int value. Need to type-cast the
   value.




2. for i in a (means you are accessing each value of array 'a' in the
   variable 'i'). Here 'i' is not use as a index variable. Need to
   update it with 'for i in range(0,len(a))' (which means iterate the
   for-loop till the length of array 'a' with the index value of 'i').

Code :

a = [10, 20, 30, 40, 50, 60]

flag = 0 

inp = int(input("Enter digit"))

for i in range(0,len(a)):

    if inp == a[i]:

        print("You found it")

        print("index = ", i)

        flag = 1

        break

if flag == 0:

    print("No found")

Output :

You are facing this problem because in for loop i get the array values and storing in i that's why showing Index Error.

Here us the code to find Input no and matching no:

a=[10,20,30,40,50,60]



inp=20



i=0



for i in a:



    c = i 



    print(c)



    if inp==i:



        print("You found it")

Code Output:

10

20

You found it

30

40

50

60 

You have to use

for i in range(len(a)):

The index error is raised by the fact that you construct your for loop as for i in a, which means that i will iterate through the contents of a. Then calling a[i] will not work because in the first step of the loop you are calling a[10].

Alternatively, you could step through the elements of a directly like this:

for a_element in a:

    if inp == a_element:

Which will compare the user input to the contents of a.