I want that a user can enter digits and if the digits matches with the list values then the program returns the entered value and it's index too.
How can i do that? I have written some line of code but it is not working....
a = [10, 20, 30, 40, 50, 60]
inp = int(input("Enter digit"))
i =0
for i in a:
if inp == a[i]:
print("You found it {}".format(a[i]))
else:
print("No found")
It is raising an IndexError.
for i in a iterates over the elements of a, not over its integer indices.
You can fix your code by using enumerate.
a = [10, 20, 30, 40, 50, 60]
inp = int(input('Enter digit: '))
for index, value in enumerate(a):
if value == inp:
print('You found it at position {}'.format(index))
break
else: # no break
print('not found')
Additionally, I changed the string input('Enter digit: ') returns to an int and break out of the loop once the target has been seen once.
See this question for solutions how to program this behavior outside of a programming exercise (TL;DR: a.index(inp)).
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)
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.
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()
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
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 :
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.
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 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))
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
no need to use for loop to search for one type of num, You can use a "Conditional Statement".
a = [10, 20, 30, 40, 50, 60]
inp = int(input("Enter digit"))
if inp in a:
index = a.index(inp)
print(f"You found it, {inp} located in list on index {index}")
else:
print("Not found")
