index() to get matching position in list

All list methods

index(x,[start[,end]])

x : input data to check
start: Optional, starting of the search position
end: Opional, End of search position

returns number of matching position.
my_list=['a','b','c','b','a','d']
print("Position of first matching  a : ", my_list.index('a'))
print("Position of first matching  b : ", my_list.index('b'))
print("Position of first matching  d : ", my_list.index('d'))
Output
Position of first matching  a :  0
Position of first matching  b :  1
Position of first matching  d :  5

Using start and end positions

my_list=['a','b','c','b','a','d','a']
print("Position of first matching  a : ", my_list.index('a',2))
print("Position of first matching  b : ", my_list.index('b',1,4))
print("Position of first matching  a : ", my_list.index('a',5))
Output
Position of first matching  a :  4
Position of first matching  b :  1
Position of first matching  a :  6
If not found then ValueError will be raised.
my_list=['a','b','c','b','a','d','a']
print("Position of first matching  b : ", my_list.index('b',5))
This code will give error as b is not available after position 5. Note that first position is 0th in the list.
All list methods Questions with solutions on List pop() to remove item based on position
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