Python 中的列表索引错误

Question

In this program, I want to search for a Number from a list. When I search a number that is in the list, it works correctly.

But if I search a number that it is not in list, it gives me this Error:

Traceback (most recent call last): File "fourth.py", line 12, in <module> if(AranaElemean==liste[i]): IndexError: list index out of range

liste=[12,23,3489,15,345,23,9,234,84];

Number=11;
i=0;
Index=0;
isWhileActive=0;
while (i<len(liste) and Number!=liste[i]):
   i=i+1;

   if(Number==liste[i]):
      Index=i;
      isWhileActive=1;
   else:
      Index=0;


if(isWhileActive==0 and i!=0):
   print("Please Enter Valid Number.");
else:
   print("Index:",Index);

Answer 1

That's because i goes from 0 to len(liste) and inside the while loop you are increasing the i by one. So when it doesn't find the desired number and i gets the value i = len(liste), you increae it by 1 in the loop so you get the error because it exceeds the range of the list.

you can use the following

while (i<len(liste)):

   if(Number==liste[i]):
      Index=i;
      isWhileActive=1;
      break
   else:
      Index=0;
   i += 1

Answer 2

Your condition should be:

while (i<len(liste)-1 and Number!=liste[i])

This is because Python list indexing begins at 0.

Therefore, for a list of length n you need to index from 0 to n-1 .