a = int(input())
l1 = []
for i in range(a):
l1[i] = 5
print(l1)
I keep getting the error:
list assignment index out of range
i will always be smaller than a so why am I getting this error? I don't wish to use append().
5 Answers
You could use this if you want to get a list of n '5' elements:
a = int(input())
l1 = [5 for _ in range(a)]
print l1
It will wait for input and save the list in the l1 list
1 Comment
Sheldore
I would replace
i by _ since it's not being used hereYou are referring to the i-th element, but your list is empty. It has to have a i-th element to write a[i] = 5. So, populate your list with a dummy value. eg, start with l1 = [None] * a instead of l1 = [].
Comments
Use insert.
a = int(input())
l1 = []
for i in range(a):
l1.insert(999999999999,5)
print(l1)
Executing:
10
[5, 5, 5, 5, 5, 5, 5, 5, 5, 5]
Comments
You might be confusing the behavior of list to dictionary. Dictionary would work.
a = int(input())
l1 = {}
for i in range(a):
l1[i] = 5
print([j for i,j in l1.items()])
You got an error because there exists no element at l1[0] to be assigned users input. If we were to make your code work, we need to pre-populate the list with dummies.
a = int(input())
l1 = (','*(a-1)).split(',')
for i in range(a):
l1[i] = 5
print(l1)
5 Comments
Vishal Singh
I don't think he is confused between Dictionary and List. He is treating List like Arrays, maybe he just started python and trying to apply other Language paradigms in Python
Prayson W. Daniel
Ah! Would not array throw IndexError?
Vishal Singh
python doesn't have arrays. If arrays were to initialize then no they won't throw error if you do array[0]=someValue
Prayson W. Daniel
I was think of numpy: I.e. l1 = np.array([]). This throws IndexError ;)
juanpa.arrivillaga
@VishalSingh Python does have arrays.
It seems to me that the simplest change is to append the elements, rather than use an index. The list doesn't need to be pre-sized and the code is easy to read for someone coming from another language. Replace "i" with "_" if you like.
a = int(input())
l1 = []
for i in range(a):
l1.append(5)
print(l1)
l1has length zero, so you cant index intol[i]for anyi. Why you don't want to useappendis beyond me.append()is the one obvious way to do it. Using anything else would make for worse code.l1 = [5 for _ in range(10)]andprint (l1)if you are so against theappend().