i wrote this program :
l = []
N = int(input("enter the size of the list"))
if N < 50:
for i in range(N):
a = int(input("add a number to the list"))
l.append(a)
for i in range(N):
del l[min(l)]
print(l)
and when i run it they say
Traceback (most recent call last):
File "<pyshell#5>", line 2, in <module>
del l[min(l)]
IndexError: list assignment index out of range
please do you have any solutions ??
Your problem is that del l[min(l)] is trying to reference the list item at index min(l). Let's assume that your list has 3 items in it:
l = [22,31,17]
Trying to delete the item at the index min(l) is referencing the index 17, which doesn't exist. Only indices 0, 1, and 2 exist.
I think what you want to do is remove the smallest item from your list sequentially. There are a number of ways to do this. The method that is closest to what you have written would be:
for i in range(N):
l.remove(min(l))
print(l)
Change
del l[min(l)]
to
del l[l.index(min(l))]
Reason : Because you want to delete element holding index of min element and not index=min element
O/P : (for input 1 2 3 4 5)
[2, 3, 4, 5]
[3, 4, 5]
[4, 5]
[5]
[]
del l[min(l)]? Consider a list[10, 20, 30]now image the execution ofdel l[min(l)]. This would bedel l[10], but there is not 10th element in the list and hence shows errordel l[min(l)]is equivalent todel l[23], which is equivalent to "delete the 24th element of l". But l doesn't have 24 elements, it has three elements.