Solving Remove Missing Deletion and Index Out of Bounds Problems in Python

Methods often miss deletions or index out-of-bounds when deleting elements Examples are as follows:

Omission of deletion:

lst=[9,25,12,36]
for i in lst:
	if i>10:
		(i)
print(lst)

>>>[9, 12]

So why was 12 omitted? The principle is actually quite simple, as shown in the figure:

The list is traversed from subscript 0. When it reaches 25, 25 is deleted and a new list is returned:

Note that the original 25 corresponded to a subscript of 1, so the system would start traversing from a subscript of 2, but in the new list, the subscript of 2 becomes 36, so 12 is skipped.

Solution:

Simply determine that if an element is deleted from the list, the list should be traversed again starting from 0.

lst=[9,25,12,36]
while True:
	for i in lst:
		if i>10:
			(i)
			# If an element is deleted, exit this traversal and start traversing the new list.
			break
	else:
		break
print(lst)

>>>[9]

Indexing crosses the line:

When we traverse the list with subscripts, we get index out of bounds, as shown in the figure:

lst=[9,25,12,36]
for i in range(len(lst)):
	# print(i)
	if lst[i]>10:
		(lst[i])
print(lst)

IndexError: list index out of range

The principle is the same, here the value of i is 0 1 2 3, when an element is deleted, the length of the new list is reduced, the index becomes 0 1 2, but i is still based on the index of the original list of values, so when i is taken to 3, the new list does not have the element, the index crosses the line.

Solution one:

lst=[25,9,12,36]
j=0
for i in range(len(lst)):
	if lst[j]>10:
		# Below is an example of j taken to 0:
		# j = 0 when the corresponding element 25 is deleted.
		The practice of # continue means to determine if element j of the new list, i.e., element 0, meets the condition
		(lst[j])
		continue
	# If the condition is not met at j = 0 of the new list, j+1 and then the next element is judged.
	j+=1
print(lst)

Solution 2 (recommended):

If you let the index traverse the list in reverse order you won't have the out-of-bounds problem.
This way, even if the length of the new list is reduced, the value of i is taken backwards, and the absence of a single element in the list has no effect on the value of i.
For example, if an element with an i of 3 is deleted, the highest index of the new list is 2 and the next value of i is exactly 2, which has no effect.

lst=[9,25,12,36]
for i in range(len(lst)-1,-1,-1):
	# i : 3 2 1 0
	if lst[i]>10:
		(lst[i])
print(lst)

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