I have this code:
delIndex = []
for a in range(0, 10):
if a not in found_index and len(found_index) > 0:
delIndex.append(a)
for index in sorted(delIndex, reverse=True):
del faces[index]
This code delete items who are not in the array found_index
Is there a way simplify the code by using a method that do this job or another thing?
I don't understand what faces is so I explain only how to remove elements by a difference of two sets of elements. In your case found_index and [a for a in range(0, 10)] . Use set built in in python, and the difference methods - documentation here:
values_to_remove = set([a for a in range(0, 10)])
found_index = set(found_index)
new_itemS = found_index.difference(values_to_remove)
# you can write also new_itemS = found_index - values_to_remove new set with elements in found_index but not in values_to_remove
You can simplify the code to:
delIndex = sorted([a for a in range(0, 10) if a not in found_index], reverse=True)
for index in delIndex:
del faces[index]
If you update your answer with more details and samples of the list faces and found_index (e.g. is it unique, does it contain duplicates), and why you do a revers-sort, I may update this answer.
If I understand correctly, that you want to filter out all elements from a list, which are contained in a second list:
[element for element in faces if element not in found_index]
You can use set for that
Try like this:-
a = [6,8,10] #to be deleted
b = list(range(11)) #compare list
ls= set(b) ^ set(a)
print(ls)#your output
I think you are looking for How to remove specific element in an array using python.
Let me know if this is what you are looking for. Alternatively provide us a sample of input and output so we can play with your code.
Update: Adding sample of code.
Sample of code:
#!/usr/bin/env python
mainArray = [1, 2, 3, 4, 5, 6]
arrayRemove = [1, 3, 6]
mainArray = [e for e in mainArray if e not in arrayRemove]
print(mainArray)
