Python doesn't remove the same amount of items from two different lists... given the same indices

Revit
,
1

I’m trying to filtering a list of elements (48 items) and a list of the corresponding parameter values (also 48 items). I have a filter loop written to filter out Null values from the elements list. And an identical filter loop to filter the Parameters list. Yet I get different list lengths depending on which list I filter first. I’ll try to explain here but I’ll also post code below.

Elements >then> list2 and OUT = [len(Elements), len(list2)]
OUT = [45, 46]
and second
list2 >then> Elements and OUT = [len(Elements), len(list2)]
OUT = [47, 46]

This really baffles me. And for reference, I tested this by swapping list2 and Elements on lines 17 and 23. I just want to remove Nulls (that I specify the indices using list2) and this is really throwing me for a loop.

Confusion
Confusion.png1920×1080 252 KB 

import clr
clr.AddReference('ProtoGeometry')
from Autodesk.DesignScript.Geometry import *
#The inputs to this node
Elements=[]
list1=IN[0]
list2=IN[0]
member=None
Elements=IN[1]
cnt=len(list1)
indices=[]
output = []

inx=0
while inx<(len(list1)) :
	if list1[inx]==None:
		list2.RemoveAt(inx)
	inx=inx+1

inx=0
while inx<(len(list1)) :
	if list1[inx]==None:
		Elements.RemoveAt(inx)
	inx=inx+1
#	Elements.RemoveAt(inx)#, 

#while(x <= (len(indices))):
#	if indices[x]==

#Elements.RemoveAt(map(indices))

#Assign your output to the OUT variable
OUT = [len(Elements), len(list2)]
#OUT = [list2[::-1], Elements]

Just plain confused

2 
list1 = IN[0]
list2 = IN[0]

This is the same as saying list1 = list2 = IN[0] which means list1 and list2 are the same list, not copies. You can use list2 = list(list1) to make an actual copy.

With all of that being said… just merge the two loops. There’s no reason you should have to duplicate a list just so you can loop through it twice. Just add Elements.RemoveAt(inx) to the first loop. I would also suggest using a for loop rather than a while loop in this case.

3

I’ll try implementing you’re way of copying. The reason why I didn’t merge the two is that it was giving the same described issue when I tried putting both of the RemoveAt commands in the same line, with a comma between.

4

They would have to be separate lines.

I just realized another issue you’re going to have is that RemoveAt is altering your list one item at a time. Which means indices won’t line up as soon as you remove an item. You would either have to get a list of indices to delete first and then delete them all at once, or create a new list with the good objects instead of removing the nulls.

1 Like
5

Oh crud, you’re right. That makes so much sense. I’ll fiddle with it and post an update here.

6

So here’s what I’ve got now. Thanks for your help!

import clr
clr.AddReference('ProtoGeometry')
from Autodesk.DesignScript.Geometry import *
import math
#The inputs to this node


lst1=IN[0]
Elements=IN[1]
inx=0
Remove=[]
Output = []
Type =[]
while inx<(len(lst1)):
	if not(lst1[inx] == None):
		Type.append(lst1[inx]), Output.append(Elements[inx])
	inx=inx+1


#Assign your output to the OUT variable
OUT = [Output, Type]