Instead of a list, build a dictionary where the keys are the list of values you are matching, and the values are what you want to return. You can then pass in the list of keys you want to pull the resulting values from the dictionary for any length/order/source of items and get the results you are after in the order you need.
Perhaps this:
# The inputs to this node will be three lists: IN[0], IN[1], and IN[2]
listA = IN[0] # List of keys
listB = IN[1] # List of values
listC = IN[2] # List of values to be matched
# Create a dictionary from listA and listB
dictAB = dict(zip(listA, listB))
# Initialize the output list
output = []
# Loop over the items in listC
for item in listC:
# If the item is found in the dictionary values, get the corresponding key
if item in dictAB.values():
# Find the key for the current item and append to the output list
key = [key for key, value in dictAB.items() if value == item]
# Assuming you want only the first match, not all possible matches
if key:
output.append(key[0])
# Assign your output to the OUT variable.
OUT = output
You will get best results by describing what you want, giving sample inputs and desired outputs for your task.
Det finns inget behov av Python här och du kommer faktiskt att sakta ner din körhastighet genom att använda den om vi har fullt sammanhang.
Du har redan din ordbok. Använd noden Dictionary.ValueAtKey.
kia ora, me whai whakaaro koe mo te reo huinga whaimana hei awhina i a koe ki te tiki tohutohu.
感謝您的建議,但我使用了你的建議,卻依然無法得到相同的筆數.我有將程式碼使用在spyder作測試,得到的結論卻是正確,我懷疑是dynamo的python script有問題.所以不知能否以dynamo的節點取代python script.謝謝.
List inputs and desired outputs for your task.
FirstIndexOf might be useful if you just want to use nodes.
Sorry, I want to ask again, I checked the value of i in listA, and the value of i that appears is consistent with the value of List.GetItemAtIndex.
However, when the execution reaches if i in d:, compared with the value of List.GetItemAtIndex, there is an extra B10, a total of 23 extra values. I don’t understand what went wrong. I hope someone can help. I understand. Thank you.
@Cloud
The forum language is english.
We don’t understand if you use your own language, in the same way that you wouldn’t understand if I wrote in Swedish.
If you don’t feel familiar with english, please use a translator before you post.
In swedish ? then we are totally sure nobody understand 
Thank you for your reminder, I have converted it into English. Thank you.
Or maybe Danish would have been even more clear
absolutly, but it isnt a language more just a sound we even dont self understand
Why not use the Dictionary value at key node?
here an solution
import sys
import clr
clr.AddReference('ProtoGeometry')
from Autodesk.DesignScript.Geometry import *
def get_key_by_value(valueTest):
return next((key_ for key_, value_ in mydict.items() if valueTest in value_), None)
listA = IN[0] # List of keys
listB = IN[1] # List of values
listC = IN[2] # List of values to be matched
# Create a dictionary from listA and listB
mydict = dict(zip(listA, listB))
OUT = [get_key_by_value(value) for value in listC]
Note:
consider providing test data (list or dyn) for forum users who read you
