New RunLLM response generated: >>>*# Question* Hi! Simple question: How do I access one particular existing file from a FlyteDirectory? I can do

FlyteDirectory.listdir

and the pick out the particular child that I want, but that is a bit cumbersome. Alternatively, I could do

dir.new_file()

, which should work, but that usage seems contradicting the method name. Shouldn't such a function exist? Thanks! # Answer To access a specific file from a

FlyteDirectory

, you can use the

listdir

method to list all files and directories, then filter for the specific file you need. There isn't a direct method to access a specific file without listing, as per the current Flyte API. The

new_file()

method is intended for creating new files, not accessing existing ones. Would you like more detail on how to implement this using

listdir

? Sources: - FlyteDirectory listdir method Slack Link https://.slack.com/archives/CP2HDHKE1/p1731513002.791699 👍 0 button 👎 0 button 💡 Hint: Mention @runllm in the thread for followups.