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“Image.open(file).to_thumb(256,256)” is not working with variable containing file name , but it is working if I am giving the file name directly

target_file = ‘/kaggle/working’/path/’forest’ i = 0 for file in target_file.iterdir(): print(file) Image.open(file).to_thumb(256,256) i+=1 if i > 5: break Output is as follows: /kaggle/working/bird_or_not/forest/9c34f24c-5db1-4a03-9ef7-60b53f6e2b2e.jpg /kaggle/working/bird_or_not/forest/e0062310-e425-4058-8bba-00a26b9b81bc.jpg /kaggle/working/bird_or_not/forest/4ceb232b-8be0-4ede-ba38-85193e1fec91.jpg /kaggle/working/bird_or_not/forest/d764afd2-8419-4de3-9372-498bb70cec65.jpeg /kaggle/working/bird_or_not/forest/9a1e05f9-d31d-4c15-b819-e91290dd1c32.jpg /kaggle/working/bird_or_not/forest/67ca71d8-6059-444f-8351-ec78eb19ae2b.jpeg And If I give the specific file name, than it is working, i.e. After running the below command, I see the image Image.open(‘bird_or_not/forest/ca274e96-48c9-46b3-9eeb-a98ddf3f665d.jpeg’).to_thumb(256,256) I want to why […]