使用“_”拆分文件名并在python中转换列表中的单词

Question

Basically what I plan to do is that I want to extract the filename from each image in my directory and store it in a list to be linked with that image itself for object detection and predicting keywords. For example, the filename is animal_cat_kowai.png, so I want my list to have [animal, cat,kowai] linked to the file. This is my code : This displays the word 'cat' only while I have 70 images in this directory. hon

Answer 1

if you just want the name of the file to be split based on "_", if filename = "animal_cat_kowai.png"

your_list = filename[:filename.index(".")].split("_")

This would result in your_lisr = ["animal","cat","kowai"]

Answer 2

I think that list is not suitable for the link. Insted, use dict .

First import required modules.

from pathlib import Path

Initiate variable.

image_dict = {}  # handle name and link
all_file = list(Path().iterdir())  # get all file in current directory

Populate image_dict .

for f in all_file:
    extension = str(f).split(".")[-1]
    filename = str(f).split("." + extension)[0]

    if extension in ["png", "jpeg"]:  # In case of there is other file that's no an image
        part_name = filename.split("_")  # Get keywords from filename

        for pn in part_name:  # Append a link from each keyword
            if image_dict.get(pn, None) is not None:
                image_dict[pn].append(f.resolve())
            else:
                image_dict[pn] = [f.resolve()]

print(image_dict)  # This is the result