目的是我想要获取文件名并以我想要的方式(对于我的所有文件)创建标签,然后将此信息保存在csv文件中
您可以将glob和pandas to_csv()
用于此任务,即:
from os import path
from glob import glob
import pandas as pd
f_filter = ["mp3", "ogg"] # a list containing the desired file extensions to be matched
m = [] # final match list
for f_path in glob('D:/museu_do_fado/mp3/**', recursive=True): # loop directory recursively
f_name = path.basename(f_path) # get the filename
f_ext = f_name.split(".")[-1].lower() # get the file extension and lower it for comparison.
if f_ext in f_filter: # filter files by f_filter
label = "Your choice"
#label = f_name[0] + f_ext[-1] # as per your example, first char of file_name and last of file_ext
m.append([f_path, f_name, f_ext, label]) # append to match list
#print(f_path, f_name, f_name, label)
df = pd.DataFrame(m, columns=['f_path', 'f_name', 'f_ext', 'label']) # create a dataframe from match list
df.to_csv("my_library.csv", index=False) # create csv from df
样品csv
:
f_path,f_name,f_ext,label
D:\museu_do_fado\mp3\MDF0001_39.mp3,MDF0001_39.mp3,mp3,Your choice
D:\museu_do_fado\mp3\MDF0001_40.mp3,MDF0001_40.mp3,mp3,Your choice
...
笔记:
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句