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python method to read file names and generate list

不言
不言Original
2018-04-27 11:20:314736browse

下面为大家分享一篇python读取文件名称生成list的方法,具有很好的参考价值,希望对大家有所帮助。一起过来看看吧

经常需要读取某个文件夹下所有的图像文件。

我使用python写了个简单的代码,读取某个文件夹下某个后缀的文件,将文件名生成为文本(csv格式)

import fnmatch
import os
import pandas as pd
import numpy as np 
import sys
InputStra = sys.argv[1]
InputStrb = sys.argv[2]
def ReadSaveAddr(Stra,Strb):
 #print(Stra)
 #print(Strb)
 print("Read :",Stra,Strb)
 a_list = fnmatch.filter(os.listdir(Stra),Strb)
 print("Find = ",len(a_list))
 df = pd.DataFrame(np.arange(len(a_list)).reshape((len(a_list),1)),columns=['Addr']) 
 df.Addr = a_list
 #print(df.head())
 df.to_csv('Get.lst',columns=['Addr'],index=False,header=False)
 print("Write To Get.lst !")
ReadSaveAddr(InputStra,InputStrb)

上面代码保存为:GetLst.py

使用时:

在cmd窗口输入:

python GetLst.py F:/train/pos *.png

发现上面代码不能深入到下一层目录,又做了点修改:

def ReadSaveAddr2(Stra,Strb):
 df = pd.DataFrame(np.arange(0).reshape(0,1),columns=['Addr']) 
 print(df)
 path = InputStra
 for dirpath,dirnames,filenames in os.walk(path):
  #for filename in filenames:
  a_list = fnmatch.filter(os.listdir(dirpath),Strb)
  if len(a_list):
   dft = pd.DataFrame(np.arange(len(a_list)).reshape((len(a_list),1)),columns=['Addr']) 
   dft.Addr = a_list
   dft.Addr = dirpath + '\\' + dft.Addr#输出绝对路径
   frames = [df,dft]
   df = pd.concat(frames)
   print(df.shape)
 df.to_csv('Get.lst',columns='Addr'],index=False,header=False)
 print("Write To Get.lst !")

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