組織和複製文件到新文件夾

Question

我想在處理它之前組織一些數據。

我所擁有的是一個原始tiff文件的文件夾（它們是來自無人機傳感器的光柵波段）。

我想將這些文件移動到新的單個文件夾中。例如IMG_001_1，IMG_001_2，IMG_001_3，IMG_001_4和IMG_001_5全部移動到標題爲IMG_001的新文件夾中。我可以改變文件的命名結構，以使代碼更簡單。

另外一個問題是文件夾中缺少一些圖像。當前的文件是IMG0016 - IMG0054（無IMG0055），IMG0056 - IMG0086（無IMG0087）和IMG0087 - IMG0161。這就是爲什麼我認爲從1-143重新命名新圖像文件夾會更簡單。

我的主要問題是實際上將文件移動到新文件夾中 - 創建文件夾非常簡單。

Answer 1

問題解決了！

import os 
import shutil 

srcpath = "C:\Users\joshuarb\Desktop\Python_Test\UnorganizedImages" 
srcfiles = os.listdir(srcpath) 

destpath = "C:\Users\joshuarb\Desktop\Python_Test\OrganizedImages" 

# extract the three letters from filenames and filter out duplicates 
destdirs = list(set([filename[0:8] for filename in srcfiles])) 


def create(dirname, destpath): 
    full_path = os.path.join(destpath, dirname) 
    os.mkdir(full_path) 
    return full_path 

def move(filename, dirpath): 
    shutil.move(os.path.join(srcpath, filename) 
      ,dirpath) 

# create destination directories and store their names along with full paths 
targets = [ 
    (folder, create(folder, destpath)) for folder in destdirs 
] 

for dirname, full_path in targets: 
    for filename in srcfiles: 
     if dirname == filename[0:8]: 
      move(filename, full_path) 

Answer 2

玩過周圍一點點，這個劇本就出來了，這應該做你想要什麼：

import os 
import shutil 
import re 

UNORG = "C:\\Users\joshuarb\Desktop\Unorganized_Images\\" 
ORG = "C:\\Users\joshuarb\Desktop\Organized_Images\\" 


def main(): 
    file_names = [os.path.join(UNORG, i) for i in get_files_of(UNORG)] 
    for count in range(0, 143): 
     current_dir = "{}IMG_{:04d}".format(ORG, count) 
     os.makedirs(current_dir) 
     move_files = get_files_to_move(file_names, count) 
     print move_files 
     for i in move_files: 
      shutil.move(i, os.path.join(current_dir, os.path.basename(i))) 


def get_files_to_move(file_names, count): 
    return [i for i in file_names if re.match('.*IMG{}_.*'.format(count), i)] 


def get_files_of(mypath): 
    (dirpath, dirnames, filenames) = os.walk(mypath).next() 
    return filenames 


if __name__ == '__main__': 
    main() 

正如你看到的，代碼沒有評論。但隨意問，如果有什麼不清楚;）