求个 idea， Python 等数量压缩文件

# -*- coding: utf-8 -*-
import os
import shutil

#path of imgr
path = 'G:/file/201904012-20190418-GKJ2/BACK/zz22'

#path of folder
folderPath = 'G:/file/201904012-20190418-GKJ2/BACK/ZZZZZZZZ'
#give the img list
file_list = sorted(os.listdir(path))
files_each_folder=1000           #设置每个文件夹里面的文件数目
folder_number = int((len(file_list)+files_each_folder-1)/files_each_folder)   #文件夹数目向上取整
sort_folder_number = [x for x in range(0,folder_number)]
prefix_folder_back='IMAGE_PATCH_BACK_PART'   #反面前缀
prefix_folder_front='IMAGE_PATCH_BACK_PART'   #正面前缀
# 创建文件夹
for number in sort_folder_number:
    new_folder_path = os.path.join(folderPath,prefix_folder_back+'%s'%number)#new_folder_path is ‘folderPath\number'

    if not os.path.exists(new_folder_path):
        os.makedirs(new_folder_path)
        print("new a floder named "+str(number)+'at the path of '+ new_folder_path)
'''define the first foloderNumber_index'''
folderNumber_index = 0
print('there are '+str(len(file_list))+' files at the path of '+path)
for index,j in enumerate(file_list):
    #for i in range(0,len(file_list)):
    old_file_path = os.path.join(path,j)
    '''define the number,it decides how many imgs each folder process'''
    #number = 1000 #int(len(file_list)/folder_number)
    if(index%files_each_folder ==0 and index!=0):
        folderNumber_index +=1
    new_file_path = os.path.join(folderPath,'%s'%(folderNumber_index))
    if not os.path.exists(new_file_path):
        print('not exist path:'+new_file_path)
        break
    shutil.copy(old_file_path,new_file_path)
    print('success move file from '+ old_file_path +' to '+new_file_path)

第 2 条附言 · 2019-05-22 13:58:22 +08:00

压缩就加几行的事情

for p in range(folder_number):
    exist_folder = os.path.join(folderPath,prefix_folder_back_Center+'%s'%p)
    shutil.make_archive(exist_folder,'zip',exist_folder)
    print('success make compress  of the '+ exist_folder )

算是完成了这个小功能了。主要是方便自己偷懒，不用手去移动压缩了。

压缩

文件夹

Python

文件

5 条回复 • 2019-05-21 20:59:59 +08:00

minami

2019-05-21 16:56:29 +08:00

获取文件名列表，然后用 tar 打包。其实 shell 脚本就能搞定

chenqh

2019-05-21 18:10:37 +08:00

```
import os
tmp_li = []
for name in os.listdir(path):
abs_path = os.path.join(path, name)
tmp_li.append(abs_path)
if len(tmp_li) == 10000:
do_with(tmp_li)
tmp_li = []
if tmp_li:
do_with(tmp_li)
```
大概这个样子

cxyfreedom

2019-05-21 18:31:13 +08:00

有序列的话，不就是切片然后批量压缩吗？要快的话用多进程。shell 或者其他语言实现都是一样的

thedrwu

2019-05-21 20:55:41 +08:00 via Android

几行 shell 就解决了：

a=`ls *.jpg` ; while [ -n "$a" ]; do zip `echo -e "$a" | head -1`.zip `echo -e "$a" | head -1000` ; a=`echo -e "$a" | sed -n '1001,$p' ` ; done

选择合适的工具做合适的事。非要 Python 的话，os.system("""…""") 最合适了（逃…

mahonejolla

2019-05-21 20:59:59 +08:00

@thedrwu #4 哭泣，是时候在学一发 shell 了？逃…<<<------轮子哥语气。