压缩所有所需文件内容的 ZipFile 脚本

Question

所以我正在处理一个脚本，该脚本需要将所有文件压缩到一个共享相同名称的文件夹中。因此，例如，文件夹结构如下所示...

001.flt
001.hdr
001.prj
002.flt
002.hdr
002.prj

。 . .

700.flt
700.hdr
700.prj

为了压缩文件，我有一个可以处理单个文件但无法识别的脚本 ["*.flt", "*.hdr", "*.prj"]

是否有解决方法可以让脚本根据名称识别文件名并按名称对它们进行分组？我希望每个单独的 zip 文件都包含文件内容，但将其压缩为 001.zip, 002.zip.....

表示 zip 文件包含不同的文件扩展名

001.zip( 001.hdr, 001.prj, 001.flt )

'''

import zipfile, sys, os, glob 


inDir = r"\test\DEM"
outDir = r"\test\DEM_out"  
filetype = "*.flt"

def zipfiletypeInDir(inDir, outDir):  
    # Check that input directory exists  
    if not os.path.exists(inDir):  
        print ("Input directory %s does not exist!" % inDir) 
        return False  

    print ("Zipping filetype(s) in folder %s to output folder %s" % (inDir, outDir))

    # Loop through "filetype" in input directory, glob will match pathnames from 
    for inShp in glob.glob(os.path.join(inDir, filetype)):  
        # Build the filename of the output zip file  
        outZip = os.path.join(outDir, os.path.splitext(os.path.basename(inShp))[0] + ".zip")  

        # Zip the "filetype"  
        zipfiletype(inShp, outZip)  
    return True  

def zipfiletype(infiletype, newZipFN):  
    print ('Starting to Zip '+(infiletype)+' to '+(newZipFN))

    # Delete output zipfile if it already exists  
    if (os.path.exists(newZipFN)):  
        print ('Deleting'+ newZipFN)
        os.remove(newZipFN)  

    # Output zipfile still exists, exit  
    if (os.path.exists(newZipFN)):  
        print ('Unable to Delete'+newZipFN)
        return False  

    # Open zip file object
    zipobj = zipfile.ZipFile(newZipFN,'w')  

    # Loop through "filetype" components  
    for infile in glob.glob( infiletype.lower().replace(filetype,"*.flt")):  
        # Skip .zip file extension  
        if os.path.splitext(infile)[1].lower() != ".zip":  
            print ("Zipping %s" % (infile)) 
            # Zip the "filetype" component  
            zipobj.write(infile,os.path.basename(infile),zipfile.ZIP_DEFLATED)  


    zipobj.close()  
    return True  


if __name__=="__main__":  


    zipfiletypeInDir(inDir, outDir)  
    print ("done!")

Answer 1

如果我提供的可能重复项没有回答您的问题....

一种方法是遍历所有文件名并制作一个字典，将所有同名文件分组。

In [54]: import collections, os, zipfile
In [55]: zips = collections.defaultdict(list)
In [55]: 
In [56]: for f in os.listdir():
    ...:     name, ext = os.path.splitext(f)
    ...:     zips[name].append(f)

然后遍历字典；为每个密钥创建一个新的 zip 文件并将每个密钥的文件添加到其中。

In [57]: outdir = r'zips'

In [58]: for k,v in zips.items():
    ...:     zname = k+'.zip'
    ...:     fpath = os.path.join(outdir,zname)
    ...:     #print(fpath)
    ...:     with zipfile.ZipFile(fpath, 'w') as z:
    ...:         for name in v:
    ...:             z.write(name)

Answer 2

我找到了我要找的东西，这个脚本识别文件的名称并使用迭代器根据名称对它们进行分组。

#group files into separate zipfolders from single directory based from
#individual file names 

import fnmatch, os, glob, zipfile

#edit data folders for in and out variables
path = r"D:\Users\in_path"
out_path = D"C:\Users\out_path"   

#create variables used in iterator
obj = os.listdir(path)
my_iterator = obj.__iter__()
##
#iterate each file name as '%s.*' 
for obj in my_iterator:

    #define name of file for rest of iterator to preform
    name = os.path.splitext(obj)[0]
    print (name)

    #create a zip folder to store data that is being compressed
    zip_path = os.path.join(out_path, name + '.zip')

    #create variable 'zip' that directs the data into the compressed folder
    zip = zipfile.ZipFile(zip_path, 'w', zipfile.ZIP_DEFLATED)
    os.chdir(path)

    #files are written to the folder with glob.glob
    for files in glob.glob('%s.*' %name):
        zip.write(os.path.join(path,files), files)

    #print each iteration of files being written
    print ('All files written to %s' %zip_path)
    zip.close()

压缩所有所需文件内容的 ZipFile 脚本

ZipFile script that zips all desired file contents

python

python-zipfile