python：解压缩错误“没有这样的文件或目录”

Question

I have zip file downloaded from website. I wanted to make script that rename zip file and before unzip, it checks how many files are in it and unzip it.

The problem is that zip file is in the directory but it keep giving me error that 'FileNotFoundError: [Errno 2] No such file or directory: 'filename.zip'' I assumed that It might be caused by file name because I use ubuntu and when I downloaded the file, the name was broken because it was not English. so I changed it into numbers (ex:20176) but still getting this error.

my script path means absolute path.

    data_type = '{}{}'.format('201706', '.zip')
    filename = [i for i in os.listdir('user/directory')]
    filename.sort(key=lambda ctime: ctime[0])
    downloaded = str(filename[0])

    old = os.path.join('user/directory', downloaded)
    new = os.path.join('user/directory', data_type)
    os.rename(old, new)

    zip = ZipFile(data_type)
    archived_files = zip.namelist()
    amount = len(archived_files)

Answer 1

Let's suppose the first filename in the sorted list is myfile.txt . Your code

old = os.path.join('user/directory', downloaded)
new = os.path.join('user/directory', data_type)
os.rename(old, new)

renames the first file in the directory listing, user/directory/myfile.txt (not, due to the considerations above, the oldest one) to user/directory/201706.zip . The next statement then tries to open 2010706.zip , which of course doesn't exist. It should work if you try

zip = ZipFile(new)

Unfortunately there's no guarantee that the file actually will be a zipfile, so the operation may fail.

Some other points to consider, perhaps in other questions:

I suspect you are misunderstanding the sorting functions: although you appear to want to sort the list of filenames on creation time, just calling a lambda's parameter ctime doesn't mean that Python will understand your needs. If only programming languages had a DWIM ("do what I mean") mode, life would be so much easier!

The key argument to sort is a function that the sort function calls once for each value to be sorted, expecting it to return a "sort key" (that is, a value that represents its place in the required ordering). Suppose I take your lambda and apply it to a filename:

In [1]: ruth_lambda = lambda ctime: ctime[0]

In [2]: ruth_lambda("MY_FILENAME.TXT")
Out[2]: 'M'

You can see that you are sorting on the first character of the filename, and I doubt that's what you really want. But we can go into that later.

A quick note on formatting: it would have been simpler to write

data_type = '{}{}'.format(zipfile_name, '.zip')

as

data_type = '{}.zip'.format(zipfile_name)

Finally, since os.listdir returns a list, instead of

filename = [i for i in os.listdir('user/directory')]

it's simpler to write

filename = os.listdir('user/directory')

though the extra computation will do no harm. As a beginner you will find that as you improve, your older code starts to look really clunky - don't worry about that, it's a common experience! Just move forwards and try not to repeat old mistakes.