如何检查文件是否存在，并在python中重命名它

Question

我正在寻找一种更仿生的方式来做我的代码目前所做的事情。我肯定还有更好的办法。我想搜索直到文件名-10，如果存在，创建一个名为filename-11的文件。

如果你能帮忙那就太好了。

编辑: 9/1/14 9:46下午

import re
import os
f=open('/Users/jakerandall/Desktop/Data Collection Python/temp.cnc', 'r')
text = re.search(r"(?<!\d)\d{4,5}(?!\d)", f.read())
JobNumber = text.string[text.start():text.end()]


if os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-10.cnc" % JobNumber):
    f=open("/Users/jakerandall/Desktop/Data Collection Python/%s-11.cnc"  % JobNumber, 'w+b')
    f.close()
    print '1'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-9.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-10.cnc' % JobNumber, 'w+b')
    f.close()
    print '2'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-8.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-9.cnc' % JobNumber, 'w+b')
    f.close()
    print '3'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-7.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-8.cnc' % JobNumber, 'w+b')
    f.close()
    print '4'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-6.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-7.cnc' % JobNumber, 'w+b')
    f.close()
    print '5'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-5.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-6.cnc' % JobNumber, 'w+b')
    f.close()
    print '6'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-4.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-5.cnc' % JobNumber, 'w+b')
    f.close()
    print '7'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-3.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-4.cnc' % JobNumber, 'w+b')
    f.close()
    print '8'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-2.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-3.cnc' % JobNumber, 'w+b')
    f.close()
    print '9'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s-1.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-2.cnc' % JobNumber, 'w+b')
    f.close()
    print '10'
elif os.path.isfile("/Users/jakerandall/Desktop/Data Collection Python/%s.cnc" % JobNumber):
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s-1.cnc' % JobNumber, 'w+b')
    f.close()
    print '11'
else:
    f=open('/Users/jakerandall/Desktop/Data Collection Python/%s.cnc' % JobNumber, 'w+b')
    f.close()
    print '12'
f.close()

Answer 1

更简单点的怎么样：

import glob

file_directory = '/Users/jakerandall/Desktop/Data Collection Python/'
files = glob.glob('{}{}*.cnc'.format(file_directory, JobNumber))

现在，files将是目录中实际存在的文件名列表，并与您的模式相匹配。

您可以检查此列表的长度，然后：

1. 如果该文件为空，则创建第一个文件，即'{}.cnc'.format(JobNumber)。
2. 如果列表的长度等于11，则需要创建文件号11 (因为模式将匹配第一个文件，即没有任何-的文件，因此11的长度意味着最后一个文件是-10.cnc)。
3. 否则，您想要的文件是1-列表的长度。因此，如果列表中有5个项，这意味着最后一个文件是-4.cnc (因为模式也将匹配第一个文件)。

您仍然需要查看是否可以打开它们，因为运行Python脚本的用户可能没有足够的权限。

下面是一个将所有这些整合在一起的例子：

import glob

file_directory = '/Users/jakerandall/Desktop/Data Collection Python/'
files = glob.glob('{}{}*.cnc'.format(file_directory, JobNumber))

# Start by assuming there are no files:
filename = '{}.cnc'.format(JobNumber)
if len(files) <= 11:
   # If there are less than 11 files, we need
   # to use the existing file, and overwrite it
   # If there are 4 files, in the directory, our
   # list will have a length of 5:
   # The original file, and then four files from -1, to -4
   # In this case, we want to use file 4, which is 1 less than
   # the length of the list:
   filename = '{}-{}.cnc'.format(JobNumber, len(files)-1)
else:
   # If we reach this point, it means
   # there were more than 10 files that match the
   # pattern. We want to use the next file,
   # which is next number higher, which is also the length
   # of the list, since it will include the first file.
   # So if the last file is -20, the list will have 20 files (from -1, to -20)
   # plus the original file, which has no - in the filename, giving
   # a length of 21, which also happens to be the number of the file
   # we want to create :)
   filename = '{}-{}.cnc'.format(JobNumber, len(files))

# Now, try to create the file
try:
   f = open(filename, 'w+b')
except IOError:
   print('Cannot create {}, check permissions?'.format(filename)) 

Answer 2

我真的写过这样的东西！不过我是从记忆开始工作的。这作为一个单独的模块非常有用，因为以这种方式备份文件是相当常见的。

# /backup_dash_one.py

import os, glob, re

def backup(full_path, num_backups=None):
    """Usage: backup('example/pathname.ext', [num_backups])
    returns: example/pathname-1.ext, advances all backups by 1

    Given example/pathname.ext, creates backups named
    example/pathname-1.ext, -2.ext, -3.ext until there are
    as many backups as num_backups, purging those older."""

    head, tail = os.path.split(full_path)
    tailname, tailext = os.path.splitext(tail)

    def find_backup_num(path):
        return int(re.search(r"-(\d+)\.[^.\\/]*", path).group(1))

    paths = sorted(glob.glob(os.path.join(head,tailname)+"-*"+tailext),
                   key=find_backup_num)
    for path in reversed(paths[:]):
        head_tail, backup_num, ext, _* = re.split(r"-(\d+)(\.[^\\./]*)$", path)
        new_path = head_tail + "-" + str(int(backup_num)+1) + ext

        with open(path) as infile, open(new_path,'w') as outfile):
            for line in infile:
                outfile.write(line)
        if new_path not in paths:
            paths.append(new_path)

    while num_backups and len(paths) > num_backups:
        os.remove(paths[-1])
        paths.pop()

就我个人而言，如果我有足够的时间来做这件事，我会做更多的研究，做一些如下的事情：

import glob, os

    class BackupFile(object):
    def __init__(self, path, mode="w", num_backups=None):
        self.num_backups = num_backups
        path_filename, ext = os.path.splitext(path)
        self.backups = glob.glob(path_filename+"-*"+ext)
        self.backups.sort(key=self.find_backup_num)
        self.backup()
        self.purge()
        with open(path_filename+"-1"+ext, 'w') as backup,\
             open(path, 'r') as original:
            for line in original:
                backup.write(line)
        self.f = open(path, mode)

    def find_backup_num(self,filename):
        return int(os.path.splitext(filename)[0].split('-')[-1])
    def backup(self):
        for path in reversed(self.backups[:]):
            head_num,ext = os.path.splitext(path)
            *head,num = head_num.split('-')
            new_path = "{}-{}{}".format('-'.join(head),
                            int(num)+1,
                            ext)
            with open(new_path, 'w') as newfile, \
                 open(path, 'r') as oldfile:
                for line in oldfile:
                    newfile.write(line)
            if new_path not in self.backups:
                self.backups.append(new_path)
    def purge(self):
        while self.num_backups and len(self.backups) > self.num_backups:
            os.remove(self.backups.pop())
    def __enter__(self):
        return self.f
    def __exit__(self, exc_type, exc_value, exc_traceback):
        self.f.close()

所以你可以这么做：

with BackupFile("path/to/file/that/needs/backups.txt", 'r+', num_backups=12) as f:
    make_change(f)
# ta-da it's backed up!

然而，我还没有多少机会去测试它，所以我猜想它有什么问题:)