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Python：如何创建连续的文件名？

发布于 2024-08-24 10:45:16 字数 101 浏览 1 评论 0原文

我希望我的程序能够以顺序格式写入文件，即：file1.txt、file2.txt、file3.txt。它仅意味着在执行代码时写入单个文件。它不能覆盖任何现有文件，并且必须创建它。我很困惑。

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却一份温柔 2024-08-31 10:45:16

两个选择：

计数器文件。
检查目录。

计数器文件。

with open("thecounter.data","r") as counter:
    count= int( counter.read() )

count += 1

每次创建新文件时，您还会使用以下内容重写计数器文件
适当的数量。非常非常快。不过，理论上是可以的
让两者不同步。如果发生碰撞事故。

您还可以通过将计数器文件制作为一小部分来使其稍微智能一些
Python 代码。

settings= {}
execfile( "thecounter.py", settings )
count = settings['count']

然后，当您更新文件时，您将编写一小段 Python 代码：count = someNumber。您可以向该文件添加注释和其他标记以简化簿记。

检查目录。

import os
def numbers( path ):
    for filename in os.listdir(path):
        name, _ = os.path.splitext()
        yield int(name[4:])
count = max( numbers( '/path/to/files' ) )

count += 1

慢点。从来没有同步问题。

Two choices:

Counter File.
Check the directory.

Counter File.

with open("thecounter.data","r") as counter:
    count= int( counter.read() )

count += 1

Each time you create a new file, you also rewrite the counter file with the
appropriate number. Very, very fast. However, it's theoretically possible to
get the two out of synch. in the event of a crash.

You can also make the counter file slightly smarter by making it a small piece of
Python code.

settings= {}
execfile( "thecounter.py", settings )
count = settings['count']

Then, when you update the file, you write a little piece of Python code: count = someNumber. You can add comments and other markers to this file to simplify your bookkeeping.

Check the directory.

import os
def numbers( path ):
    for filename in os.listdir(path):
        name, _ = os.path.splitext()
        yield int(name[4:])
count = max( numbers( '/path/to/files' ) )

count += 1

Slower. Never has a synchronization problem.

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_蜘蛛 2024-08-31 10:45:16

或者您可以附加当前系统时间来制作唯一的文件名......

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情魔剑神 2024-08-31 10:45:16

这是我实现此方法的方式：

import os
import glob
import re

#we need natural sort to avoid having the list sorted as such:
#['./folder1.txt', './folder10.txt', './folder2.txt', './folder9.txt']
def sorted_nicely(strings):
    "Sort strings the way humans are said to expect."
    return sorted(strings, key=natural_sort_key)

def natural_sort_key(key):
    import re
    return [int(t) if t.isdigit() else t for t in re.split(r'(\d+)', key)]

#check if folder.txt exists
filename = "folder.txt" #default file name

#if it does find the last count
if(os.path.exists(filename)):
        result = sorted_nicely( glob.glob("./folder[0-9]*.txt"))
        if(len(result)==0):
                filename="folder1.txt"
        else:
                last_result = result[-1]
                number = re.search( "folder([0-9]*).txt",last_result).group(1)
                filename="folder%i.txt"%+(int(number)+1)

感谢 Darius Bacon 的自然排序函数（请参阅他的答案：https://stackoverflow.com/a /341730)

抱歉，如果上面的代码很糟糕

Here is the way I implemented this:

import os
import glob
import re

#we need natural sort to avoid having the list sorted as such:
#['./folder1.txt', './folder10.txt', './folder2.txt', './folder9.txt']
def sorted_nicely(strings):
    "Sort strings the way humans are said to expect."
    return sorted(strings, key=natural_sort_key)

def natural_sort_key(key):
    import re
    return [int(t) if t.isdigit() else t for t in re.split(r'(\d+)', key)]

#check if folder.txt exists
filename = "folder.txt" #default file name

#if it does find the last count
if(os.path.exists(filename)):
        result = sorted_nicely( glob.glob("./folder[0-9]*.txt"))
        if(len(result)==0):
                filename="folder1.txt"
        else:
                last_result = result[-1]
                number = re.search( "folder([0-9]*).txt",last_result).group(1)
                filename="folder%i.txt"%+(int(number)+1)

Thanks to Darius Bacon for the natural sort functions(see his answer here: https://stackoverflow.com/a/341730)

Sorry if the above code is fugly

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