在 Python 中的相对位置打开文件

发布于 2024-11-30 22:34:15 字数 296 浏览 8 评论 0原文

假设我的 python 代码在名为 main 的目录中执行，并且应用程序需要访问 main/2091/data.txt。

我应该如何使用open(location)？参数location 应该是什么？

我发现下面的简单代码可以工作..它有什么缺点吗？

file = "\2091\sample.txt"
path = os.getcwd()+file
fp = open(path, 'r+');

原文

Suppose my python code is executed a directory called main and the application needs to access main/2091/data.txt.

how should I use open(location)? what should the parameter location be?

I found that below simple code will work.. does it have any disadvantages?

file = "\2091\sample.txt"
path = os.getcwd()+file
fp = open(path, 'r+');

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

可爱暴击 2024-12-07 22:34:15

对于这种类型的事情，您需要小心您的实际工作目录是什么。例如，您可能无法从文件所在的目录运行脚本。在这种情况下，您不能仅使用相对路径本身。

如果您确定所需的文件位于脚本实际所在的子目录中，则可以使用 __file__ 来帮助您。 __file__ 是您正在运行的脚本所在位置的完整路径。

所以你可以摆弄这样的事情：

import os
script_dir = os.path.dirname(__file__) #<-- absolute dir the script is in
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

With this type of thing you need to be careful what your actual working directory is. For example, you may not run the script from the directory the file is in. In this case, you can't just use a relative path by itself.

If you are sure the file you want is in a subdirectory beneath where the script is actually located, you can use __file__ to help you out here. __file__ is the full path to where the script you are running is located.

So you can fiddle with something like this:

import os
script_dir = os.path.dirname(__file__) #<-- absolute dir the script is in
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

回复收藏 0 原文

送舟行 2024-12-07 22:34:15

这段代码工作正常：

import os

def read_file(file_name):
    file_handle = open(file_name)
    print file_handle.read()
    file_handle.close()

file_dir = os.path.dirname(os.path.realpath('__file__'))
print file_dir

#For accessing the file in the same folder
file_name = "same.txt"
read_file(file_name)

#For accessing the file in a folder contained in the current folder
file_name = os.path.join(file_dir, 'Folder1.1/same.txt')
read_file(file_name)

#For accessing the file in the parent folder of the current folder
file_name = os.path.join(file_dir, '../same.txt')
read_file(file_name)

#For accessing the file inside a sibling folder.
file_name = os.path.join(file_dir, '../Folder2/same.txt')
file_name = os.path.abspath(os.path.realpath(file_name))
print file_name
read_file(file_name)

This code works fine:

import os

def read_file(file_name):
    file_handle = open(file_name)
    print file_handle.read()
    file_handle.close()

file_dir = os.path.dirname(os.path.realpath('__file__'))
print file_dir

#For accessing the file in the same folder
file_name = "same.txt"
read_file(file_name)

#For accessing the file in a folder contained in the current folder
file_name = os.path.join(file_dir, 'Folder1.1/same.txt')
read_file(file_name)

#For accessing the file in the parent folder of the current folder
file_name = os.path.join(file_dir, '../same.txt')
read_file(file_name)

#For accessing the file inside a sibling folder.
file_name = os.path.join(file_dir, '../Folder2/same.txt')
file_name = os.path.abspath(os.path.realpath(file_name))
print file_name
read_file(file_name)

回复收藏 0 原文

深爱不及久伴 2024-12-07 22:34:15

我创建了一个帐户，只是为了澄清我认为在 Russ 最初的回复中发现的差异。

作为参考，他最初的答案是：

import os
script_dir = os.path.dirname(__file__)
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

这是一个很好的答案，因为它试图动态创建所需文件的绝对系统路径。

Cory Mawhorter 注意到 __file__ 是一个相对路径（在我的系统上也是如此），并建议使用 os.path.abspath(__file__) 。然而，os.path.abspath返回当前脚本的绝对路径（即/path/to/dir/foobar.py）

要使用此方法（以及如何我最终让它工作了）你必须从路径末尾删除脚本名称：

import os
script_path = os.path.abspath(__file__) # i.e. /path/to/dir/foobar.py
script_dir = os.path.split(script_path)[0] #i.e. /path/to/dir/
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

生成的abs_file_path（在本例中）变为：/path/to/dir/2091/data.txt

I created an account just so I could clarify a discrepancy I think I found in Russ's original response.

For reference, his original answer was:

import os
script_dir = os.path.dirname(__file__)
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

This is a great answer because it is trying to dynamically creates an absolute system path to the desired file.

Cory Mawhorter noticed that __file__ is a relative path (it is as well on my system) and suggested using os.path.abspath(__file__). os.path.abspath, however, returns the absolute path of your current script (i.e. /path/to/dir/foobar.py)

To use this method (and how I eventually got it working) you have to remove the script name from the end of the path:

import os
script_path = os.path.abspath(__file__) # i.e. /path/to/dir/foobar.py
script_dir = os.path.split(script_path)[0] #i.e. /path/to/dir/
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

The resulting abs_file_path (in this example) becomes: /path/to/dir/2091/data.txt

回复收藏 0 原文

愿得七秒忆 2024-12-07 22:34:15

这取决于您使用的操作系统。如果您想要一个与 Windows 和 *nix 兼容的解决方案，例如：

from os import path

file_path = path.relpath("2091/data.txt")
with open(file_path) as f:
    <do stuff>

应该可以正常工作。

path 模块能够为其运行的任何操作系统格式化路径。另外，只要您拥有正确的权限，Python 就可以很好地处理相对路径。

编辑：

正如kindall在评论中提到的，python无论如何都可以在unix风格和windows风格路径之间转换，所以更简单的代码也可以工作：

with open("2091/data/txt") as f:
    <do stuff>

话虽这么说，path 模块仍然有一些有用的功能。

It depends on what operating system you're using. If you want a solution that is compatible with both Windows and *nix something like:

from os import path

file_path = path.relpath("2091/data.txt")
with open(file_path) as f:
    <do stuff>

should work fine.

The path module is able to format a path for whatever operating system it's running on. Also, python handles relative paths just fine, so long as you have correct permissions.

Edit:

As mentioned by kindall in the comments, python can convert between unix-style and windows-style paths anyway, so even simpler code will work:

with open("2091/data/txt") as f:
    <do stuff>

That being said, the path module still has some useful functions.

回复收藏 0 原文

桃扇骨 2024-12-07 22:34:15

我花了很多时间来发现为什么我的代码找不到在 Windows 系统上运行 Python 3 的文件。所以我补充道。之前/并且一切正常：

import os

script_dir = os.path.dirname(__file__)
file_path = os.path.join(script_dir, './output03.txt')
print(file_path)
fptr = open(file_path, 'w')

I spend a lot time to discover why my code could not find my file running Python 3 on the Windows system. So I added . before / and everything worked fine:

import os

script_dir = os.path.dirname(__file__)
file_path = os.path.join(script_dir, './output03.txt')
print(file_path)
fptr = open(file_path, 'w')

回复收藏 0 原文

似最初 2024-12-07 22:34:15

试试这个：

from pathlib import Path

data_folder = Path("/relative/path")
file_to_open = data_folder / "file.pdf"

f = open(file_to_open)

print(f.read())

Python 3.4 引入了一个新的标准库，用于处理文件和路径，称为 pathlib。这对我有用！

Try this:

from pathlib import Path

data_folder = Path("/relative/path")
file_to_open = data_folder / "file.pdf"

f = open(file_to_open)

print(f.read())

Python 3.4 introduced a new standard library for dealing with files and paths called pathlib. It works for me!

回复收藏 0 原文

GRAY°灰色天空 2024-12-07 22:34:15

代码：

import os
script_path = os.path.abspath(__file__) 
path_list = script_path.split(os.sep)
script_directory = path_list[0:len(path_list)-1]
rel_path = "main/2091/data.txt"
path = "/".join(script_directory) + "/" + rel_path

说明：

导入库：

import os

使用__file__获取当前脚本的路径：

script_path = os.path.abspath(__file__)

将脚本路径分成多项：

path_list = script_path.split(os.sep)

删除列表中的最后一项（实际的脚本文件） ):

script_directory = path_list[0:len(path_list)-1]

添加相对文件的路径：

rel_path = "main/2091/data.txt

加入列表项，并添加相对路径的文件：

path = "/".join(script_directory) + "/" + rel_path

现在您可以对文件执行任何您想要的操作，例如：

file = open(path)

Code:

import os
script_path = os.path.abspath(__file__) 
path_list = script_path.split(os.sep)
script_directory = path_list[0:len(path_list)-1]
rel_path = "main/2091/data.txt"
path = "/".join(script_directory) + "/" + rel_path

Explanation:

Import library:

import os

Use __file__ to attain the current script's path:

script_path = os.path.abspath(__file__)

Separates the script path into multiple items:

path_list = script_path.split(os.sep)

Remove the last item in the list (the actual script file):

script_directory = path_list[0:len(path_list)-1]

Add the relative file's path:

rel_path = "main/2091/data.txt

Join the list items, and addition the relative path's file:

path = "/".join(script_directory) + "/" + rel_path

Now you are set to do whatever you want with the file, such as, for example:

file = open(path)

回复收藏 0 原文

平安喜乐 2024-12-07 22:34:15

import os
def file_path(relative_path):
    dir = os.path.dirname(os.path.abspath(__file__))
    split_path = relative_path.split("/")
    new_path = os.path.join(dir, *split_path)
    return new_path

with open(file_path("2091/data.txt"), "w") as f:
    f.write("Powerful you have become.")

import os
def file_path(relative_path):
    dir = os.path.dirname(os.path.abspath(__file__))
    split_path = relative_path.split("/")
    new_path = os.path.join(dir, *split_path)
    return new_path

with open(file_path("2091/data.txt"), "w") as f:
    f.write("Powerful you have become.")

回复收藏 0 原文

肥爪爪 2024-12-07 22:34:15

获取父文件夹的路径，然后os.join您的相关文件到最后。

# get parent folder with `os.path`
import os.path

BASE_DIR = os.path.dirname(os.path.abspath(__file__))

# now use BASE_DIR to get a file relative to the current script
os.path.join(BASE_DIR, "config.yaml")

与 pathlib 相同：

# get parent folder with `pathlib`'s Path
from pathlib import Path

BASE_DIR = Path(__file__).absolute().parent

# now use BASE_DIR to get a file relative to the current script
BASE_DIR / "config.yaml"

Get the path of the parent folder, then os.join your relative files to the end.

# get parent folder with `os.path`
import os.path

BASE_DIR = os.path.dirname(os.path.abspath(__file__))

# now use BASE_DIR to get a file relative to the current script
os.path.join(BASE_DIR, "config.yaml")

The same thing with pathlib:

# get parent folder with `pathlib`'s Path
from pathlib import Path

BASE_DIR = Path(__file__).absolute().parent

# now use BASE_DIR to get a file relative to the current script
BASE_DIR / "config.yaml"

回复收藏 0 原文

就此别过 2024-12-07 22:34:15

如果该文件位于您的父文件夹中，例如。 follower.txt，您只需使用 open('../follower.txt', 'r').read()

回复收藏 0 原文

暖风昔人 2024-12-07 22:34:15

Python 只是将您提供的文件名传递给操作系统，然后操作系统打开它。如果您的操作系统支持像 main/2091/data.txt 这样的相对路径（提示：确实如此），那么就可以正常工作。

您可能会发现回答此类问题的最简单方法就是尝试一下，看看会发生什么。

回复收藏 0 原文

不气馁 2024-12-07 22:34:15

不确定这是否在所有地方都有效。

我在 ubuntu 中使用 ipython。

如果您想读取当前文件夹子目录中的文件：

/current-folder/sub-directory/data.csv

您的脚本位于当前文件夹中
只需尝试这个：

import pandas as pd
path = './sub-directory/data.csv'
pd.read_csv(path)

Not sure if this work everywhere.

I'm using ipython in ubuntu.

If you want to read file in current folder's sub-directory:

/current-folder/sub-directory/data.csv

your script is in current-folder
simply try this:

import pandas as pd
path = './sub-directory/data.csv'
pd.read_csv(path)

回复收藏 0 原文

萧瑟寒风 2024-12-07 22:34:15

在 Python 3.4 (PEP 428) 中，pathlib 被引入，允许您以面向对象的方式处理文件时尚：

from pathlib import Path

working_directory = Path(os.getcwd())
path = working_directory / "2091" / "sample.txt"
with path.open('r+') as fp:
    # do magic

with 关键字也将确保即使出现问题（例如未处理的Exception、sigint 或类似情况），您的资源也会正确关闭

In Python 3.4 (PEP 428) the pathlib was introduced, allowing you to work with files in an object oriented fashion:

from pathlib import Path

working_directory = Path(os.getcwd())
path = working_directory / "2091" / "sample.txt"
with path.open('r+') as fp:
    # do magic

The with keyword will also ensure that your resources get closed properly, even if you get something goes wrong (like an unhandled Exception, sigint or similar)

回复收藏 0 原文

两相知 2024-12-07 22:34:15

当我还是一个初学者时，我发现这些描述有点令人生畏。一开始我会尝试
对于 Windows

f= open('C:\Users\chidu\Desktop\Skipper New\Special_Note.txt','w+')
print(f)

，这会引发语法错误。我曾经经常感到困惑。然后在谷歌上浏览了一下。找到了发生错误的原因。 为初学者编写此内容

这是因为要以 Unicode 读取路径，您只需在启动文件路径时添加 \

f= open('C:\\Users\chidu\Desktop\Skipper New\Special_Note.txt','w+')
print(f)

现在只需添加 \在启动目录之前。

When I was a beginner I found these descriptions a bit intimidating. As at first I would try
For Windows

f= open('C:\Users\chidu\Desktop\Skipper New\Special_Note.txt','w+')
print(f)

and this would raise an syntax error. I used get confused alot. Then after some surfing across google. found why the error occurred. Writing this for beginners

It's because for path to be read in Unicode you simple add a \ when starting file path

f= open('C:\\Users\chidu\Desktop\Skipper New\Special_Note.txt','w+')
print(f)

And now it works just add \ before starting the directory.

回复收藏 0 原文

~没有更多了~

关于作者

晌融

暂无简介

文章

28 人气

关注发私信

友情链接

文江博客

在 Python 中的相对位置打开文件

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

评论（14）

关于作者

相关话题

热门标签

推荐作者

櫻之舞

弥枳

m2429

寻找一个思念的角度

野却迷人

我怀念的。

友情链接

在 Python 中的相对位置打开文件

如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

评论（14）

关于作者

相关话题

热门标签

推荐作者

櫻之舞

弥枳

m2429

寻找一个思念的角度

野却迷人

我怀念的。

友情链接

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。