blocks|key|585660|text|Python只是将您给它的文件名传递给操作系统，操作系统将打开它。如果你的操作系统支持像main/2091/data.txt这样的相对路径(提示:它支持)，那么它将工作得很好。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|585661|你可能会发现，回答这样的问题最简单的方法就是尝试一下，看看会发生什么。|585662|entityMap^0|18|I|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

Python just passes the filename you give it to the operating system, which opens it. If your operating system supports relative paths like <code>main/2091/data.txt</code> (hint: it does), then that will work fine.

You may find that the easiest way to answer a question like this is to try it and see what happens.

blocks|key|801737|text|在这种情况下，你需要注意你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，您不能只使用相对路径本身。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801738|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__来帮助您解决此问题。__file__是您正在运行的脚本所在位置的完整路径。|offset|length|style|CODE|801739|所以你可以摆弄这样的东西：|801740|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)|code-block|syntax|javascript|801741|entityMap^0|0|T|8|1B|8|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|X|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

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 <code>__file__</code> to help you out here. <code>__file__</code> is the full path to where the script you are running is located.

So you can fiddle with something like this:

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

blocks|key|801821|text|不确定这是否适用于所有地方。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801822|我在ubuntu中使用ipython。|801823|如果要读取当前文件夹子目录中的文件：|801824|/current-folder/sub-directory/data.csv|code-block|syntax|javascript|801825|您的脚本在当前文件夹中，只需尝试以下命令：|801826|import+pandas+as+pd
path+=+'./sub-directory/data.csv'
pd.read_csv(path)|801827|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Not sure if this work everywhere.

I'm using ipython in ubuntu.

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

<pre><code>/current-folder/sub-directory/data.csv
</code></pre>

your script is in current-folder
simply try this:

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

blocks|key|801912|text|我创建了一个帐户，只是为了澄清我在Russ最初的回复中发现的一个差异。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801913|作为参考，他最初的回答是：|801914|import+os
script_dir+=+os.path.dirname(__file__)
rel_path+=+"2091/data.txt"
abs_file_path+=+os.path.join(script_dir,+rel_path)|code-block|syntax|javascript|801915|这是一个很好的答案，因为它试图动态地创建到所需文件的绝对系统路径。|801916|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__)。但是，os.path.abspath会返回当前脚本(即/path/to/dir/foobar.py)的绝对路径|offset|length|style|CODE|801917|要使用此方法(以及我最终是如何使其工作的)，您必须从路径的末尾删除脚本名称：|801918|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)|801919|生成的abs_file_path+(在本例中)变为：/path/to/dir/2091/data.txt|801920|entityMap^0|0|0|0|0|H|8|1E|P|27|F|2V|M|0|0|0|Q|Q|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|12|8|@$M|13|N|14|O|P]|$M|15|N|16|O|P]|$M|17|N|18|O|P]|$M|19|N|1A|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|1C|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1D|8|@$M|1E|N|1F|O|P]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|X|$]]

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:

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

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

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

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

<pre><code>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)
</code></pre>

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

blocks|key|801938|text|下面的代码运行良好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801939|import+os


def+readFile(filename):
++++filehandle+=+open(filename)
++++print+filehandle.read()
++++filehandle.close()



fileDir+=+os.path.dirname(os.path.realpath('__file__'))
print+fileDir

#For+accessing+the+file+in+the+same+folder
filename+=+"same.txt"
readFile(filename)

#For+accessing+the+file+in+a+folder+contained+in+the+current+folder
filename+=+os.path.join(fileDir,+'Folder1.1/same.txt')
readFile(filename)

#For+accessing+the+file+in+the+parent+folder+of+the+current+folder
filename+=+os.path.join(fileDir,+'../same.txt')
readFile(filename)

#For+accessing+the+file+inside+a+sibling+folder.
filename+=+os.path.join(fileDir,+'../Folder2/same.txt')
filename+=+os.path.abspath(os.path.realpath(filename))
print+filename
readFile(filename)|code-block|syntax|javascript|801940|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This code works fine:

<pre><code>import os


def readFile(filename):
 filehandle = open(filename)
 print filehandle.read()
 filehandle.close()



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

#For accessing the file in the same folder
filename = "same.txt"
readFile(filename)

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

#For accessing the file in the parent folder of the current folder
filename = os.path.join(fileDir, '../same.txt')
readFile(filename)

#For accessing the file inside a sibling folder.
filename = os.path.join(fileDir, '../Folder2/same.txt')
filename = os.path.abspath(os.path.realpath(filename))
print filename
readFile(filename)
</code></pre>

blocks|key|585994|text|如果文件在您的父文件夹中，例如。follower.txt，您可以简单地使用open('../follower.txt',+'r').read()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|585995|entityMap^0|11|Z|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

If the file is in your parent folder, eg. follower.txt, you can simply use <code>open('../follower.txt', 'r').read()</code>

blocks|key|802021|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|802022|from+pathlib+import+Path

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

f+=+open(file_to_open)

print(f.read())|code-block|syntax|javascript|802023|Python3.4引入了一个新的标准库，用于处理文件和路径，称为pathlib。这对我很有效！|802024|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Try this:

<pre><code>from pathlib import Path

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

f = open(file_to_open)

print(f.read())
</code></pre>

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

blocks|key|802080|text|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Python3的文件。所以我加了一句。之前/和一切都运行正常：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|802081|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')|code-block|syntax|javascript|802082|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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:

<pre><code>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')
</code></pre>

blocks|key|586103|text|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.")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|586104|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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.")
</code></pre>

blocks|key|586122|text|获取父文件夹的路径，然后对相关文件执行os.join操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|586123|#+get+parent+folder+with+`os.path`
import+os.path
BASE_DIR+=+os.path.dirname(os.path.abspath(__file__))
file+=+"config.yaml"
final_path+=+os.path.join(BASE_DIR,+file)

#+get+parent+folder+with+`pathlib`'s+Path
from+pathlib+import+Path
BASE_DIR+=+Path(__file__).absolute().parent
file+=+"config.yaml"
final_path+=+BASE_DIR+/+file|code-block|syntax|javascript|586124|entityMap^0|J|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Get the path of the parent folder, then <code>os.join</code> your relative files to the end.
<pre class="lang-py prettyprint-override"><code># get parent folder with `os.path`
import os.path
BASE_DIR = os.path.dirname(os.path.abspath(__file__))
file = &quot;config.yaml&quot;
final_path = os.path.join(BASE_DIR, file)

# get parent folder with `pathlib`'s Path
from pathlib import Path
BASE_DIR = Path(__file__).absolute().parent
file = &quot;config.yaml&quot;
final_path = BASE_DIR / file
</code></pre>

blocks|key|586151|text|在Python3.4+(PEP+428)中引入了pathlib，允许您以面向对象的方式处理文件：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|586152|from+pathlib+import+Path

working_directory+=+Path(os.getcwd())
path+=+working_directory+/+"2091"+/+"sample.txt"
with+path.open('r%2B')+as+fp:
++++#+do+magic|code-block|syntax|javascript|586153|with关键字还将确保您的资源正确关闭，即使您遇到一些错误(如未处理的Exception、sigint或类似的)|586154|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0428/|1|https://docs.python.org/3/library/pathlib.html|2|https://www.python.org/dev/peps/pep-0343/^0|O|7|C|7|0|O|7|1|0|0|0|4|Z|9|0|4|2|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]|$9|14|A|15|1|16]]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|M|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]|W|$5|P|Q|R|E|$S|X]]]]

In Python 3.4 (<a href="https://www.python.org/dev/peps/pep-0428/" rel="nofollow noreferrer">PEP 428</a>) the <a href="https://docs.python.org/3/library/pathlib.html" rel="nofollow noreferrer"><code>pathlib</code></a> was introduced, allowing you to work with files in an object oriented fashion:
<pre><code>from pathlib import Path

working_directory = Path(os.getcwd())
path = working_directory / &quot;2091&quot; / &quot;sample.txt&quot;
with path.open('r+') as fp:
 # do magic
</code></pre>
The <a href="https://www.python.org/dev/peps/pep-0343/" rel="nofollow noreferrer"><code>with</code></a> keyword will also ensure that your resources get closed properly, even if you get something goes wrong (like an unhandled <code>Exception</code>, sigint or similar)

Suppose my python code is executed a directory called <code>main</code> and the application needs to access <code>main/2091/data.txt</code>.
how should I use <code>open(location)</code>? what should the parameter <code>location</code> be?
Edit:
I found that below simple code will work.. does it have any disadvantages?
<pre><code>file = &quot;\2091\sample.txt&quot;
path = os.getcwd()+file
fp = open(path, 'r+');
</code></pre>

Open file in a relative location in Python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设我的python代码是在一个名为main的目录中执行的，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？编辑：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.ge...

问在Python中的相对位置打开文件
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