blocks|key|1296993|text|import+shutil

shutil.move('a.txt',+'b.kml')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1296994|这将用于重命名或移动文件。|unstyled|1296995|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>import shutil

shutil.move('a.txt', 'b.kml')
</code></pre>

This will work to rename or move a file.

blocks|key|1297001|text|使用os.rename|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1297002|import+os

os.rename('a.txt',+'b.kml')|code-block|syntax|javascript|1297003|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/os.html#os.rename^0|2|9|2|9|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Use <a href="http://docs.python.org/library/os.html#os.rename" rel="noreferrer"><code>os.rename</code></a>:

<pre><code>import os

os.rename('a.txt', 'b.kml')
</code></pre>

blocks|key|1297074|text|os.rename(old,+new)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1297075|这可以在Python文档中找到：http://docs.python.org/library/os.html|1297076|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/os.html^0|0|J|0|G|12|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@$9|T|A|U|1|V]]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

<code>os.rename(old, new)</code>

This is found in the Python docs: <a href="http://docs.python.org/library/os.html" rel="noreferrer">http://docs.python.org/library/os.html</a>

blocks|key|1082099|text|文件可能在目录中，在这种情况下请指定路径：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1082100|import+os
old_file+=+os.path.join("directory",+"a.txt")
new_file+=+os.path.join("directory",+"b.kml")
os.rename(old_file,+new_file)|code-block|syntax|javascript|1082101|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|$]]

File may be inside a directory, in that case specify the path:

<pre><code>import os
old_file = os.path.join("directory", "a.txt")
new_file = os.path.join("directory", "b.kml")
os.rename(old_file, new_file)
</code></pre>

blocks|key|1082133|text|您可以使用os.system调用终端来完成此任务：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1082134|os.system('mv+oldfile+newfile')|code-block|syntax|javascript|1082135|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|$]]

You can use os.system to invoke terminal to accomplish the task:

<pre><code>os.system('mv oldfile newfile')
</code></pre>

blocks|key|1082199|text|从Python3.4开始，我们可以使用pathlib模块来解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1082200|如果您使用的是较旧的版本，则可以使用here的后端口版本|1082201|假设您不在想要重命名的根路径中(只是为了增加一点难度)，并且必须提供完整路径，我们可以这样看：|1082202|some_path+=+'a/b/c/the_file.extension'|code-block|syntax|javascript|1082203|因此，您可以采用您的路径并从中创建一个Path对象：|style|CODE|1082204|from+pathlib+import+Path
p+=+Path(some_path)|1082205|只是为了提供一些关于这个对象的信息，我们可以从中提取一些东西。例如，如果出于某种原因，我们想通过将文件名从the_file修改为the_file_1来重命名文件，那么我们可以获得文件名部分：|1082206|name_without_extension+=+p.stem|1082207|并且仍然手持这个扩展：|1082208|ext+=+p.suffix|1082209|我们可以通过一个简单的字符串操作来执行修改：|1082210|Python+3.6及更高版本使用f字符串！|1082211|new_file_name+=+f"{name_without_extension}_1"|1082212|否则：|1082213|new_file_name+=+"{}_{}".format(name_without_extension,+1)|1082214|现在，我们可以通过在创建的path对象上调用rename方法并附加ext来完成所需的正确重命名结构，从而执行重命名：|1082215|p.rename(Path(p.parent,+new_file_name+%2B+ext))|1082216|更简短地展示它的简单性：|1082217|Python+3.6%2B：|1082218|from+pathlib+import+Path
p+=+Path(some_path)
p.rename(Path(p.parent,+f"{p.stem}_1_{p.suffix}"))|1082219|低于Python+3.6的版本改用字符串格式方法：|1082220|from+pathlib+import+Path
p+=+Path(some_path)
p.rename(Path(p.parent,+"{}_{}_{}".format(p.stem,+1,+p.suffix))|1082221|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/pathlib.html|1|https://pypi.python.org/pypi/pathlib2/^0|J|7|0|0|I|4|1|0|0|0|J|4|0|0|1H|8|1S|A|0|0|0|0|0|0|0|0|0|M|6|X|3|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1Y|8|@]|9|@$A|1Z|B|20|1|21]]|C|$]]|$1|D|3|E|5|6|7|22|8|@]|9|@$A|23|B|24|1|25]]|C|$]]|$1|F|3|G|5|6|7|26|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|27|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|28|8|@$A|29|B|2A|O|P]]|9|@]|C|$]]|$1|Q|3|R|5|J|7|2B|8|@]|9|@]|C|$K|L]]|$1|S|3|T|5|6|7|2C|8|@$A|2D|B|2E|O|P]|$A|2F|B|2G|O|P]]|9|@]|C|$]]|$1|U|3|V|5|J|7|2H|8|@]|9|@]|C|$K|L]]|$1|W|3|X|5|6|7|2I|8|@]|9|@]|C|$]]|$1|Y|3|Z|5|J|7|2J|8|@]|9|@]|C|$K|L]]|$1|10|3|11|5|6|7|2K|8|@]|9|@]|C|$]]|$1|12|3|13|5|6|7|2L|8|@]|9|@]|C|$]]|$1|14|3|15|5|J|7|2M|8|@]|9|@]|C|$K|L]]|$1|16|3|17|5|6|7|2N|8|@]|9|@]|C|$]]|$1|18|3|19|5|J|7|2O|8|@]|9|@]|C|$K|L]]|$1|1A|3|1B|5|6|7|2P|8|@$A|2Q|B|2R|O|P]|$A|2S|B|2T|O|P]]|9|@]|C|$]]|$1|1C|3|1D|5|J|7|2U|8|@]|9|@]|C|$K|L]]|$1|1E|3|1F|5|6|7|2V|8|@]|9|@]|C|$]]|$1|1G|3|1H|5|6|7|2W|8|@]|9|@]|C|$]]|$1|1I|3|1J|5|J|7|2X|8|@]|9|@]|C|$K|L]]|$1|1K|3|1L|5|6|7|2Y|8|@]|9|@]|C|$]]|$1|1M|3|1N|5|J|7|2Z|8|@]|9|@]|C|$K|L]]|$1|1O|3|-4|5|6|7|30|8|@]|9|@]|C|$]]]|1P|$1Q|$5|1R|1S|1T|C|$1U|1V]]|1W|$5|1R|1S|1T|C|$1U|1X]]]]

As of Python 3.4 one can use the <a href="https://docs.python.org/3/library/pathlib.html" rel="noreferrer">pathlib</a> module to solve this.

If you happen to be on an older version, you can use the backported version found <a href="https://pypi.python.org/pypi/pathlib2/" rel="noreferrer">here</a>

Let's assume you are not in the root path (just to add a bit of difficulty to it) you want to rename, and have to provide a full path, we can look at this: 

<pre><code>some_path = 'a/b/c/the_file.extension'
</code></pre>

So, you can take your path and create a <code>Path</code> object out of it:

<pre><code>from pathlib import Path
p = Path(some_path)
</code></pre>

Just to provide some information around this object we have now, we can extract things out of it. For example, if for whatever reason we want to rename the file by modifying the filename from <code>the_file</code> to <code>the_file_1</code>, then we can get the filename part:

<pre><code>name_without_extension = p.stem
</code></pre>

And still hold the extension in hand as well: 

<pre><code>ext = p.suffix
</code></pre>

We can perform our modification with a simple string manipulation:

Python 3.6 and greater make use of f-strings!

<pre><code>new_file_name = f"{name_without_extension}_1"
</code></pre>

Otherwise:

<pre><code>new_file_name = "{}_{}".format(name_without_extension, 1)
</code></pre>

And now we can perform our rename by calling the <code>rename</code> method on the path object we created and appending the <code>ext</code> to complete the proper rename structure we want:

<pre><code>p.rename(Path(p.parent, new_file_name + ext))
</code></pre>

More shortly to showcase its simplicity: 

Python 3.6+:

<pre><code>from pathlib import Path
p = Path(some_path)
p.rename(Path(p.parent, f"{p.stem}_1_{p.suffix}"))
</code></pre>

Versions less than Python 3.6 use the string format method instead:

<pre><code>from pathlib import Path
p = Path(some_path)
p.rename(Path(p.parent, "{}_{}_{}".format(p.stem, 1, p.suffix))
</code></pre>

blocks|key|1297293|text|import+shutil
import+os

files+=+os.listdir("./pics/")+

for+key+in+range(0,+len(files)):
+++print+files[key]
+++shutil.move("./pics/"+%2B+files[key],"./pics/img"+%2B+str(key)+%2B+".jpeg")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1297294|这个应该可以了。python+3%2B|unstyled|1297295|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>import shutil
import os

files = os.listdir("./pics/") 

for key in range(0, len(files)):
 print files[key]
 shutil.move("./pics/" + files[key],"./pics/img" + str(key) + ".jpeg")
</code></pre>

This should do it. python 3+

blocks|key|1082254|text|使用os.rename。但您必须将这两个文件的完整路径传递给函数。如果我有一个文件a.txt在我的桌面上，我会这样做，我也必须给全部重命名的文件。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1082255|os.rename('C:\\Users\\Desktop\\a.txt',+'C:\\Users\\Desktop\\b.kml')|code-block|syntax|javascript|1082256|entityMap^0|2|9|15|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Use <code>os.rename</code>. But you have to pass full path of both files to the function. If I have a file <code>a.txt</code> on my desktop so I will do and also I have to give full of renamed file too.
<pre><code>os.rename('C:\\Users\\Desktop\\a.txt', 'C:\\Users\\Desktop\\b.kml')
</code></pre>

blocks|key|1297361|text|import+os

#+Set+the+path
path+=+'a\\b\\c'++
#+save+current+working+directory
saved_cwd+=+os.getcwd()
#+change+your+cwd+to+the+directory+which+contains+files
os.chdir(path)
os.rename('a.txt',+'b.klm')
#+moving+back+to+the+directory+you+were+in+
os.chdir(saved_cwd)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1297362|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

# Set the path
path = 'a\\b\\c' 
# save current working directory
saved_cwd = os.getcwd()
# change your cwd to the directory which contains files
os.chdir(path)
os.rename('a.txt', 'b.klm')
# moving back to the directory you were in 
os.chdir(saved_cwd)
</code></pre>

blocks|key|1082291|text|import+os
import+re
from+pathlib+import+Path

for+f+in+os.listdir(training_data_dir2):
++for+file+in+os.listdir(+training_data_dir2+%2B+'/'+%2B+f):
++++oldfile=+Path(training_data_dir2+%2B+'/'+%2B+f+%2B+'/'+%2B+file)
++++newfile+=+Path(training_data_dir2+%2B+'/'+%2B+f+%2B+'/'+%2B+file[49:])
++++p=oldfile
++++p.rename(newfile)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1082292|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
import re
from pathlib import Path

for f in os.listdir(training_data_dir2):
 for file in os.listdir( training_data_dir2 + '/' + f):
 oldfile= Path(training_data_dir2 + '/' + f + '/' + file)
 newfile = Path(training_data_dir2 + '/' + f + '/' + file[49:])
 p=oldfile
 p.rename(newfile)
</code></pre>

blocks|key|1082362|text|使用Pathlib库的Path.rename而不是os.rename：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1082363|import+pathlib

original_path+=+pathlib.Path('a.txt')
new_path+=+original_path.rename('b.kml')|code-block|syntax|javascript|1082364|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/pathlib.html#pathlib.Path.rename^0|B|B|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Using the Pathlib library's <a href="https://docs.python.org/3/library/pathlib.html#pathlib.Path.rename" rel="nofollow noreferrer">Path.rename</a> instead of os.rename:
<pre class="lang-py prettyprint-override"><code>import pathlib

original_path = pathlib.Path('a.txt')
new_path = original_path.rename('b.kml')
</code></pre>

blocks|key|1297458|text|Os.chdir(r“D：\文件夹1\文件夹2”)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1297459|Os.rename(源，dst)|1297460|#src和dst应该在Folder2内部|1297461|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

os.chdir(r&quot;D:\Folder1\Folder2&quot;) 
os.rename(src,dst) 
#src and dst should be inside Folder2

blocks|key|1082412|text|这里需要注意的重要一点是，我们应该检查是否存在具有新文件名的文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1082413|假设如果b.kml文件存在，则用相同的文件名重命名其他文件会导致删除现有的b.kml。|1082414|import+os

if+not+os.path.exists('b.kml'):
++++os.rename('a.txt','b.kml')|code-block|syntax|javascript|1082415|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

One important point to note here, we should check if any files exists with the new filename.
suppose if b.kml file exists then renaming other file with the same filename leads to deletion of existing b.kml.
<pre><code>import os

if not os.path.exists('b.kml'):
 os.rename('a.txt','b.kml')
</code></pre>

blocks|key|1082449|text|从Python3.3版和更高版本开始，通常首选使用os.replace而不是os.rename，因此如果目标文件已经存在，则不会引发FileExistsError。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1082450|assert+os.path.isfile('old.txt')
assert+os.path.isfile('new.txt')

os.rename('old.txt',+'new.txt')
#+Raises+FileExistsError
os.replace('old.txt',+'new.txt')
#+Does+not+raise+exception

assert+not+os.path.isfile('old.txt')
assert+os.path.isfile('new.txt')|code-block|syntax|javascript|1082451|请参阅documentation。|1082452|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/os.html#os.replace^0|P|A|12|9|1U|F|0|0|3|D|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

As of Python version 3.3 and later, it is generally preferred to use <code>os.replace</code> instead of <code>os.rename</code> so <code>FileExistsError</code> is not raised if the destination file already exists.
<pre><code>assert os.path.isfile('old.txt')
assert os.path.isfile('new.txt')

os.rename('old.txt', 'new.txt')
# Raises FileExistsError
os.replace('old.txt', 'new.txt')
# Does not raise exception

assert not os.path.isfile('old.txt')
assert os.path.isfile('new.txt')
</code></pre>
See the <a href="https://docs.python.org/3/library/os.html#os.replace" rel="noreferrer">documentation</a>.

blocks|key|1082509|text|下面是一个仅使用pathlib而不使用os的示例，它基于字符串replace操作更改目录中所有文件的名称，而不使用字符串连接：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1082510|from+pathlib+import+Path

path+=+Path('/talend/studio/plugins/org.talend.designer.components.bigdata_7.3.1.20200214_1052\components/tMongoDB44Connection')

for+p+in+path.glob("tMongoDBConnection*"):
++++new_name+=+p.name.replace("tMongoDBConnection",+"tMongoDB44Connection")
++++new_name+=+p.parent/new_name
++++p.rename(new_name)|code-block|syntax|javascript|1082511|entityMap^0|8|7|J|2|V|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Here is an example using <code>pathlib</code> only without touching <code>os</code> which changes the names of all files in a directory, based on a string <code>replace</code> operation without using also string concatenation:
<pre><code>from pathlib import Path

path = Path('/talend/studio/plugins/org.talend.designer.components.bigdata_7.3.1.20200214_1052\components/tMongoDB44Connection')

for p in path.glob(&quot;tMongoDBConnection*&quot;):
 new_name = p.name.replace(&quot;tMongoDBConnection&quot;, &quot;tMongoDB44Connection&quot;)
 new_name = p.parent/new_name
 p.rename(new_name)
</code></pre>

I want to change <code>a.txt</code> to <code>b.kml</code>.

问如何使用Python重命名文件
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