如何从Python路径中获取不带扩展名的文件名?-Python 实用宝典

如何从Python路径中获取不带扩展名的文件名?

如何从Python路径中获取不带扩展名的文件名?

问题:如何从Python路径中获取不带扩展名的文件名?

如何从Python路径中获取不带扩展名的文件名?

How to get the filename without the extension from a path in Python?


回答 0

获取不带扩展名的文件名:

import os
print(os.path.splitext("/path/to/some/file.txt")[0])

印刷品:

/path/to/some/file

的文档os.path.splitext

重要说明:如果文件名有多个点,则仅删除最后一个扩展名之后的扩展名。例如:

import os
print(os.path.splitext("/path/to/some/file.txt.zip.asc")[0])

印刷品:

/path/to/some/file.txt.zip

如果您需要处理这种情况,请参见下面的其他答案。

Getting the name of the file without the extension:

import os
print(os.path.splitext("/path/to/some/file.txt")[0])

Prints:

/path/to/some/file

Documentation for os.path.splitext.

Important Note: If the filename has multiple dots, only the extension after the last one is removed. For example:

import os
print(os.path.splitext("/path/to/some/file.txt.zip.asc")[0])

Prints:

/path/to/some/file.txt.zip

See other answers below if you need to handle that case.


回答 1

您可以使用以下方法制作自己的:

>>> import os
>>> base=os.path.basename('/root/dir/sub/file.ext')
>>> base
'file.ext'
>>> os.path.splitext(base)
('file', '.ext')
>>> os.path.splitext(base)[0]
'file'

重要说明:如果.文件名中有多个,则仅删除最后一个。例如:

/root/dir/sub/file.ext.zip -> file.ext

/root/dir/sub/file.ext.tar.gz -> file.ext.tar

请参阅下面的其他答案。

You can make your own with:

>>> import os
>>> base=os.path.basename('/root/dir/sub/file.ext')
>>> base
'file.ext'
>>> os.path.splitext(base)
('file', '.ext')
>>> os.path.splitext(base)[0]
'file'

Important note: If there is more than one . in the filename, only the last one is removed. For example:

/root/dir/sub/file.ext.zip -> file.ext

/root/dir/sub/file.ext.tar.gz -> file.ext.tar

See below for other answers that address that.


回答 2

使用pathlib在Python 3.4+

from pathlib import Path

Path('/root/dir/sub/file.ext').stem

将返回

'file'

Using pathlib in Python 3.4+

from pathlib import Path

Path('/root/dir/sub/file.ext').stem

will return

'file'

回答 3

>>> print(os.path.splitext(os.path.basename("hemanth.txt"))[0])
hemanth
>>> print(os.path.splitext(os.path.basename("hemanth.txt"))[0])
hemanth

回答 4

在Python 3.4+中,您可以使用pathlib解决方案

from pathlib import Path

print(Path(your_path).resolve().stem)

In Python 3.4+ you can use the pathlib solution

from pathlib import Path

print(Path(your_path).resolve().stem)

回答 5

https://docs.python.org/3/library/os.path.html

在python 3 pathlib中,“ pathlib模块提供了高级路径对象。” 所以,

>>> from pathlib import Path
>>> p = Path("/a/b/c.txt")
>>> print(p.with_suffix(''))
\a\b\c
>>> print(p.stem)
c

https://docs.python.org/3/library/os.path.html

In python 3 pathlib "The pathlib module offers high-level path objects." so,

>>> from pathlib import Path
>>> p = Path("/a/b/c.txt")
>>> print(p.with_suffix(''))
\a\b\c
>>> print(p.stem)
c

回答 6

如果您想保留文件的路径,然后删除扩展名

>>> file = '/root/dir/sub.exten/file.data.1.2.dat'
>>> print ('.').join(file.split('.')[:-1])
/root/dir/sub.exten/file.data.1.2

If you want to keep the path to the file and just remove the extension

>>> file = '/root/dir/sub.exten/file.data.1.2.dat'
>>> print ('.').join(file.split('.')[:-1])
/root/dir/sub.exten/file.data.1.2

回答 7

如果扩展名中包含多个点,则os.path.splitext()不起作用。

例如images.tar.gz

>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> print os.path.splitext(file_name)[0]
images.tar

您可以只在基本名称中找到第一个点的索引,然后对基本名称进行切片以仅获取不带扩展名的文件名。

>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> index_of_dot = file_name.index('.')
>>> file_name_without_extension = file_name[:index_of_dot]
>>> print file_name_without_extension
images

os.path.splitext() won't work if there are multiple dots in the extension.

For example, images.tar.gz

>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> print os.path.splitext(file_name)[0]
images.tar

You can just find the index of the first dot in the basename and then slice the basename to get just the filename without extension.

>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> index_of_dot = file_name.index('.')
>>> file_name_without_extension = file_name[:index_of_dot]
>>> print file_name_without_extension
images

回答 8

@IceAdor引用了@ user2902201解决方案的注释中的rsplit。rsplit是支持多个期间的最简单的解决方案。

在这里说明:

file = 'my.report.txt'
print file.rsplit('.', 1)[0]

我的报告

@IceAdor's refers to rsplit in a comment to @user2902201's solution. rsplit is the simplest solution that supports multiple periods.

Here it is spelt out:

file = 'my.report.txt'
print file.rsplit('.', 1)[0]

my.report


回答 9

但是即使导入os,也无法将其称为path.basename。是否可以直接将其作为基名来调用?

import os,然后使用 os.path.basename

importing os并不意味着您os.foo无需参考即可使用os

But even when I import os, I am not able to call it path.basename. Is it possible to call it as directly as basename?

import os, and then use os.path.basename

importing os doesn't mean you can use os.foo without referring to os.


回答 10

我以为我会使用os.path.splitext的变体不需要使用数组索引的情况的使用进行修改。

该函数始终返回一(root, ext)对,因此可以安全使用:

root, ext = os.path.splitext(path)

例:

>>> import os
>>> path = 'my_text_file.txt'
>>> root, ext = os.path.splitext(path)
>>> root
'my_text_file'
>>> ext
'.txt'

Thought I would throw in a variation to the use of the os.path.splitext without the need to use array indexing.

The function always returns a (root, ext) pair so it is safe to use:

root, ext = os.path.splitext(path)

Example:

>>> import os
>>> path = 'my_text_file.txt'
>>> root, ext = os.path.splitext(path)
>>> root
'my_text_file'
>>> ext
'.txt'

回答 11

其他方法不会删除多个扩展名。有些文件名也没有扩展名。此代码段同时处理这两个实例,并且在Python 2和3中均可使用。它从路径中获取基本名称,将值分割为点,然后返回第一个(即文件名的初始部分)。

import os

def get_filename_without_extension(file_path):
    file_basename = os.path.basename(file_path)
    filename_without_extension = file_basename.split('.')[0]
    return filename_without_extension

这是一组要运行的示例:

example_paths = [
    "FileName", 
    "./FileName",
    "../../FileName",
    "FileName.txt", 
    "./FileName.txt.zip.asc",
    "/path/to/some/FileName",
    "/path/to/some/FileName.txt",
    "/path/to/some/FileName.txt.zip.asc"
]

for example_path in example_paths:
    print(get_filename_without_extension(example_path))

在每种情况下,打印出的值是:

FileName

The other methods don't remove multiple extensions. Some also have problems with filenames that don't have extensions. This snippet deals with both instances and works in both Python 2 and 3. It grabs the basename from the path, splits the value on dots, and returns the first one which is the initial part of the filename.

import os

def get_filename_without_extension(file_path):
    file_basename = os.path.basename(file_path)
    filename_without_extension = file_basename.split('.')[0]
    return filename_without_extension

Here's a set of examples to run:

example_paths = [
    "FileName", 
    "./FileName",
    "../../FileName",
    "FileName.txt", 
    "./FileName.txt.zip.asc",
    "/path/to/some/FileName",
    "/path/to/some/FileName.txt",
    "/path/to/some/FileName.txt.zip.asc"
]

for example_path in example_paths:
    print(get_filename_without_extension(example_path))

In every case, the value printed is:

FileName

回答 12

import os

filename = C:\\Users\\Public\\Videos\\Sample Videos\\wildlife.wmv

这将返回filenameextension(C:\用户\公用\视频\样品影片\野生动物)

temp = os.path.splitext(filename)[0]  

现在,您可以filename通过

os.path.basename(temp)   #this returns just the filename (wildlife)

import os

filename = C:\\Users\\Public\\Videos\\Sample Videos\\wildlife.wmv

This returns the filename without the extension(C:\Users\Public\Videos\Sample Videos\wildlife)

temp = os.path.splitext(filename)[0]  

Now you can get just the filename from the temp with

os.path.basename(temp)   #this returns just the filename (wildlife)

回答 13

多扩展识别过程。适用于strunicode路径。适用于Python 2和3。

import os

def file_base_name(file_name):
    if '.' in file_name:
        separator_index = file_name.index('.')
        base_name = file_name[:separator_index]
        return base_name
    else:
        return file_name

def path_base_name(path):
    file_name = os.path.basename(path)
    return file_base_name(file_name)

行为:

>>> path_base_name('file')
'file'
>>> path_base_name(u'file')
u'file'
>>> path_base_name('file.txt')
'file'
>>> path_base_name(u'file.txt')
u'file'
>>> path_base_name('file.tar.gz')
'file'
>>> path_base_name('file.a.b.c.d.e.f.g')
'file'
>>> path_base_name('relative/path/file.ext')
'file'
>>> path_base_name('/absolute/path/file.ext')
'file'
>>> path_base_name('Relative\\Windows\\Path\\file.txt')
'file'
>>> path_base_name('C:\\Absolute\\Windows\\Path\\file.txt')
'file'
>>> path_base_name('/path with spaces/file.ext')
'file'
>>> path_base_name('C:\\Windows Path With Spaces\\file.txt')
'file'
>>> path_base_name('some/path/file name with spaces.tar.gz.zip.rar.7z')
'file name with spaces'

A multiple extension aware procedure. Works for str and unicode paths. Works in Python 2 and 3.

import os

def file_base_name(file_name):
    if '.' in file_name:
        separator_index = file_name.index('.')
        base_name = file_name[:separator_index]
        return base_name
    else:
        return file_name

def path_base_name(path):
    file_name = os.path.basename(path)
    return file_base_name(file_name)

Behavior:

>>> path_base_name('file')
'file'
>>> path_base_name(u'file')
u'file'
>>> path_base_name('file.txt')
'file'
>>> path_base_name(u'file.txt')
u'file'
>>> path_base_name('file.tar.gz')
'file'
>>> path_base_name('file.a.b.c.d.e.f.g')
'file'
>>> path_base_name('relative/path/file.ext')
'file'
>>> path_base_name('/absolute/path/file.ext')
'file'
>>> path_base_name('Relative\\Windows\\Path\\file.txt')
'file'
>>> path_base_name('C:\\Absolute\\Windows\\Path\\file.txt')
'file'
>>> path_base_name('/path with spaces/file.ext')
'file'
>>> path_base_name('C:\\Windows Path With Spaces\\file.txt')
'file'
>>> path_base_name('some/path/file name with spaces.tar.gz.zip.rar.7z')
'file name with spaces'

回答 14

import os
path = "a/b/c/abc.txt"
print os.path.splitext(os.path.basename(path))[0]
import os
path = "a/b/c/abc.txt"
print os.path.splitext(os.path.basename(path))[0]

回答 15

在Windows系统上,我也使用drivername前缀,例如:

>>> s = 'c:\\temp\\akarmi.txt'
>>> print(os.path.splitext(s)[0])
c:\temp\akarmi

因此,由于不需要驱动器号或目录名,因此使用:

>>> print(os.path.splitext(os.path.basename(s))[0])
akarmi

On Windows system I used drivername prefix as well, like:

>>> s = 'c:\\temp\\akarmi.txt'
>>> print(os.path.splitext(s)[0])
c:\temp\akarmi

So because I do not need drive letter or directory name, I use:

>>> print(os.path.splitext(os.path.basename(s))[0])
akarmi

回答 16

为了方便起见,一个简单的函数包装了以下两种方法os.path

def filename(path):
  """Return file name without extension from path.

  See https://docs.python.org/3/library/os.path.html
  """
  import os.path
  b = os.path.split(path)[1]  # path, *filename*
  f = os.path.splitext(b)[0]  # *file*, ext
  #print(path, b, f)
  return f

经过Python 3.5测试。

For convenience, a simple function wrapping the two methods from os.path :

def filename(path):
  """Return file name without extension from path.

  See https://docs.python.org/3/library/os.path.html
  """
  import os.path
  b = os.path.split(path)[1]  # path, *filename*
  f = os.path.splitext(b)[0]  # *file*, ext
  #print(path, b, f)
  return f

Tested with Python 3.5.


回答 17

解决此问题的最简单方法是

import ntpath 
print('Base name is ',ntpath.basename('/path/to/the/file/'))

这样可以节省您的时间和计算成本。

the easiest way to resolve this is to

import ntpath 
print('Base name is ',ntpath.basename('/path/to/the/file/'))

this saves you time and computation cost.


回答 18

非常非常非常简单没有其他模块!

import os
p = r"C:\Users\bilal\Documents\face Recognition python\imgs\northon.jpg"

# Get the filename only from the initial file path.
filename = os.path.basename(p)

# Use splitext() to get filename and extension separately.
(file, ext) = os.path.splitext(filename)

# Print outcome.
print("Filename without extension =", file)
print("Extension =", ext)

Very very very simpely no other modules !!!

import os
p = r"C:\Users\bilal\Documents\face Recognition python\imgs\northon.jpg"

# Get the filename only from the initial file path.
filename = os.path.basename(p)

# Use splitext() to get filename and extension separately.
(file, ext) = os.path.splitext(filename)

# Print outcome.
print("Filename without extension =", file)
print("Extension =", ext)

回答 19

我们可以做一些简单的split/ pop因为在这里看到(魔术https://stackoverflow.com/a/424006/1250044),提取文件名(尊重Windows和POSIX差异)。

def getFileNameWithoutExtension(path):
  return path.split('\\').pop().split('/').pop().rsplit('.', 1)[0]

getFileNameWithoutExtension('/path/to/file-0.0.1.ext')
# => file-0.0.1

getFileNameWithoutExtension('\\path\\to\\file-0.0.1.ext')
# => file-0.0.1

We could do some simple split / pop magic as seen here (https://stackoverflow.com/a/424006/1250044), to extract the filename (respecting the windows and POSIX differences).

def getFileNameWithoutExtension(path):
  return path.split('\\').pop().split('/').pop().rsplit('.', 1)[0]

getFileNameWithoutExtension('/path/to/file-0.0.1.ext')
# => file-0.0.1

getFileNameWithoutExtension('\\path\\to\\file-0.0.1.ext')
# => file-0.0.1

回答 20

import os
list = []
def getFileName( path ):
for file in os.listdir(path):
    #print file
    try:
        base=os.path.basename(file)
        splitbase=os.path.splitext(base)
        ext = os.path.splitext(base)[1]
        if(ext):
            list.append(base)
        else:
            newpath = path+"/"+file
            #print path
            getFileName(newpath)
    except:
        pass
return list

getFileName("/home/weexcel-java3/Desktop/backup")
print list
import os
list = []
def getFileName( path ):
for file in os.listdir(path):
    #print file
    try:
        base=os.path.basename(file)
        splitbase=os.path.splitext(base)
        ext = os.path.splitext(base)[1]
        if(ext):
            list.append(base)
        else:
            newpath = path+"/"+file
            #print path
            getFileName(newpath)
    except:
        pass
return list

getFileName("/home/weexcel-java3/Desktop/backup")
print list

回答 21

import os文件名,file_extension = os.path.splitext('/ d1 / d2 / example.cs')文件名是'/ d1 / d2 / example'file_extension是'.cs'

import os filename, file_extension = os.path.splitext('/d1/d2/example.cs') filename is '/d1/d2/example' file_extension is '.cs'


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