Question: I have a lot of strings like this 

```
/home/build_targets/JavaFiles/Commonfeature/CM/src/com/ems/ClusterGroupTreePanel.java

```

I need to get the file name ClusterGroupTreePanel.java out.

Is there an efficient way to achieve this in **Python**?

Answer: 
use [`os.path.basename`](http://docs.python.org/2/library/os.path.html#os.path.basename)

```
>>> import os.path
>>> os.path.basename('/home/build_targets/JavaFiles/Commonfeature/CM/src/com/ems/ClusterGroupTreePanel.java')
'ClusterGroupTreePanel.java'

```

Generally, if your question goes something like "How do I manipulate this filesystem path name in an interesting way?". The answer is usually "use some combination of functions in the `os.path` module". There's lots of great stuff in there and it would be well worth your time to familiarize yourself with the different functions.
os.path.[basename](http://docs.python.org/2/library/os.path.html#os.path.basename)(path) will give you the file name, if you want to leave it out you can use [os.path.split](http://docs.python.org/2/library/os.path.html#os.path.split) and take the first part

```
os.path.split(full_path)[0]

```