Re: [galaxy-dev] Getting tool dirpath in a Python code file
Hi Leandro,
Leandro Hermida wrote, On 04/29/2011 06:17 AM:
But I have one complication I guess. My tool is already running a Python
code file to dynamically create a drop-down menu using the dynamic_options
attribute a la:
param type=select dynamic_options=my_options()
...
code file=my_options.py/
In this function my_options I would need the tool dir path. How would I be
able to combine what you did with this? Which function runs at the right
time before my_options() is executed is it exec_before_process?
This just gets messier and messier :)
The following hack will get you the tool's path inside the dynamic option
code, but it is SO ugly, I would highly recommend against it.
A better way would be to add some code to ./galaxy/lib/parameters/basic.py
inside get_options() to add the 'tool' object to the other_values object,
so that custom code functions would be able to access it (or something similar
and as clean).
That being said, here goes:
= dynamic_options_path.xml ==
tool id=cshl_dynamic_optinos_path_test name=dynamic_options_path
description=
commandecho '$input1' gt; '$output'/command
inputs
param name=input1 type=text value=10 label=Dummy/
param name=mylist type=select label=Dynamic Options
dynamic_options=get_my_options(code_ref=get_my_options) display=radio /
/inputs
code file=dynamic_options_code.py /
outputs
data name=output format=txt /
/outputs
/tool
===
== dynamic_options_code.py ==
import os
import os.path
import sys
def get_my_options(code_ref):
root_dir=os.getcwd()
sys.stderr.write(Galaxy root dir = %s\n % ( root_dir ) )
sys.stderr.write(Custom code file = %s\n % (
code_ref.__code__.co_filename) )
tool_path = root_dir + '/' + os.path.dirname(
code_ref.__code__.co_filename)
sys.stderr.write(tool path = %s\n % ( tool_path ) )
res = [('Hello','World',False),]
return res
==
The output (STDERR) when viewing the tool's form is:
Galaxy root dir = /home/gordon/projects/galaxy_dev
Custom code file = ./tools/cshl_tests/dynamic_options_code.py
tool path = /home/gordon/projects/galaxy_dev/./tools/cshl_tests
The trick is that the dynamic function itself (get_my_options()) is a valid
object (which is luckily accessible from inside the code), and python keeps
track of the file that contained the function.
Good luck (and don't try this at home :) ),
-gordon
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Re: [galaxy-dev] Getting tool dirpath in a Python code file
Hi galaxy developers, Just want to double-check, there is no way to import some kind of galaxy tool context info into python code you are running for a tool? best, leandro On Fri, Apr 15, 2011 at 8:03 PM, Leandro Hermida soft...@leandrohermida.com wrote: On Fri, Apr 15, 2011 at 7:27 PM, Peter Cock p.j.a.c...@googlemail.comwrote: On Thu, Apr 14, 2011 at 2:56 PM, Leandro Hermida soft...@leandrohermida.com wrote: On Thu, Apr 14, 2011 at 3:17 PM, Peter Cock p.j.a.c...@googlemail.com wrote: For standard Python tools in Galaxy, I'm using os.path.split(sys.argv[0])[0] to get the path, which on reflection probably should be written as os.path.dirname(sys.argv[0]) as you suggest. What do __file__ and sys.argv[0] give you? The simplest way to debug this is to add a print statement, since Galaxy will show the stdout. Hi Peter, __file__ throws an error: global name '__file__' is not defined I guess the script is being loaded as a string, and run with eval(...) or something like that. It would also explain why sys.argv[0] would be one of the Galaxy script files. os.path.abspath(os.path.dirname(sys.argv[0])) gives me /path/to/galaxy/scripts directory which is two levels up from what the tool directory I want for example /path/to/galaxy/tools/mytool So combine that with ../tools/mytool/ and you're done? OK, you have to know the name of the folder your tool *should* be in... so not a perfect solution. Thanks Peter, yes that's the same idea I did as a quick fix also don't like the fact that my tool directory is hard-coded but oh well. There must be a way within Python in Galaxy importing something from Galaxy that has the current tool directory path, it would seem that Galaxy needs to know this and would store it anyway? best, Leandro ___ Please keep all replies on the list by using reply all in your mail client. To manage your subscriptions to this and other Galaxy lists, please use the interface at: http://lists.bx.psu.edu/
Re: [galaxy-dev] Getting tool dirpath in a Python code file
Leandro Hermida wrote, On 04/28/2011 08:55 AM: Hi galaxy developers, Just want to double-check, there is no way to import some kind of galaxy tool context info into python code you are running for a tool? Nothing is impossible... just depends on how messy you want to get :) for me, the following works: === python_path.xml === tool id=cshl_python_path_test name=pythonpath description= commandecho '$input1' gt; '$output'/command inputs param name=input1 type=text value=10 label=Dummy/ /inputs code file=python_path_code.py / outputs data name=output format=txt / /outputs /tool == python_path_code.py = from os import path import sys def exec_after_process(app, inp_data, out_data, param_dict, tool, stdout, stderr): tool_path = path.abspath(tool.tool_dir); sys.stderr.write(! path = %s\n % (tool_path)) When the tool runs, the following line is printed to STDERR: == ! path = /home/gordon/projects/galaxy_dev/tools/cshl_tests == Help this helps, -gordon best, leandro On Fri, Apr 15, 2011 at 8:03 PM, Leandro Hermida soft...@leandrohermida.com mailto:soft...@leandrohermida.com wrote: On Fri, Apr 15, 2011 at 7:27 PM, Peter Cock p.j.a.c...@googlemail.com mailto:p.j.a.c...@googlemail.com wrote: On Thu, Apr 14, 2011 at 2:56 PM, Leandro Hermida soft...@leandrohermida.com mailto:soft...@leandrohermida.com wrote: On Thu, Apr 14, 2011 at 3:17 PM, Peter Cock p.j.a.c...@googlemail.com mailto:p.j.a.c...@googlemail.com wrote: For standard Python tools in Galaxy, I'm using os.path.split(sys.argv[0])[0] to get the path, which on reflection probably should be written as os.path.dirname(sys.argv[0]) as you suggest. What do __file__ and sys.argv[0] give you? The simplest way to debug this is to add a print statement, since Galaxy will show the stdout. Hi Peter, __file__ throws an error: global name '__file__' is not defined I guess the script is being loaded as a string, and run with eval(...) or something like that. It would also explain why sys.argv[0] would be one of the Galaxy script files. os.path.abspath(os.path.dirname(sys.argv[0])) gives me /path/to/galaxy/scripts directory which is two levels up from what the tool directory I want for example /path/to/galaxy/tools/mytool So combine that with ../tools/mytool/ and you're done? OK, you have to know the name of the folder your tool *should* be in... so not a perfect solution. Thanks Peter, yes that's the same idea I did as a quick fix also don't like the fact that my tool directory is hard-coded but oh well. There must be a way within Python in Galaxy importing something from Galaxy that has the current tool directory path, it would seem that Galaxy needs to know this and would store it anyway? best, Leandro ___ Please keep all replies on the list by using reply all in your mail client. To manage your subscriptions to this and other Galaxy lists, please use the interface at: http://lists.bx.psu.edu/
Re: [galaxy-dev] Getting tool dirpath in a Python code file
On Thu, Apr 14, 2011 at 2:56 PM, Leandro Hermida soft...@leandrohermida.com wrote: On Thu, Apr 14, 2011 at 3:17 PM, Peter Cock p.j.a.c...@googlemail.com wrote: For standard Python tools in Galaxy, I'm using os.path.split(sys.argv[0])[0] to get the path, which on reflection probably should be written as os.path.dirname(sys.argv[0]) as you suggest. What do __file__ and sys.argv[0] give you? The simplest way to debug this is to add a print statement, since Galaxy will show the stdout. Hi Peter, __file__ throws an error: global name '__file__' is not defined I guess the script is being loaded as a string, and run with eval(...) or something like that. It would also explain why sys.argv[0] would be one of the Galaxy script files. os.path.abspath(os.path.dirname(sys.argv[0])) gives me /path/to/galaxy/scripts directory which is two levels up from what the tool directory I want for example /path/to/galaxy/tools/mytool So combine that with ../tools/mytool/ and you're done? OK, you have to know the name of the folder your tool *should* be in... so not a perfect solution. Peter ___ Please keep all replies on the list by using reply all in your mail client. To manage your subscriptions to this and other Galaxy lists, please use the interface at: http://lists.bx.psu.edu/
Re: [galaxy-dev] Getting tool dirpath in a Python code file
On Thu, Apr 14, 2011 at 2:08 PM, Leandro Hermida soft...@leandrohermida.com wrote: Hi, I have a tool with a code file=my_script.py/ tag and in that code file I'm trying to get the tool dirpath where that script and the tool XML exist. I've tried: os.path.abspath(os.path.dirname(sys.argv[0])) os.path.abspath(os.path.dirname(__file__)) And both don't work as expected. Is there a galaxy class I could import which will have the tool directory path? regards, Leandro For standard Python tools in Galaxy, I'm using os.path.split(sys.argv[0])[0] to get the path, which on reflection probably should be written as os.path.dirname(sys.argv[0]) as you suggest. What do __file__ and sys.argv[0] give you? The simplest way to debug this is to add a print statement, since Galaxy will show the stdout. Peter ___ Please keep all replies on the list by using reply all in your mail client. To manage your subscriptions to this and other Galaxy lists, please use the interface at: http://lists.bx.psu.edu/
