The easiest answer is probably to change your working directory, then call the second .py file from where it is:

python a.py && cd testA && python ../b.py

Of course you might find it even easier to write a script that does it all for you, like so:

Save this as runTests.sh in the same directory as a.py is:

#!/bin/sh
python a.py
cd testA
python ../b.py

Make it executable:

chmod +x ./runTests.sh

Then you can simply enter your directory and run it:

./runTests.sh

No you don't need to use:

cd home/directoryA/directoryB/directoryC/DirectoryD
./somefile 

You can simply run the command by prefixing it with its path:

/home/directoryA/directoryB/directoryC/DirectoryD/somefile

Because you are already in the /home/directoryA you can use the current directory shortcut . and run the command like this:

./directoryB/directoryC/DirectoryD/somefile

I noticed OP has expanded scope via comments under other answers. Here is some additional information:

• To find out where somefile is located use: locate somefile.
• If somefile was added today you need to first update the locate database by running sudo updatedb.
• When there are multiple versions of somefile located in the PATH you can find out which one is executed first use which somefile.
• If you want to run somefile without specifying a directory name in front put it in the path. To check the path use echo $PATH. Common path locations to put somefile are /usr/local/bin (if it uses sudo powers) and /home/your_user_name/bin (you might have to create the directory first).
• You can also add /home/directoryA/directoryB/directoryC/DirectoryD/ to your path but that would be highly unusual. However you could then simply type somefile no matter what directory you are in and it will run.
• Of course somefile must be executable which you set with the command: chmod a+x /home/directoryA/directoryB/directoryC/DirectoryD/somefile

You can use os.path.dirname() and __file__ to get absolute paths like this:

#!/usr/bin/python

import os  # We need this module

# Get path of the current dir, then use it to create paths:
CURRENT_DIR = os.path.dirname(__file__)
file_path = os.path.join(CURRENT_DIR, 'test.txt')

# Then work using the absolute paths:
f = open(file_path,'w')
f.write('testing the script')

This way the script will work on files placed only in the same directory as the script, regardless of the place from which you execute it.

sh /path/to/script will spawn a new shell and run she script independent of your current shell. The source (.) command will call all the commands in the script in the current shell. If the script happens to call exit for example, then you'll lose the current shell. Because of this it is usually safer to call scripts in a separate shell with sh or to execute them as binaries using either the full (starting with /) or relative path (./). If called as binaries, they will be executed with the specified interpreter (#!/bin/bash, for example).

As for knowing whether or not a script will find the files it needs, there is no good answer, other than looking at the script to see what it does. As an option, you could always go to the script's folder in a sub-process without leaving your current folder:

(cd /wherever/ ; sh script.sh)

If the problem is just "one line":

cd foo; python ../awesome_script.py; cd ..

will do

From the error message:

/usr/local/Cellar/python/2.7.12/Frameworks/Python.framework/Versions/2.7/Resources/Python.app/Contents/MacOS/Python: can't open file 'Projects/sym.py': [Errno 2] No such file or directory

you can see that your ~/Scripts/script.py tries to reference another file using a relative path Projects/sym.py.

Most likely you are calling the func function when your current directory is set to one that does not contain Projects/sym.py.

If your intention is to include a file from ~/Projects/sym.py you need to use an absolute path in the Python code or change the current directory to ~ inside the func prior to calling python.

You can call your script using only the whole path, without the dot .:

/path/to/script

sudo also works fine:

sudo /path/to/script

Will they usually find all the stuff they need to run properly?

Do you mean like, "will my script find the files which are in the same folder?" That depends of your code. For example, if you have the script /tmp/test.sh:

#!/bin/bash
ls

If you invoke it from your home folder, it will run ls in your home:

test@ubuntu:~:/tmp/test.sh
Desktop     Dropbox         Imagens  NetBeansProjects     Público     
Documentos  Modelos    R          Vídeos

In this situation, dirname is your ally:

#!/bin/bash
current=$(dirname $0)
ls $current

Running it from your home folder, it gives:

test@ubuntu:~:/tmp/test.sh
acroread_1000_1000     hiRF7yLSOD              pulse-2L9K88eMlGn7   
unity_support_test.0    clementine-art-jt5332.jpg

which is the content of my /tmp/ folder.

Quick and dirty:

In your start up script instead of just executing the python script, use cd first.

#!/bin/sh

cd /home/username/projectname &&
python ./scriptname.py

If you want to run your Python script (let me call it myScript.py) on each *.grd file in all subfolders (recursively) of a given folder, then find is the right command to do it:

find . -name '*.grd' -exec python myScript.py {} \;

This command recursively finds each file in the given path¹ with the name matching *.grd and calls python myScript.py <found .grd file>² for each found file. You can test this by adding echo right after -exec, then it will just print the commands.

¹ I used . in my example and you can use it, too, if you cd to the specified folder first. Otherwise, you can use any appropriate relative or absolute path here.

² If your script is executable and has the correct shebang, you can leave out python and call myScript.py directly.

See also

• How to Search for Files Recursively into Subdirectories
• How can I recursively find a directory by name and delete its contents (including all sub-directories and files) while keeping the directory itself?

Do you have a shebang? Your first line of the script should point to the actual interpreter, e.g.

#!/usr/bin/python3

Also, make sure the permissions are set correctly, e.g. you'll probably want something like 775.

Here's to show it works just fine:

$ pwd
/tmp/test
$ ls
test.py
$ head -3 test.py 
#!/usr/bin/python3
# SPDX-License-Identifier: GPL-2.0 OR GPL-3.0
#
$ export PATH=/tmp/test:$PATH
$ test.py
usb1              1d6b:0002 09 1IF  [USB 2.00,   480 Mbps,   0mA] (xhci-hcd 0000:04:00.1) hub
usb2              1d6b:0003 09 1IF  [USB 3.10, 10000 Mbps,   0mA] (xhci-hcd 0000:04:00.1) hub
usb3              1d6b:0002 09 1IF  [USB 2.00,   480 Mbps,   0mA] (xhci-hcd 0000:04:00.3) hub
...
skipped

i.e. everything works.

The actual answer after getting all the info: your program is in /home/pol/bin/Python while $PATH only contains /home/pol/bin. Move the program to /home/pol/bin or add /home/pol/bin/Python to $PATH.