Storing strings into list from Python regex over a file

Question

I am using Python's regex to extract paths from a file and then trying to append the paths into a single list. The paths are specified in different lines in the file (options) as follows:

EXE_INC = \
    -I$(LIB_SRC)/me/bMesh/lnInclude \
    -I$(LIB_SRC)/mTools/lnInclude \
    -I$(LIB_SRC)/dynamicM/lnInclude

The function that I have is:

def libinclude():
with open('options', 'r') as options:
    for lines in options:
        if 'LIB_SRC' in lines:
            result = []
            lib_src_path = re.search(r'\s*-I\$\(LIB_SRC\)(?P<lpath>\/.*)', lines.strip())
            lib_path = lib_src_path.group(1).split()
            result.append(lib_path[0])
            print result
return (result)

The output that I am getting is:

['/me/bMesh/lnInclude']
['/mTools/lnInclude']
['/dynamicM/lnInclude']

However, as you can see I get three different lists, each obtained for one of the lines in the file, options. Is there a way to get these three paths in one list so that the values are available outside the function.

Greg · Accepted Answer · 2015-02-19 02:37:56Z

2

For sure, you're creating a new list every time you find a match. So only create one list and just continue to append to it each time you find a match

def libinclude():
with open('options', 'r') as options:
    result = []
    for lines in options:
        if 'LIB_SRC' in lines:
            lib_src_path = re.search(r'\s*-I\$\(LIB_SRC\)(?P<lpath>\/.*)', lines.strip())
            lib_path = lib_src_path.group(1).split()
            result.append(lib_path[0])
            print result

answered Feb 19, 2015 at 2:37

Greg

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1 Comment

hypersonics Over a year ago

Thanks @Greg. Indeed trivial!

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