Check regex result in a list ,if-else python

Question

I have a list of words (even 2 or 3 words in 1 object) . I have compiled a regex to match any of the words in the list to a random string . If it matches , that object of the list is returned ,else it will execute another regex on the string to return an expression.
my code:
"layouts" is the list of words and "layout_re" uses the words in list to match in the string "str". If ms is present in list , return ms as it is otherwise use expression of p to return the pattern.

def layout_corr(str):

    ms = layout_re.search(str)
    if ms in layouts:
       ms=ms

    else:
       p = re.compile(r'(?:\w+\s+){1,2}layout')
       ms = re.findall(p,str)
    return ms

The output i get is that of the else statement .
eg:- str = " no 804 8th main 2nd c cross hrbr layout 1st block"
the list contains "hrbr layout" hence hrbr layout must be returned. But instead , the function returns "cross hrbr layout" which is the ms calculated by else statement.. What is the problem with the if statement??

re.search returns a match object (or None), while re.findall returns a list. Neither one returns purely object matched. — Joel Cornett
– Joel Cornett, Commented Mar 6, 2014 at 5:42

Jerry · Accepted Answer · 2014-03-06 06:15:44Z

1

re.search returns a match object, not a string, so that:

ms in layouts

will always return false. Use this instead:

def layout_corr(str):
    ms = layout_re.search(str)
    if ms:
       return ms.group()    
    else:
       p = re.compile(r'((?:\w+\s+){1,2})layout')
       ms = re.findall(p,str)
       return ms

answered Mar 6, 2014 at 6:15

Jerry

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2 Comments

Jerry Over a year ago

You could use (\w+(?:\s+\w+)?)\s+layout for the second regex, if you want to trim spaces.

Jerry Over a year ago

Also, you should avoid using str since it's a function name (already told you in chat, but mentioning it here too)

Nir Alfasi · Accepted Answer · 2014-03-06 05:56:17Z

0

1) change ms=ms to return ms

2) In order to "catch" the string "hrbr layout" - change the line:

p = re.compile(r'(?:\w+\s+){1,2}layout')

to:

p = re.compile(r' (\w+? layout)')

Demo:

p = re.compile(r' (\w+? layout)')
ms = re.findall(p,"no  804  8th main 2nd c cross hrbr layout 1st block")
print(ms[0]) // prints: "hrbr layout"

edited Mar 6, 2014 at 5:56

answered Mar 6, 2014 at 5:47

Nir Alfasi

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2 Comments

Hypothetical Ninja Over a year ago

hrbr is in list. if i use this expression then it will extract 1 word before layout always.. what i want is to check presence in the list called"layouts" and return the same if present.. If not , then return 2 words before layout.. Hrbr is in the layouts list , so i want the 1 st statement to return the exprssion

Nir Alfasi Over a year ago

@Sword in that case change ms=ms to return ms

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