Re.findall help

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January 26th, 2013, 06:16 PM

taeBaby

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Re.findall help

Hi, so I'm fairly new to python but I've been practicing coding on my own.

Recently, I'm working with the re.match, re.findall, re.ignorecase.

This is my code:

Code:

def mission(s):
  match = re.findall(r'\d+', s, re.IGNORECASE)
  return match

It gives me this output:

Code:

['1', '0']
['20', '500']
['3']
['20', '1']

From this:

Code:

mission('Recon Mission 1 accomplished. Enemy found: 0.') 
mission('recon mission 20 accomplished. enemies found: 500.') 
mission("Recon Mission 3 failed.")
mission("I have 20 carrots and 1 mushroom.")

But I need to get this output:

Code:

(1, 0)
(20, 500)
(3, failed)
fourth line returns empty b/c it does not contain mission or status

I'm not sure how to extract mission number and status (either 'failed' or number of enemies found)

Any suggestions on how to do this correctly?

January 26th, 2013, 07:27 PM

b49P23TIvg

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1) convert the number strings to integers.
2) Use simple regular expressions. Don't try to parse English with a single re.
This code might be closer to what you want. Could be more restrictive about accepting interesting sentences.

Code:

# python -m doctest -v p.py

import re

def mission(s):
    '''
        >>> (1, 0) == mission('Recon Mission 1 accomplished. Enemy found: 0.')
        True
        >>> (20, 500) == mission('recon mission 20 accomplished. enemies found: 500.')
        True
        >>> (3, 'failed') == mission("Recon Mission 3 failed.")
        True
        >>> () == mission("I have 20 carrots and 1 mushroom.")
        True
    '''
    match = list(map(int,re.findall(r'\d+', s)))
    sl = s.lower()
    if 'mission' in sl:
        if ('accomplished' in sl) and ('found' in sl):
            return tuple(match)
        if match and ('failed' in sl):
            return (match[0],'failed',)
    return ()

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January 26th, 2013, 07:56 PM

taeBaby

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Thanks for that @b49P23TIvg

what does map() do?

But while I was away, I actually figured out how to do it another way, the only problem I have now is how to remove the brackets and quotation marks from the numbers to just have parentheses.

This is my code now:

Code:

def mission(s):
  first = re.findall(r'Recon Mission (\d+)', s, re.IGNORECASE)
  second = re.findall(r'found: (\d+)', s)
  if first and second:
    s = first, second
  elif first and second !=None:
    second = 'failed'
    s = first, second
  else:
    s = None
  return s

My output:

Code:

(['1'], ['0'])
(['2'], ['500'])
(['3'], 'failed')
None

How to code to get my (['1'], ['0']) to output as (1, 0)?
Or is it not possible to do so unless I do it your way or another way?

January 26th, 2013, 08:43 PM

b49P23TIvg

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Using your mission function, we can clean the output into strings matching your requirement.

Code:

import re

def mission(s):
    first = re.findall(r'Recon Mission (\d+)', s, re.IGNORECASE)
    second = re.findall(r'found: (\d+)', s)
    if first and second:
        s = first, second
    elif first and second !=None:
        second = 'failed'
        s = first, second
    else:
        s = None
    return s

def clean(s):
    '''
        >>> print(clean(mission('Recon Mission 1 accomplished. Enemy found: 0.')))
        (1, 0)
        >>> print(clean(mission('recon mission 20 accomplished. enemies found: 500.')))
        (20, 500)
        >>> print(clean(mission("Recon Mission 3 failed.")))
        (3, failed)
        >>> print(clean(mission("I have 20 carrots and 1 mushroom.")))
        None
    '''
    return re.sub("[][']",'',str(s)) # as a regular expression

    return str(s).replace('[','').replace(']','').replace("'",'')   # with builtin string methods

map(function,iterable)
Suppose function(object_x) returns object_y .
Then if iterating iterable gives objects_x
map(function,objects_x)
is an iterable that, when iterated gives objects_y . I've stated this confusingly, obtusely, but generally. map changed between python 2 and 3. In python 3 map uses lazy evaluation, it avoids calling the function until the result is needed. map isn't so hard.
For example, in python 2 and 3

Code:

>>> def double(a): return 2*a
... 

>>> list(map(double,[3,'ho']))
[6, 'hoho']
>>>

January 26th, 2013, 09:07 PM

taeBaby

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thank you, I understand the map function now

Also, thanks for the other code, I've implemented it and although it removed the brackets and double quotes, it still seems to have the single quotes, at least it's replaced it all with the single quote.

My output is closer to what I want but instead of (1,0) I now get '(1,0)'

Is there a way to get rid of ' around the parentheses? Or is there a way to replace ' with an empty str?

thank you again!

January 26th, 2013, 09:21 PM

b49P23TIvg

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>>> print( 'string' )
string
>>> 'string'
'string'
>>>

January 26th, 2013, 09:37 PM

taeBaby

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Hmm...I understand the print ('string') but I think it's different from what I actually want b/c that's the word itself: string
I'm not sure that it will work in my situation...sry if I'm a bit slow and not really explaining myself well...

I guess I'll continue to play around with the code and google some more.

this is my code now:

Code:

def mission(s):
  first = re.findall(r'Recon Mission (\d+)', s, re.IGNORECASE)
  second = re.findall(r'found: (\d+)', s)
  if first and second:
    s = first, second
    x = str(s).replace('[','').replace(']','').replace("'",'')
  elif first and second !=None:
    second = 'failed'
    t = first, second
    x = str(t).replace('[','').replace(']','').replace("'",'')
  else:
    x = None
  return x

My output:

Code:

'(1, 0)'
'(2, 500)'
'(3, failed)'
None

I think I'm suppose to use int() but when I tried implementing that, it didn't really work, I kept getting an error.

January 26th, 2013, 09:48 PM

b49P23TIvg

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We've come full circle, using integers then removing characters from strings to make them look like numbers.

If you truly want
(3, failed)
then it's a string, not a number and a piece that doesn't look like a number or a string.

January 26th, 2013, 10:08 PM

taeBaby

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yeah, I feel like I'm making things worse and confusing myself more. Sry about that.

I went back and decided to just implement your earlier code so now I have this:

Code:

def mission(s):
  match = map(int,re.findall(r'\d+', s))
  sl = s.lower()
  if 'mission' in sl:
    if ('accomplished' in sl) and ('found' in sl):
      s = tuple(match)
    if match and ('failed' in sl):
      s = (match[0],'failed',)
  else:
    s = None
  return s

It works the way I want it to now so thank you.

Just one last question, if I wanted to implement re.Ignorecase instead of using s.lower(), how would I go about doing that?

#10

January 27th, 2013, 12:47 PM

b49P23TIvg

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Posts: 3,364

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I removed Ignorecase because it had nothing to do with digits. Maybe with these.

Code:

  if re.search('mission',s,re.Ignorecase):
    if re.search('accomplished.*found',s,re.Ignorecase):
    if match and (re.search('failed',s,re.Ignorecase)):

untested.

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