String separation in required format, Pythonic way

Question

I have a string in the format:

t='@abc @def Hello this part is text'

I want to get this:

l=["abc", "def"] 
s='Hello this part is text'

I did this:

a=t[t.find(' ',t.rfind('@')):].strip()
s=t[:t.find(' ',t.rfind('@'))].strip()
b=a.split('@')
l=[i.strip() for i in b][1:]

It works for the most part, but it fails when the text part has the '@'. Eg, when:

t='@abc @def My email is [e-mail]'

it fails. The @names are there in the beginning and there can be text after @names, which may possibly contain @.

Clearly I can append initally with a space and find out first word without '@'. But that doesn't seem an elegant solution.

What is a pythonic way of solving this?

Answer 1

[edit: this is implementing what was suggested by Osama above]

This will create L based on the @ variables from the beginning of the string, and then once a non @ var is found, just grab the rest of the string.

t = '@one @two @three some text   afterward with @ symbols@ meow@meow'

words = t.split(' ')         # split into list of words based on spaces
L = []
s = ''
for i in range(len(words)):  # go through each word
    word = words[i]
    if word[0] == '@':       # grab @'s from beginning of string
        L.append(word[1:])
        continue
    s = ' '.join(words[i:])  # put spaces back in
    break                    # you can ignore the rest of the words

You can refactor this to be less code, but I'm trying to make what is going on obvious.

Answer 2

t='@abc @def Hello this part is text'

words = t.split(' ')

names = []
while words:
    w = words.pop(0)
    if w.startswith('@'):
        names.append(w[1:])
    else:
        break

text = ' '.join(words)

print names
print text

Answer 3

 [i.strip('@') for i in t.split(' ', 2)[:2]]     # for a fixed number of @def
 a = [i.strip('@') for i in t.split(' ') if i.startswith('@')]
 s = ' '.join(i for i in t.split(' ') if not i.startwith('@'))

Answer 4

You might also use regular expressions:

import re
rx = re.compile("@([\w]+) @([\w]+) (.*)")
t='@abc @def Hello this part is text and my email is foo@ba.r'
a,b,s = rx.match(t).groups()

But this all depends on how your data can look like. So you might need to adjust it. What it does is basically creating group via () and checking for what's allowed in them.

Answer 5

Here's just another variation that uses split() and no regexpes:

t='@abc @def My email is [e-mail]'
tags = []
words = iter(t.split())

# iterate over words until first non-tag word
for w in words:
  if not w.startswith("@"):
    # join this word and all the following
    s = w + " " + (" ".join(words))
    break
  tags.append(w[1:])
else:
  s = "" # handle string with only tags

print tags, s

Here's a shorter but perhaps a bit cryptic version that uses a regexp to find the first space followed by a non-@ character:

import re
t = '@abc @def My email is [e-mail] @extra bye'
m = re.search(r"\s([^@].*)$", t)
tags = [tag[1:] for tag in t[:m.start()].split()]
s = m.group(1)
print tags, s # ['abc', 'def'] My email is [e-mail] @extra bye

This doesn't work properly if there are no tags or no text. The format is underspecified. You'll need to provide more test cases to validate.

Answer 6

How about this:

1. Splitting by space.

2. foreach word, check

   2.1. if word starts with @ then Push to first list

   2.2. otherwise just join the remaining words by spaces.