get python dictionary from string containing key value pairs

Question

i have a python string in the format:

str = "name: srek age :24 description: blah blah"

is there any way to convert it to dictionary that looks like

{'name': 'srek', 'age': '24', 'description': 'blah blah'}  

where each entries are (key,value) pairs taken from string. I tried splitting the string to list by

str.split()  

and then manually removing :, checking each tag name, adding to a dictionary. The drawback of this method is: this method is nasty, I have to manually remove : for each pair and if there is multi word 'value' in string (for example, blah blah for description), each word will be a separate entry in a list which is not desirable. Is there any Pythonic way of getting the dictionary (using python 2.7) ?

Answer 1

without re:

r = "name: srek age :24 description: blah blah cat: dog stack:overflow"
lis=r.split(':')
dic={}
try :
 for i,x in enumerate(reversed(lis)):
    i+=1
    slast=lis[-(i+1)]
    slast=slast.split()
    dic[slast[-1]]=x

    lis[-(i+1)]=" ".join(slast[:-1])
except IndexError:pass    
print(dic)

{'age': '24', 'description': 'blah blah', 'stack': 'overflow', 'name': 'srek', 'cat': 'dog'}

Answer 2

>>> r = "name: srek age :24 description: blah blah"
>>> import re
>>> regex = re.compile(r"\b(\w+)\s*:\s*([^:]*)(?=\s+\w+\s*:|$)")
>>> d = dict(regex.findall(r))
>>> d
{'age': '24', 'name': 'srek', 'description': 'blah blah'}

Explanation:

\b           # Start at a word boundary
(\w+)        # Match and capture a single word (1+ alnum characters)
\s*:\s*      # Match a colon, optionally surrounded by whitespace
([^:]*)      # Match any number of non-colon characters
(?=          # Make sure that we stop when the following can be matched:
 \s+\w+\s*:  #  the next dictionary key
|            # or
 $           #  the end of the string
)            # End of lookahead