将冒号分隔的列表转换为字典？

Question

I wrote something like this to convert comma separated list to a dict.

def list_to_dict( rlist ) :
    rdict = {}
    i = len (rlist)
    while i:
        i = i - 1
        try :
            rdict[rlist[i].split(":")[0].strip()] = rlist[i].split(":")[1].strip()
        except :
            print rlist[i] + ' Not a key value pair'
            continue


    return rdict

Isn't there a way to

for i, row = enumerate rlist
    rdict = tuple ( row ) 

or something?

Answer 1

If I understand your requirements correctly, then you can use the following one-liner.

def list_to_dict(rlist):
    return dict(map(lambda s : s.split(':'), rlist))

Example:

>>> list_to_dict(['alpha:1', 'beta:2', 'gamma:3'])
{'alpha': '1', 'beta': '2', 'gamma': '3'}

You might want to strip() the keys and values after splitting in order to trim white-space.

return dict(map(lambda s : map(str.strip, s.split(':')), rlist))

Answer 2

You can do:

>>> li=['a:1', 'b:2', 'c:3']
>>> dict(e.split(':') for e in li)
{'a': '1', 'c': '3', 'b': '2'}

Answer 3

You mention both colons and commas so perhaps you have a string with key/values pairs separated by commas, and with the key and value in turn separated by colons, so:

def list_to_dict(rlist):
    return {k.strip():v.strip() for k,v in (pair.split(':') for pair in rlist.split(','))}

>>> list_to_dict('a:1,b:10,c:20')
{'a': '1', 'c': '20', 'b': '10'}
>>> list_to_dict('a:1, b:10, c:20')
{'a': '1', 'c': '20', 'b': '10'}
>>> list_to_dict('a   :    1       , b:    10, c:20')
{'a': '1', 'c': '20', 'b': '10'}

This uses a dictionary comprehension iterating over a generator expression to create a dictionary containing the key/value pairs extracted from the string. strip() is called on the keys and values so that whitespace will be handled.