Converting colon separated list into a dict?
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?
Answers:
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))
You can do:
>>> li=['a:1', 'b:2', 'c:3']
>>> dict(e.split(':') for e in li)
{'a': '1', 'c': '3', 'b': '2'}
If the list of strings require stripping, you can do:
>>> li=["a:1n", "b:2n", "c:3n"]
>>> dict(t.split(":") for t in map(str.strip, li))
{'a': '1', 'b': '2', 'c': '3'}
Or, also:
>>> dict(t.split(":") for t in (s.strip() for s in li))
{'a': '1', 'b': '2', 'c': '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.
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?
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))
You can do:
>>> li=['a:1', 'b:2', 'c:3']
>>> dict(e.split(':') for e in li)
{'a': '1', 'c': '3', 'b': '2'}
If the list of strings require stripping, you can do:
>>> li=["a:1n", "b:2n", "c:3n"]
>>> dict(t.split(":") for t in map(str.strip, li))
{'a': '1', 'b': '2', 'c': '3'}
Or, also:
>>> dict(t.split(":") for t in (s.strip() for s in li))
{'a': '1', 'b': '2', 'c': '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.