python: what is go keyword in generator expression while operating dictionary -

this simple query while checking out various discussions on dictionary comprehension on here. did this:

newlist=[{v:k} k,v in dicnew.items() if v==value] >>> newlist [{2: 'a'}, {2: 'ac'}, {2: 'b'}, {2: 'love'}] 

what did next this:

newlist.setdefault(v,[]).append((v,k) k,v in dicnew.items() if v==value) >>> newlist {'go': [<generator object <genexpr> @ 0x01e98d50>]} 

what happened? 'go'?

you called .setdefault() a value of v:

newlist.setdefault(v,[]) 

and v must have been defined , set 'go'. had run code in new interpreter, or executed del v first, python have raised nameerror exception instead:

>>> newlist = {} >>> newlist.setdefault(v,[]).append((v,k) k,v in dicnew.items() if v==value) traceback (most recent call last):   file "<stdin>", line 1, in <module> nameerror: name 'v' not defined >>> v = 'go' >>> newlist.setdefault(v, []) [] >>> newlist {'go': []} 

the setdefault() part executed before .append() method executed. v name used in generator expression inside .append() method has nothing v name used in .setdefault() call.

in python 2.7 , earlier, list comprehension variables 'leak' parent scope:

>>> [foo foo in range(3)] [0, 1, 2] >>> foo 2 

as such, v set in previous loop, , last value assigned 'go'.

if wanted 'invert' dictionary collecting list of keys each value, use:

from collections import defaultdict  keys_for_value = defaultdict(list) key, value in original_dict.iteritems():     keys_for_value[value].append(key) 

if must insist on one-liner, use itertools.groupby , sorting:

from itertools import groupby operator import itemgetter  v = itemgetter(1) keys_for_value = {value: [k k, v in items] value, items in groupby(sorted(original_dict.iteritems(), key=v, key=v)} 

this going slower, need sort dictionary items first (cost o(n log n)) before looping on sorted results (itself o(n), total of o(n) + o(n log n)), opposed simple o(n) complexity of using defaultdict , for loop.