On 11/11/2011 02:31 PM, macm wrote:
> > Hi Folks
> >
> > I pass a nested dictionary to a function.
> >
> > def Dicty( dict[k1][k2] ):
> > print k1
> > print k2
> >
> > There is a fast way (trick) to get k1 and k2 as string.
> >
> > Whithout loop all dict. Just it!
> >
> > Regards
> >
> > macm
If my guessing was correct is this what you are looking for?
nesteddict = { 'a': { 'A' : 'value1 a_A' , 'B' : 'value2 a_B' },
'b': { 'A' : 'value3 b_A' , 'B' : 'value4 b_B' },
'c': { 'A' : 'value3 b_A' , 'B' : 'value4 b_B' },
}
def find_in_nested_dict(adict, avalue):
results = []
for key1, sub_dict in adict.items():
for key2, value in sub_dict.items():
if avalue == value:
results.append( (key1, key2) )
return results
def mk_lookup(adict):
lookup = {}
for key1, sub_dict in adict.items():
for key2, value in sub_dict.items():
entry = lookup.get(value, [])
entry.append( (key1, key2) )
lookup[value] = entry
return lookup
# good if you just want so search one value
value = nesteddict['c']['B']
keys = find_in_nested_dict(nesteddict, value)
print "found %r in %r" % (value, keys)
# if you need many lookups perhaps better to 'precalculate a
# 'reversed' dict
lookup = mk_lookup(nesteddict)
keys = lookup[value]
print "found %r in %r" % (value, keys)
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