dict invert - learning question

[dave]

Hello,

here is a piece of code I wrote to check the frequency of values and
switch them around to keys in a new dictionary. Just to measure how
many times a certain key occurs:

def invert(d):
inv = {}
for key in d:
val = d[key]
if val not in inv:
inv.setdefault( val, [key])
else:
inv[val].append(key)
return inv
Using the methods above (I'm just a beginner) could I have written it
more concisely? Any criticism/critique in the code would be greatly
appreciated.

Many Thanks,

dave

[Duncan Booth]

dave <sq************ *@y1a2hoo3.comw rote:

Hello,

here is a piece of code I wrote to check the frequency of values and
switch them around to keys in a new dictionary. Just to measure how
many times a certain key occurs:

def invert(d):
inv = {}
for key in d:
val = d[key]
if val not in inv:
inv.setdefault( val, [key])
else:
inv[val].append(key)
return inv
Using the methods above (I'm just a beginner) could I have written it
more concisely? Any criticism/critique in the code would be greatly
appreciated.

The obvious change I'd make is:
if val not in inv:
inv[val] = key
else:
... etc.

The setdefault method is only appropriate if you don't know
whether or not the key is in the dictionary. since you've already tests
that you should just go ahead and set the value.
Otherwise it all seems fine.

[mrkafk]

Assuming all the values are unique:

>>a={1:'a', 2:'b', 3:'c'}

>>dict(zip(a.ke ys(), a.values()))

{1: 'a', 2: 'b', 3: 'c'}

The problem is you obviously can't assume that in most cases.

Still, zip() is very useful function.

[mrkafk]

On 4 Maj, 01:27, [e-mail] m wrote:

>a={1:'a', 2:'b', 3:'c'}

Oops, it should obviously be:

>>dict(zip(a.va lues(), a.keys()))

{'a': 1, 'c': 3, 'b': 2}

[Jerry Hill]

On Sat, May 3, 2008 at 4:29 PM, dave <sq************ *@y1a2hoo3.comw rote:

here is a piece of code I wrote to check the frequency of values and switch
them around to keys in a new dictionary. Just to measure how many times a
certain key occurs:

def invert(d):
inv = {}
for key in d:
val = d[key]
if val not in inv:
inv.setdefault( val, [key])
else:
inv[val].append(key)
return inv
Using the methods above (I'm just a beginner) could I have written it more
concisely? Any criticism/critique in the code would be greatly appreciated.

If you're using python 2.5, the collections module has the defaultdict
type that is useful for things like this.

from collections import defaultdict

def invert2(d):
inv = defaultdict(lis t)
for key, value in d.items():
inv[value] += key
return dict(inv)

If you don't mind returning a defaultdict instead of a regular
dictionary, you could just 'return inv'. The defaultdict is useful
for building up the frequency counts in the first place too:

def wordcount(wordl ist):
wc = defaultdict(int )
for word in wordlist:
wc[word] += 1
return dict(wc)
--
Jerry