Dictionary sorting problem

[JerryB]

Hi,
I have a dictionary for counting ocurrences of strings in a document.
The dictionary looks like this:

'hello':135
'goodbye':30
'lucy':4
'sky':55
'diamonds':239843
'yesterday':4

I want to print the dictionary so I see most common words first:

'diamonds':239843
'hello':135
'sky':55
'goodbye':30
'lucy':4
'yesterday':4

How do I do this? Notice I can't 'swap' the dictionary (making keys
values and values keys) and sort because I have values like lucy &
yesterday which have the same number of occurrences.

Thanks.

[Irmen de Jong]

Don't try to 'swap' the dict, just sort a list based on the items in
the dict. Try this:

original= {
'hello':135,
'goodbye':30,
'lucy':4,
'sky':55,
'diamonds':239843,
'yesterday':4 }

items = sorted( (v,k) for (k,v) in original.iteritems() )
items.reverse() # depending on what order you want
print items

The result is:
[(239843, 'diamonds'), (135, 'hello'), (55, 'sky'), (30, 'goodbye'), (4, 'yesterday'),
(4, 'lucy')]

--Irmen

[Jason Mobarak]

You can't sort dictionaries (as implemented by hash tables), they are
unordered data types, so by definition there's no way to force an order
on them.

http://en.wikipedia.org/wiki/Hash_tables

[Bengt Richter]

or tell sorted what to do ;-)

>>> original= {
... 'hello':135,
... 'goodbye':30,
... 'lucy':4,
... 'sky':55,
... 'diamonds':239843,
... 'yesterday':4 }
>>> list(sorted(original.iteritems(), None, lambda t:t[1], True))
[('diamonds', 239843), ('hello', 135), ('sky', 55), ('goodbye', 30), ('yesterday', 4), ('lucy',4)]

Regards,
Bengt Richter

[Duncan Booth]

Bengt Richter wrote:

> or tell sorted what to do ;-)
>
> >>> original= {
> ... 'hello':135,
> ... 'goodbye':30,
> ... 'lucy':4,
> ... 'sky':55,
> ... 'diamonds':239843,
> ... 'yesterday':4 }
> >>> list(sorted(original.iteritems(), None, lambda t:t[1], True))
> [('diamonds', 239843), ('hello', 135), ('sky', 55), ('goodbye', 30),
> ('yesterday', 4), ('lucy',4)]

or a slight variation on this theme which just gives you the keys in value
order rather than the tuples:

>>> for k in sorted(original, key=original.get, reverse=True):
print k, original[k]

diamonds 239843
hello 135
sky 55
goodbye 30

yesterday 4
lucy 4

[Bengt Richter]

On 17 Sep 2005 11:01:41 GMT, Duncan Booth <duncan...@invalid.invalid> wrote:

>Bengt Richter wrote:
>
>> or tell sorted what to do ;-)
>>
>> >>> original= {
>> ... 'hello':135,
>> ... 'goodbye':30,
>> ... 'lucy':4,
>> ... 'sky':55,
>> ... 'diamonds':239843,
>> ... 'yesterday':4 }
>> >>> list(sorted(original.iteritems(), None, lambda t:t[1], True))
>> [('diamonds', 239843), ('hello', 135), ('sky', 55), ('goodbye', 30),
>> ('yesterday', 4), ('lucy',4)]
>
>or a slight variation on this theme which just gives you the keys in value
>order rather than the tuples:
>
>>>> for k in sorted(original, key=original.get, reverse=True):
> print k, original[k]
>

Nice. I like the keyword usage too. Much clearer than my hastypaste ;-)

>
>diamonds 239843
>hello 135
>sky 55
>goodbye 30
>yesterday 4
>lucy 4

Regards,
Bengt Richter