Velocity Reviews > [dictionary] how to get key by item

# [dictionary] how to get key by item

Egor Bolonev
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Posts: n/a

 12-14-2004
saluton al ciuj

i know how to get item by key

==================
dict = {10 : 50, 2 : 12, 4 : 43}

print dict[2]

>> 12

but i wonder how to get key by item

print dict[12]

>> 2

==================

is there a more fast way than that one (my dictionary is really big)
==================
dict = {10 : 50, 2 : 12, 4 : 43}
item = 12
for key in dict.keys():
if dict[key] == item:
print key
break
==================

Skip Montanaro
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Posts: n/a

 12-14-2004

Egor> i know how to get item by key
...
Egor> but i wonder how to get key by item

Assuming your dictionary defines a one-to-one mapping, just invert it:

>>> forward = {10 : 50, 2 : 12, 4 : 43}
>>> reverse = dict([(v,k) for (k,v) in forward.iteritems()])
>>> print forward

{10: 50, 4: 43, 2: 12}
>>> print reverse

{50: 10, 43: 4, 12: 2}

That doubles your storage, so you'll have to trade that off against the
speed gain of not having to loop over the entire dictionary.

Skip

Roy Smith
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Posts: n/a

 12-14-2004
In article <(E-Mail Removed)>,
Skip Montanaro <(E-Mail Removed)> wrote:

> Egor> i know how to get item by key
> ...
> Egor> but i wonder how to get key by item
>
> Assuming your dictionary defines a one-to-one mapping, just invert it:
>
> >>> forward = {10 : 50, 2 : 12, 4 : 43}
> >>> reverse = dict([(v,k) for (k,v) in forward.iteritems()])
> >>> print forward

> {10: 50, 4: 43, 2: 12}
> >>> print reverse

> {50: 10, 43: 4, 12: 2}
>
> That doubles your storage, so you'll have to trade that off against the
> speed gain of not having to loop over the entire dictionary.

Well, you *do* loop over the entire dictionary, but you only do it once,
when you create the reverse dict. If you are only going to do a single
lookup, it's no gain, but if you amortize the cost over many lookups,
it's almost certainly a big win.

This raises an interesting question. Let's assume that you add all the
entries to the dictionary before you do any lookups, and you then need
to lookup things up in both directions. Which is faster, to
simultaneously build both the forward and reverse dicts, or to just
build the forward one and when you're done doing that, build the reverse
one in a single shot with the above list comprehension?

BTW, does Python really build the intermediate list and throw it away
after using it to initialize the dictionary, or is it smart enough to
know that it doesn't really need to build the whole list in memory?

Skip Montanaro
Guest
Posts: n/a

 12-14-2004

>> That doubles your storage, so you'll have to trade that off against
>> the speed gain of not having to loop over the entire dictionary.

Roy> Well, you *do* loop over the entire dictionary, but you only do it
Roy> once, when you create the reverse dict. If you are only going to
Roy> do a single lookup, it's no gain, but if you amortize the cost over
Roy> many lookups, it's almost certainly a big win.

Sure, but the OP said his dictionary was big. It's up to him to decide
whether the space-time tradeoff is worth it (or even possible).

Roy> BTW, does Python really build the intermediate list and throw it
Roy> away after using it to initialize the dictionary, or is it smart
Roy> enough to know that it doesn't really need to build the whole list
Roy> in memory?

That's why I called .iteritems() in my example. It won't generate the
entire list of tuples as .items() would.

Skip

Roy Smith
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Posts: n/a

 12-14-2004
Skip Montanaro <(E-Mail Removed)> wrote:

> Roy> BTW, does Python really build the intermediate list and throw it
> Roy> away after using it to initialize the dictionary, or is it smart
> Roy> enough to know that it doesn't really need to build the whole list
> Roy> in memory?
>
> That's why I called .iteritems() in my example. It won't generate the
> entire list of tuples as .items() would.

I know it won't generate the list of items from the forward dict, but I
was thinking of the list generated by the list comprehension, passed as
the argument to the reverse dict constructor. That's the throw-away
list I was thinking of (see Tim Delaney's response to my post).

Aahz
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Posts: n/a

 12-14-2004
In article <(E-Mail Removed)>,
Skip Montanaro <(E-Mail Removed)> wrote:
>
>Assuming your dictionary defines a one-to-one mapping, just invert it:
>
> >>> forward = {10 : 50, 2 : 12, 4 : 43}
> >>> reverse = dict([(v,k) for (k,v) in forward.iteritems()])
> >>> print forward

> {10: 50, 4: 43, 2: 12}
> >>> print reverse

> {50: 10, 43: 4, 12: 2}
>
>That doubles your storage, so you'll have to trade that off against the
>speed gain of not having to loop over the entire dictionary.

To be precise, it doubles the storage of the *dictionary*, but it does
*NOT* double the storage of the keys and items. Depending on how big
those are, the cost of building a second dict might be mostly lost in the
noise.
--
Aahz ((E-Mail Removed)) <*> http://www.pythoncraft.com/

"19. A language that doesn't affect the way you think about programming,
is not worth knowing." --Alan Perlis

Keith Dart
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Posts: n/a

 12-14-2004
Skip Montanaro wrote:
> Egor> i know how to get item by key
> ...
> Egor> but i wonder how to get key by item
>
> Assuming your dictionary defines a one-to-one mapping, just invert it:
>
> >>> forward = {10 : 50, 2 : 12, 4 : 43}
> >>> reverse = dict([(v,k) for (k,v) in forward.iteritems()])
> >>> print forward

> {10: 50, 4: 43, 2: 12}
> >>> print reverse

> {50: 10, 43: 4, 12: 2}
>
> That doubles your storage, so you'll have to trade that off against the
> speed gain of not having to loop over the entire dictionary.
>
> Skip

But beware that all the items in the original dictionary must be
hashable. The example shows just integers, so I assume they are in this
case. But generally, this may not work.

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Keith Dart <(E-Mail Removed)>
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Fredrik Lundh
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Posts: n/a

 12-14-2004
Skip Montanaro wrote:

careful: it creates another dictionary structure with the same size as the first
one, but it doesn't copy the objects in the dictionary.

so whether it doubles the actual memory usage depends on what data you
have in the dictionary (last time I checked, ints and dictionary slots were the
same size, but I cannot think of any other object that isn't larger...)

(but you knew that, of course)

</F>

Ola Natvig
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Posts: n/a

 12-14-2004
Skip Montanaro wrote:
> Egor> i know how to get item by key
> ...
> Egor> but i wonder how to get key by item
>
> Assuming your dictionary defines a one-to-one mapping, just invert it:
>
> >>> forward = {10 : 50, 2 : 12, 4 : 43}
> >>> reverse = dict([(v,k) for (k,v) in forward.iteritems()])
> >>> print forward

> {10: 50, 4: 43, 2: 12}
> >>> print reverse

> {50: 10, 43: 4, 12: 2}
>
> That doubles your storage, so you'll have to trade that off against the
> speed gain of not having to loop over the entire dictionary.
>
> Skip

If some keys has the same value as the item this will cause problems
because keys in your result dictionary can be overwritten. Could it be a
option to build the result dictionary as a dictionary with the values
as the keys, and lists of keys as the value. Perhaps you need to use a
loop for this.

--
--------------------------------------
Ola Natvig <(E-Mail Removed)>
infoSense AS / development

Nick Coghlan
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Posts: n/a

 12-14-2004
Ola Natvig wrote:
> If some keys has the same value as the item this will cause problems
> because keys in your result dictionary can be overwritten. Could it be a
> option to build the result dictionary as a dictionary with the values
> as the keys, and lists of keys as the value. Perhaps you need to use a
> loop for this.
>

<<Python 2.4>>

..>>> d = dict(foo=1, bar=1, bob=7, jane=42, mary=16, fred=16)
..>>> from itertools import groupby
..>>> val = d.__getitem__
..>>> grouped = groupby(sorted(d.iterkeys(), key=val), val)
..>>> r = dict((value, list(keys)) for value, keys in grouped)
..>>> r
{16: ['mary', 'fred'], 1: ['bar', 'foo'], 42: ['jane'], 7: ['bob']}

Cheers,
Nick.

--
Nick Coghlan | http://www.velocityreviews.com/forums/(E-Mail Removed) | Brisbane, Australia
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