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get keys with the same values

P: n/a
Hello,

I have a dictionary and will get all keys which have the same values.

d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}

I will something as :

d.keys(where their values are the same)

With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']

Would somebody tell me how I can do it?

Regards,
Nader
Jun 27 '08 #1
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13 Replies


P: n/a
Nader:
d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}
I will something as :
d.keys(where their values are the same)
That's magic.

With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']
Would somebody tell me how I can do it?
You can create a new dict where the keys are the values of the input
dict and the values are a list of the keys of the original dict. So
scanning the keys, values of the input dict, you can fill the second
dict. Then you can scan the second dict, and create a list that
contains only value lists longer than one.

Bye,
bearophile
Jun 27 '08 #2

P: n/a
On Thu, 12 Jun 2008 03:58:53 -0700 (PDT), Nader <n.*****@gmail.com>
wrote:
>Hello,

I have a dictionary and will get all keys which have the same values.

d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}
That's not a dictionary, it's a syntax error. If you actually
have a dictionary you could say

d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}

dd = {}

for key, value in d.items():
try:
dd[value].append(key)
except KeyError:
dd[value] = [key]

Possibly dd is now what you really want; if you really
want what you said you want you could use

[l for l in dd.values() if len(l) 1]
>I will something as :

d.keys(where their values are the same)

With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']

Would somebody tell me how I can do it?

Regards,
Nader
David C. Ullrich
Jun 27 '08 #3

P: n/a
On Jun 12, 1:35 pm, bearophileH...@lycos.com wrote:
Nader:
d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}
I will something as :
d.keys(where their values are the same)

That's magic.
With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']
Would somebody tell me how I can do it?

You can create a new dict where the keys are the values of the input
dict and the values are a list of the keys of the original dict. So
scanning the keys, values of the input dict, you can fill the second
dict. Then you can scan the second dict, and create a list that
contains only value lists longer than one.

Bye,
bearophile
Is it niet possible with one or two statement, maybe with list
comprehension. For exmple:

l = [(k,v) for k in d.keys() for v in d.values() | en here we need
some extra logic (v = 1)]

I don;t konw how we can define a logic statement in a list
comprehension.
It will be very compact, if it would possible.

Nader
Jun 27 '08 #4

P: n/a
On Jun 12, 1:41 pm, David C. Ullrich <dullr...@sprynet.comwrote:
On Thu, 12 Jun 2008 03:58:53 -0700 (PDT), Nader <n.em...@gmail.com>
wrote:
Hello,
I have a dictionary and will get all keys which have the same values.
d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}

That's not a dictionary, it's a syntax error. If you actually
have a dictionary you could say

d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}

dd = {}

for key, value in d.items():
try:
dd[value].append(key)
except KeyError:
dd[value] = [key]

Possibly dd is now what you really want; if you really
want what you said you want you could use

[l for l in dd.values() if len(l) 1]
I will something as :
d.keys(where their values are the same)
With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']
Would somebody tell me how I can do it?
Regards,
Nader

David C. Ullrich
Thank for your type about the syntax error. This an example example,
the keys of my dictionary are tuples:
d = {(37.75, 42.22): 1 , (37.51, 40.02): 3 (45.55, 24.27): 4 (47.08,
30.99) : 1}

But what I will is to get all keys which has the same valus. And not
the keys that have value more than 1!

Nader
Jun 27 '08 #5

P: n/a
On Jun 12, 1:48*pm, Nader <n.em...@gmail.comwrote:
On Jun 12, 1:35 pm, bearophileH...@lycos.com wrote:
Nader:
d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}
I will something as :
d.keys(where their values are the same)
That's magic.
With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']
Would somebody tell me how I can do it?
You can create a new dict where the keys are the values of the input
dict and the values are a list of the keys of the original dict. So
scanning the keys, values of the input dict, you can fill the second
dict. Then you can scan the second dict, and create a list that
contains only value lists longer than one.
Bye,
bearophile

Is it niet possible with one or two statement, maybe with list
comprehension. For exmple:

l = [(k,v) for k in d.keys() for v in d.values() | en here we need
some extra logic (v = 1)]

I don;t konw how we can define a logic statement in a list
comprehension.
It will be very compact, if it would possible.

Nader
If you are going to use this reverse look-up alot you'd be better off
building another dictionary with the original values being keys and
the original keys being values, if it is used infrequently enough you
can search for it with result_list = [k for k,v in dictionary.items()
if v == search_value]
Jun 27 '08 #6

P: n/a
On Jun 12, 2:05 pm, Chris <cwi...@gmail.comwrote:
On Jun 12, 1:48 pm, Nader <n.em...@gmail.comwrote:
On Jun 12, 1:35 pm, bearophileH...@lycos.com wrote:
Nader:
d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}
I will something as :
d.keys(where their values are the same)
That's magic.
With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']
Would somebody tell me how I can do it?
You can create a new dict where the keys are the values of the input
dict and the values are a list of the keys of the original dict. So
scanning the keys, values of the input dict, you can fill the second
dict. Then you can scan the second dict, and create a list that
contains only value lists longer than one.
Bye,
bearophile
Is it niet possible with one or two statement, maybe with list
comprehension. For exmple:
l = [(k,v) for k in d.keys() for v in d.values() | en here we need
some extra logic (v = 1)]
I don;t konw how we can define a logic statement in a list
comprehension.
It will be very compact, if it would possible.
Nader

If you are going to use this reverse look-up alot you'd be better off
building another dictionary with the original values being keys and
the original keys being values, if it is used infrequently enough you
can search for it with result_list = [k for k,v in dictionary.items()
if v == search_value]
Thank you! It is the anwser which I was looking for. [(k,v) for k,v
in d.items() if v is pattern].
But I don't understand what tou mean of "reverse look-up a lot"! I
have to read some informations inclusive (latitudes and longitudes)
form a file and after some processing to save part of this information
to other file.
Why do I make a new dictionary?

Nader
Jun 27 '08 #7

P: n/a
Nader <n.*****@gmail.comwrote:
I have a dictionary and will get all keys which have the same values.

d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}

I will something as :

d.keys(where their values are the same)

With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']

Would somebody tell me how I can do it?
Here's one way:
>>import itertools, functools, operator
d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}
get = functools.partial(operator.getitem, d)
for (k,v) in itertools.groupby(sorted(d, key=get), key=get):
print k, list(v)
1 ['a', 'e']
2 ['c']
3 ['b', 'd']
4 ['f']

or if you want a dictionary:
>>dict((k,list(v)) for (k,v) in
itertools.groupby(sorted(d, key=get), key=get))
{1: ['a', 'e'], 2: ['c'], 3: ['b', 'd'], 4: ['f']}

--
Duncan Booth http://kupuguy.blogspot.com
Jun 27 '08 #8

P: n/a
On Jun 12, 2:15*pm, Nader <n.em...@gmail.comwrote:
On Jun 12, 2:05 pm, Chris <cwi...@gmail.comwrote:
On Jun 12, 1:48 pm, Nader <n.em...@gmail.comwrote:
On Jun 12, 1:35 pm, bearophileH...@lycos.com wrote:
Nader:
d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}
I will something as :
d.keys(where their values are the same)
That's magic.
With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']
Would somebody tell me how I can do it?
You can create a new dict where the keys are the values of the input
dict and the values are a list of the keys of the original dict. So
scanning the keys, values of the input dict, you can fill the second
dict. Then you can scan the second dict, and create a list that
contains only value lists longer than one.
Bye,
bearophile
Is it niet possible with one or two statement, maybe with list
comprehension. For exmple:
l = [(k,v) for k in d.keys() for v in d.values() | en here we need
some extra logic (v = 1)]
I don;t konw how we can define a logic statement in a list
comprehension.
It will be very compact, if it would possible.
Nader
If you are going to use this reverse look-up alot you'd be better off
building another dictionary with the original values being keys and
the original keys being values, if it is used infrequently enough you
can search for it with result_list = [k for k,v in dictionary.items()
if v == search_value]

Thank you! It is the anwser which I was looking for. [(k,v) for k,v
in *d.items() if v is pattern].
But I don't understand what tou mean of "reverse look-up a lot"! I
have to read some informations inclusive (latitudes and longitudes)
form a file and after some processing to save part of this information
to other file.
Why do I make a new dictionary?

Nader
If you are just going to perform the lookup once or twice then it's
fine to traverse (step through) your original dictionary. If you are
going to look up data often though it might be a better idea to build
another dictionary with the reverse of your original dictionary as it
will yield faster results.

For example, your original dictionary of values is built and then you
perform a handful of operations and move on, then use the list
comprehension to get your data and move on. If, on the other hand,
you build your dictionary and then maybe iterate over a file and need
to look-up the information for every line in the file it would be
better suited to build a new dictionary that transposed the key and
value pairs for less resource intensive and faster operation. eg:

reverse_dict = {}
for k,v in original_dict:
if v in reverse_dict:
reverse_dict[v].append(k)
else:
reverse_dict[v] = [k]

Then once that is built and you want to find which keys in the
original dictionary have the value of "1" you can just do
"list_of_keys = reverse_dict[1]".

Essentially if you are going to do alot of searching for values that
match values found in a dictionary you would be better off to create
the new data structure.

Hope that helps.
Jun 27 '08 #9

P: n/a
On Jun 12, 6:41*am, David C. Ullrich <dullr...@sprynet.comwrote:
On Thu, 12 Jun 2008 03:58:53 -0700 (PDT), Nader <n.em...@gmail.com>
wrote:
Hello,
I have a dictionary and will get all keys which have the same values.
<snip>
>
d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}

dd = {}

for key, value in d.items():
* try:
* * dd[value].append(key)
* except KeyError:
* * dd[value] = [key]
<snip>

Instead of all that try/except noise, just use the new defaultdict:
>>from collections import defaultdict

d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}

dd = defaultdict(list)
for key, value in d.items():
... dd[value].append(key)
...
>>for k,v in dd.items():
... print k,':',v
...
1 : ['a', 'e']
2 : ['c']
3 : ['b', 'd']
4 : ['f']

-- Paul
Jun 27 '08 #10

P: n/a
In article
<d4**********************************@p25g2000hsf. googlegroups.com>,
Nader <n.*****@gmail.comwrote:
On Jun 12, 1:41 pm, David C. Ullrich <dullr...@sprynet.comwrote:
On Thu, 12 Jun 2008 03:58:53 -0700 (PDT), Nader <n.em...@gmail.com>
wrote:
>Hello,
>I have a dictionary and will get all keys which have the same values.
>d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}
That's not a dictionary, it's a syntax error. If you actually
have a dictionary you could say

d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}

dd = {}

for key, value in d.items():
try:
dd[value].append(key)
except KeyError:
dd[value] = [key]

Possibly dd is now what you really want; if you really
want what you said you want you could use

[l for l in dd.values() if len(l) 1]
>I will something as :
>d.keys(where their values are the same)
>With this statement I can get two lists for this example:
>l1= ['a','e']
>l2=['b','d']
>Would somebody tell me how I can do it?
>Regards,
>Nader
David C. Ullrich

Thank for your type about the syntax error. This an example example,
the keys of my dictionary are tuples:
d = {(37.75, 42.22): 1 , (37.51, 40.02): 3 (45.55, 24.27): 4 (47.08,
30.99) : 1}

But what I will is to get all keys which has the same valus.
That's exactly what the code I posted does.
And not
the keys that have value more than 1!
It doesn't do that. Or if you prefer, it doesn't do that!

Instead of looking at it and trying to figure out what it does,
which of course can be tricky, try _running_ the code.
Then print dd. Then print [l for l in dd.values() if len(l) 1] .
>
Nader
--
David C. Ullrich
Jun 27 '08 #11

P: n/a
In article
<3f**********************************@e39g2000hsf. googlegroups.com>,
Paul McGuire <pt***@austin.rr.comwrote:
On Jun 12, 6:41*am, David C. Ullrich <dullr...@sprynet.comwrote:
On Thu, 12 Jun 2008 03:58:53 -0700 (PDT), Nader <n.em...@gmail.com>
wrote:
>Hello,
>I have a dictionary and will get all keys which have the same values.
<snip>

d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}

dd = {}

for key, value in d.items():
* try:
* * dd[value].append(key)
* except KeyError:
* * dd[value] = [key]
<snip>

Instead of all that try/except noise, just use the new defaultdict:
That's certainly much better. The machine where I am in
the early morning is still stuck with Python 1.5.3...
>from collections import defaultdict

d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}

dd = defaultdict(list)
for key, value in d.items():
... dd[value].append(key)
...
>for k,v in dd.items():
... print k,':',v
...
1 : ['a', 'e']
2 : ['c']
3 : ['b', 'd']
4 : ['f']

-- Paul
--
David C. Ullrich
Jun 27 '08 #12

P: n/a
Paul McGuire <pt***@austin.rr.comwrote:
Instead of all that try/except noise, just use the new defaultdict:
>from collections import defaultdict

d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}

dd = defaultdict(list)
for key, value in d.items():
... dd[value].append(key)
...
>for k,v in dd.items():
... print k,':',v
...
1 : ['a', 'e']
2 : ['c']
3 : ['b', 'd']
4 : ['f']
Or use the little understood dict.setdefault method which has been
with us from time immemorial...
>>d = {'a' : 1, 'b' : 3, 'c' : 2,'d' : 3,'e' : 1,'f' : 4}
dd = {}
for k, v in d.items():
.... dd.setdefault(v, []).append(k)
....
>>dd
{1: ['a', 'e'], 2: ['c'], 3: ['b', 'd'], 4: ['f']}
>>>
--
Nick Craig-Wood <ni**@craig-wood.com-- http://www.craig-wood.com/nick
Jun 27 '08 #13

P: n/a
You could use my mseqdict implementation of a sorted dict.

http://home.arcor.de/wolfgang.grafen...s/Modules.html

swap:
This method can only be applied when all values of the dictionary are
immutable. The Python dictionary cannot hold mutable keys! So swap
doesn't work if only one of the values has the type list or dictionary.
Tuples and instances of classes are save as long as they don't emulate
lists or dictionaries.

from
http://home.arcor.de/wolfgang.grafen...Mseqdict.html:
>>x=seqdict.mseqdict(dict)
x['Bild']='painting'
x['Ziel']='goal'
x['Tor'] ='goal'
x # A small German - English dictionary
mseqdict(
['gewinnen', 'deshalb', 'Abend', 'aber', 'Bild', 'Erkennung',
'Fl\366te', 'Ziel', 'Tor'],
{'Tor': ['goal'], 'Ziel': ['goal'], 'gewinnen': ['gain'], 'deshalb':
['therefore'], 'Abend': ['evening'], 'aber': ['but'], 'Bild':
['picture', 'painting'], 'Erkennung': ['recognition'], 'Fl\366te':
['flute']})
>>x.swap()
x # A small English - German dictionary
mseqdict(
['gain', 'therefore', 'evening', 'but', 'picture', 'painting',
'recognition', 'flute', 'goal'],
{'but': ['aber'], 'recognition': ['Erkennung'], 'painting': ['Bild'],
'flute': ['Fl\366te'], 'gain': ['gewinnen'], 'goal': ['Ziel', 'Tor'],
'evening': ['Abend'], 'therefore': ['deshalb'], 'picture': ['Bild']})

Best regards

Wolfgang
Nader schrieb:
Hello,

I have a dictionary and will get all keys which have the same values.

d = {('a' : 1), ('b' : 3), ('c' : 2),('d' : 3),('e' : 1),('f' : 4)}

I will something as :

d.keys(where their values are the same)

With this statement I can get two lists for this example:
l1= ['a','e']
l2=['b','d']

Would somebody tell me how I can do it?

Regards,
Nader
Jun 27 '08 #14

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