How to use list as key of dictionary?

[Davy]

Hi all,

We know that list cannot be used as key of dictionary. So, how to work
around it?

For example, there is random list like l=[1,323,54,67].

Any suggestions are welcome!

Best regards,
Davy

[Boris Borcic]

Davy wrote:

Hi all,

We know that list cannot be used as key of dictionary.

Yeah, but do we know why ?

So, how to work
around it?

That's a subsidiary question.

>
For example, there is random list like l=[1,323,54,67].

Don't use 1owercase L as a variab1e name, p1ease !

>
Any suggestions are welcome!

>>{tuple([1,323,54,67]):666}

{(1, 323, 54, 67): 666}

[Matimus]

On Nov 5, 10:53 pm, Davy <zhushe...@gmai l.comwrote:

Hi all,

We know that list cannot be used as key of dictionary. So, how to work
around it?

For example, there is random list like l=[1,323,54,67].

Any suggestions are welcome!

Best regards,
Davy

Use a tuple instead.

>>d = {}
d[tuple([1,2,3,4])] = 'hello world'
d

{(1, 2, 3, 4): 'hello world'}

>>d[1,2,3,4]

'hello world'

Matt

[Davy]

Hi Matimus and Boris,

Thank you :)

And a further question about vector above rank 1, how can I use it as
the key of dictionary?

For example, if I have list like L=[[1,2,3],[4,5,6,7]],
Then I do L_tuple = tuple(L)

>>L_tuple = ([1,2,3],[4,5,6,7])

But {L_tuple:'hello '} cause an error?

Best regards,
Davy

On Nov 6, 3:09 pm, Matimus <mccre...@gmail .comwrote:

On Nov 5, 10:53 pm, Davy <zhushe...@gmai l.comwrote:

Hi all,

We know that list cannot be used as key of dictionary. So, how to work
around it?

For example, there is random list like l=[1,323,54,67].

Any suggestions are welcome!

Best regards,
Davy

Use a tuple instead.

>d = {}
d[tuple([1,2,3,4])] = 'hello world'
d

{(1, 2, 3, 4): 'hello world'}>>d[1,2,3,4]

'hello world'

Matt

[Wildemar Wildenburger]

Davy wrote:

Hi Matimus and Boris,

Thank you :)

And a further question about vector above rank 1, how can I use it as
the key of dictionary?

For example, if I have list like L=[[1,2,3],[4,5,6,7]],
Then I do L_tuple = tuple(L)

>>>L_tuple = ([1,2,3],[4,5,6,7])

But {L_tuple:'hello '} cause an error?

Yes, because your key still contains mutable elements. That should not
surprise you. If it does, please (re-)read
<URL:http://docs.python.org/tut/node7.html#SECT ION007500000000 000000000>
and <URL:http://docs.python.org/lib/typesmapping.ht ml>.

maybe something like this could help:

def tupleize(non_tu ple):
try:
return tuple(tupleize( thing) for thing in non_tuple)
except TypeError:
# non_tuple is not iterable
return non_tuple

/W

[Duncan Booth]

Wildemar Wildenburger <la*********@kl apptsowieso.net wrote:

maybe something like this could help:

def tupleize(non_tu ple):
try:
return tuple(tupleize( thing) for thing in non_tuple)
except TypeError:
# non_tuple is not iterable
return non_tuple

Just don't try passing that a string or anything containing a string.

[Dustan]

On Nov 6, 3:58 am, Duncan Booth <duncan.bo...@i nvalid.invalidw rote:

Wildemar Wildenburger <lasses_w...@kl apptsowieso.net wrote:

maybe something like this could help:

def tupleize(non_tu ple):
try:
return tuple(tupleize( thing) for thing in non_tuple)
except TypeError:
# non_tuple is not iterable
return non_tuple

Just don't try passing that a string or anything containing a string.

Untested

def tupleize(non_tu ple):
if isinstance(non_ tuple, str):
return non_tuple
try:
return tuple(tupleize( thing) for thing in non_tuple)
except TypeError:
# non_tuple is not iterable
return non_tuple

[Michael Wronna]

On Tue, 6 Nov 2007, Boris Borcic wrote:

>We know that list cannot be used as key of dictionary.

Yeah, but do we know why ?

I think, because lists are mutable and a key of a dictionary
MUST be unmutable, not to crash the dictionary by accidently
changing one of its keys!

Mike

[Davy]

And there may be more complex list(vector like 3 or 4 dimentional data
structure), is there any easy method to tackle this problem?

Any suggestions are welcome!

Best regards,
Davy

On Nov 6, 4:50 pm, Davy <zhushe...@gmai l.comwrote:

Hi Matimus and Boris,

Thank you :)

And a further question about vector above rank 1, how can I use it as
the key of dictionary?

For example, if I have list like L=[[1,2,3],[4,5,6,7]],
Then I do L_tuple = tuple(L)>>L_tup le = ([1,2,3],[4,5,6,7])

But {L_tuple:'hello '} cause an error?

Best regards,
Davy

On Nov 6, 3:09 pm, Matimus <mccre...@gmail .comwrote:

On Nov 5, 10:53 pm, Davy <zhushe...@gmai l.comwrote:

Hi all,

We know that list cannot be used as key of dictionary. So, how to work
around it?

For example, there is random list like l=[1,323,54,67].

Any suggestions are welcome!

Best regards,
Davy

Use a tuple instead.

>>d = {}
>>d[tuple([1,2,3,4])] = 'hello world'
>>d

{(1, 2, 3, 4): 'hello world'}>>d[1,2,3,4]

'hello world'

Matt- Hide quoted text -

- Show quoted text -

[Paul McGuire]

On Nov 6, 4:08 am, Dustan <DustanGro...@g mail.comwrote:

On Nov 6, 3:58 am, Duncan Booth <duncan.bo...@i nvalid.invalidw rote:

Wildemar Wildenburger <lasses_w...@kl apptsowieso.net wrote:

maybe something like this could help:

def tupleize(non_tu ple):
try:
return tuple(tupleize( thing) for thing in non_tuple)
except TypeError:
# non_tuple is not iterable
return non_tuple

Just don't try passing that a string or anything containing a string.

Untested

def tupleize(non_tu ple):
if isinstance(non_ tuple, str):
return non_tuple
try:
return tuple(tupleize( thing) for thing in non_tuple)
except TypeError:
# non_tuple is not iterable
return non_tuple

isinstance(x,ba sestring)

is preferred over

isinstance(x,st r)

in case x is a unicode.

-- Paul