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Sorting a set works, sorting a dictionary fails ?

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  Sorting a set works, sorting a dictionary fails ? Νικόλαος Κούρας <nikos.gr33k@gmail.com> - 2013-06-10 01:04 -0700
    Re: Sorting a set works, sorting a dictionary fails ? Νικόλαος Κούρας <nikos.gr33k@gmail.com> - 2013-06-10 01:16 -0700
      Re: Sorting a set works, sorting a dictionary fails ? Νικόλαος Κούρας <nikos.gr33k@gmail.com> - 2013-06-10 01:18 -0700
        Re: Sorting a set works, sorting a dictionary fails ? Νικόλαος Κούρας <nikos.gr33k@gmail.com> - 2013-06-10 01:29 -0700
          Re: Sorting a set works, sorting a dictionary fails ? Fábio Santos <fabiosantosart@gmail.com> - 2013-06-10 10:27 +0100
            Re: Sorting a set works, sorting a dictionary fails ? Νικόλαος Κούρας <nikos.gr33k@gmail.com> - 2013-06-10 02:32 -0700
              Re: Sorting a set works, sorting a dictionary fails ? Νικόλαος Κούρας <nikos.gr33k@gmail.com> - 2013-06-10 02:48 -0700
                Re: Sorting a set works, sorting a dictionary fails ? Fábio Santos <fabiosantosart@gmail.com> - 2013-06-10 11:13 +0100
                Re: Sorting a set works, sorting a dictionary fails ? Ulrich Eckhardt <ulrich.eckhardt@dominolaser.com> - 2013-06-10 15:11 +0200
          Re: Sorting a set works, sorting a dictionary fails ? Ulrich Eckhardt <ulrich.eckhardt@dominolaser.com> - 2013-06-10 11:40 +0200
            Re: Sorting a set works, sorting a dictionary fails ? Νικόλαος Κούρας <nikos.gr33k@gmail.com> - 2013-06-10 03:20 -0700
            Re: Sorting a set works, sorting a dictionary fails ? Νικόλαος Κούρας <nikos.gr33k@gmail.com> - 2013-06-10 03:42 -0700
              Re: Sorting a set works, sorting a dictionary fails ? Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2013-06-10 11:47 +0000
              Re: Sorting a set works, sorting a dictionary fails ? Denis McMahon <denismfmcmahon@gmail.com> - 2013-06-11 02:14 +0000
          Re: Sorting a set works, sorting a dictionary fails ? Larry Hudson <orgnut@yahoo.com> - 2013-06-11 02:16 -0700
    Re: Sorting a set works, sorting a dictionary fails ? Ulrich Eckhardt <ulrich.eckhardt@dominolaser.com> - 2013-06-10 11:19 +0200

#47529 — Sorting a set works, sorting a dictionary fails ?

sets and dicts are unordered. 

================
Yo order the a set i use:
names = set() #the i fill it with data

for name in sorted( names ): 
================

Now for the dictionary:

================
months = { 'Ιανουάριος':1, 'Φεβρουάριος':2, 'Μάρτιος':3, 'Απρίλιος':4, 'Μάϊος':5, 'Ιούνιος':6, \
	   'Ιούλιος':7, 'Αύγουστος':8, 'Σεπτέμβριος':9, 'Οκτώβριος':10, 'Νοέμβριος':11, 'Δεκέμβριος':12 }

for key in sorted( months.keys() ):
================

I'm having trouble ordering a dictionary though.

But how come i was able to sort the set names() and not being able to sort the dictionary keys as well with the sorted function?= i used?

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#47531

What if i wanted to sort it out if alphabetically and not by the values?

Thsi worked:

for item in sorted(months.items(),key=lambda num : num[1]):

but this failed:

for item in sorted(months.items()):

why?

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#47533

Τη Δευτέρα, 10 Ιουνίου 2013 11:16:37 π.μ. UTC+3, ο χρήστης Νικόλαος Κούρας έγραψε:
> What if i wanted to sort it out if alphabetically and not by the values?
> 
> 
> 
> Thsi worked:
> 
> 
> 
> for item in sorted(months.items(),key=lambda num : num[1]):
> 
> 
> 
> but this failed:
> 
> 
> 
> for item in sorted(months.items()):
> 
> 
> 
> why?

sorry what i was tryign to say was why not as: for item in sorted(months.values()):

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#47534

Trying this:

months = { 'Ιανουάριος':1, 'Φεβρουάριος':2, 'Μάρτιος':3, 'Απρίλιος':4, 'Μάϊος':5, 'Ιούνιος':6, \
	   'Ιούλιος':7, 'Αύγουστος':8, 'Σεπτέμβριος':9, 'Οκτώβριος':10, 'Νοέμβριος':11, 'Δεκέμβριος':12 }

for key in sorted( months.values() ):
	print('''
		<option value="%s"> %s </option>
	''' % (months[key], key) )


output this:

[Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173]   File "/home/nikos/public_html/cgi-bin/pelatologio.py", line 310, in <module>, referer: http://superhost.gr/
[Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173]     ''' % (months[key], key) ), referer: http://superhost.gr/
[Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173] KeyError: 1, referer: http://superhost.gr/

KeyError 1 ??!! All i did was to tell python to sort the dictionary values, which are just integers.

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#47536

[Multipart message — attachments visible in raw view] — view raw

On 10 Jun 2013 09:34, "Νικόλαος Κούρας" <nikos.gr33k@gmail.com> wrote:
>
> Trying this:
>
> months = { 'Ιανουάριος':1, 'Φεβρουάριος':2, 'Μάρτιος':3, 'Απρίλιος':4,
'Μάϊος':5, 'Ιούνιος':6, \
>            'Ιούλιος':7, 'Αύγουστος':8, 'Σεπτέμβριος':9, 'Οκτώβριος':10,
'Νοέμβριος':11, 'Δεκέμβριος':12 }
>
> for key in sorted( months.values() ):
>         print('''
>                 <option value="%s"> %s </option>
>         ''' % (months[key], key) )
>
>
> output this:
>
> [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173]   File
"/home/nikos/public_html/cgi-bin/pelatologio.py", line 310, in <module>,
referer: http://superhost.gr/
> [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173]     ''' %
(months[key], key) ), referer: http://superhost.gr/
> [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173] KeyError: 1,
referer: http://superhost.gr/
>
> KeyError 1 ??!! All i did was to tell python to sort the dictionary
values, which are just integers.
> --
> http://mail.python.org/mailman/listinfo/python-list

KeyError: 1 means that there is no int(1) key. I think you meant to do "for
key in sorted(yourdict.keys())"

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#47537

Τη Δευτέρα, 10 Ιουνίου 2013 12:27:14 μ.μ. UTC+3, ο χρήστης Fábio Santos έγραψε:
> On 10 Jun 2013 09:34, "Νικόλαος Κούρας" <nikos...@gmail.com> wrote:
> 
> >
> 
> > Trying this:
> 
> >
> 
> > months = { 'Ιανουάριος':1, 'Φεβρουάριος':2, 'Μάρτιος':3, 'Απρίλιος':4, 'Μάϊος':5, 'Ιούνιος':6, \
> 
> >            'Ιούλιος':7, 'Αύγουστος':8, 'Σεπτέμβριος':9, 'Οκτώβριος':10, 'Νοέμβριος':11, 'Δεκέμβριος':12 }
> 
> >
> 
> > for key in sorted( months.values() ):
> 
> >         print('''
> 
> >                 <option value="%s"> %s </option>
> 
> >         ''' % (months[key], key) )
> 
> >
> 
> >
> 
> > output this:
> 
> >
> 
> > [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173]   File "/home/nikos/public_html/cgi-bin/pelatologio.py", line 310, in <module>, referer: http://superhost.gr/
> 
> 
> > [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173]     ''' % (months[key], key) ), referer: http://superhost.gr/
> 
> > [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173] KeyError: 1, referer: http://superhost.gr/
> 
> >
> 
> > KeyError 1 ??!! All i did was to tell python to sort the dictionary values, which are just integers.
> 
> > --
> 
> > http://mail.python.org/mailman/listinfo/python-list
> 
> KeyError: 1 means that there is no int(1) key. I think you meant to do "for key in sorted(yourdict.keys())"


both 
for key in sorted( months.keys() ):
for key in sorted( months.values() ):


fail. meybe this is abug andsorted function doesnt work as expected.

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#47538

After many tried this did the job:

for key in sorted(months.items(),key=lambda num : num[1]):
	print('''
		<option value="%s"> %s </option>
	''' % (key[1], key[0]) )


but its really frustrating not being able to:

for key in sorted( months.values() ): 
        print(''' 
                <option value="%s"> %s </option> 
        ''' % (months[key], key) ) 

Which seemed to be an abivous way to do it.
names set() was able to order like this why not the dictionary too?

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#47540

[Multipart message — attachments visible in raw view] — view raw

On 10 Jun 2013 10:53, "Νικόλαος Κούρας" <nikos.gr33k@gmail.com> wrote:
>
> After many tried this did the job:
>
> for key in sorted(months.items(),key=lambda num : num[1]):
>         print('''
>                 <option value="%s"> %s </option>
>         ''' % (key[1], key[0]) )
>
>
> but its really frustrating not being able to:
>
> for key in sorted( months.values() ):
>         print('''
>                 <option value="%s"> %s </option>
>         ''' % (months[key], key) )
>
> Which seemed to be an abivous way to do it.
> names set() was able to order like this why not the dictionary too?

Why not

for key, value in sorted(d.items()):

Tuples are totally sortable.

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#47553

Am 10.06.2013 11:48, schrieb Νικόλαος Κούρας:
> After many tried this did the job:
>
> for key in sorted(months.items(),key=lambda num : num[1]):
> 	print('''
> 		<option value="%s"> %s </option>
> 	''' % (key[1], key[0]) )

This code is still sending a misleading message. What you are referring 
to as "key" here is in fact a (key, value) tuple. I'd use Fábio's 
suggestion and use the automatic splitting:

     for name, idx in sorted(months.items(), key=lambda num : num[1]):
         print('month #{} is {}'.format(idx, name))


> but its really frustrating not being able to:
>
> for key in sorted( months.values() ):
>          print('''
>                  <option value="%s"> %s </option>
>          ''' % (months[key], key) )
>
> Which seemed to be an abivous way to do it.

You are composing three things:

1. months.values() - gives you a sequence with the month numbers
2. sorted() - gives you a sorted sequence
3. for-iteration - iterates over a sequence

At which point is Python doing anything non-obvious? Also, have you 
considered reversing the dictionary mapping or creating a second one 
with the reversed mapping? Or maybe take a look at collections.OrderedDict?


> names set() was able to order like this why not the dictionary too?

Well, why don't you use a set then, if it solves your problem? An in 
which place does anything behave differently? Sorry to bring you the 
news, but your expectations are not fulfilled because your assumptions 
about how things should work are already flawed, I'm afraid.


Uli

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#47541

Am 10.06.2013 10:29, schrieb Νικόλαος Κούρας:
> for key in sorted( months.values() ):
       ^^^                   ^^^^^^

> KeyError 1 ??!! All i did was to tell python to sort the dictionary values, which are just integers.

...and which you then proceed to use as key, which is obviously wrong.

Uli

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#47543

Τη Δευτέρα, 10 Ιουνίου 2013 12:40:01 μ.μ. UTC+3, ο χρήστης Ulrich Eckhardt έγραψε:
> Am 10.06.2013 10:29, schrieb Νικόλαος Κούρας:
> 
> > for key in sorted( months.values() ):
> 
>        ^^^                   ^^^^^^
> 
> 
> 
> > KeyError 1 ??!! All i did was to tell python to sort the dictionary values, which are just integers.
> 
> 
> 
> ...and which you then proceed to use as key, which is obviously wrong.

How hsould have i written it then?

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#47544

months = { '@@@@@@@@@@':0, 'Ιανουάριος':1, 'Φεβρουάριος':2, 'Μάρτιος':3, 'Απρίλιος':4, 'Μάϊος':5, 'Ιούνιος':6, \
	   'Ιούλιος':7, 'Αύγουστος':8, 'Σεπτέμβριος':9, 'Οκτώβριος':10, 'Νοέμβριος':11, 'Δεκέμβριος':12 }

for key in sorted( months.values() ): 
	print('''
		<option value="%s"> %s </option>
	''' % (months[key], key) )
==============

please tell me Uli why this dont work as expected to.

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#47549

On Mon, 10 Jun 2013 03:42:38 -0700, Νικόλαος Κούρας wrote:

> for key in sorted( months.values() ):

> please tell me Uli why this dont work as expected to.

Because values are not keys. You are looking at the values, and trying to 
use them as keys.

months = {'Φεβρουάριος':2, 'Ιανουάριος':1}
print("==Values==")
for x in sorted(months.values()):
    print(x)

print("==Keys==")
for x in sorted(months.keys()):
    print(x)


prints:


==Values==
1
2
==Keys==
Ιανουάριος
Φεβρουάριος



-- 
Steven

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#47627

On Mon, 10 Jun 2013 03:42:38 -0700, Νικόλαος Κούρας wrote:


> for key in sorted( months.values() ):
> 	print('''
> 		<option value="%s"> %s </option>
> 	''' % (months[key], key) )
> ==============
> 
> please tell me Uli why this dont work as expected to.

Because inside the for loop, your value 'key' is an item from the sorted 
list of the values part of months. When you use months[key], you're 
trying to look up in months based on the value part, not the key part. 
Calling it key is confusing you.

Try this:

things = { "a":1, "b":3, "c":2 }

for thing in sorted( things.values() ):
	print thing

>> Prints 1, 2, 3

for thing in sorted( things.keys() ):
	print thing

>> Prints a, b, c

Now although things["b"] is a valid reference because "b" is a key in a 
key - value pair, things[3] is not a valid reference, because 3 is not a 
key in a key - value pair.

-- 
Denis McMahon, denismfmcmahon@gmail.com

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#47650

On 06/10/2013 01:29 AM, Νικόλαος Κούρας wrote:
> Trying this:
>
> months = { 'Ιανουάριος':1, 'Φεβρουάριος':2, 'Μάρτιος':3, 'Απρίλιος':4, 'Μάϊος':5, 'Ιούνιος':6, \
> 	   'Ιούλιος':7, 'Αύγουστος':8, 'Σεπτέμβριος':9, 'Οκτώβριος':10, 'Νοέμβριος':11, 'Δεκέμβριος':12 }
>
> for key in sorted( months.values() ):
> 	print('''
> 		<option value="%s"> %s </option>
> 	''' % (months[key], key) )
>
>
> output this:
>
> [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173]   File "/home/nikos/public_html/cgi-bin/pelatologio.py", line 310, in <module>, referer: http://superhost.gr/
> [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173]     ''' % (months[key], key) ), referer: http://superhost.gr/
> [Mon Jun 10 11:25:11 2013] [error] [client 79.103.41.173] KeyError: 1, referer: http://superhost.gr/
>
> KeyError 1 ??!! All i did was to tell python to sort the dictionary values, which are just integers.
>
Come on now, wake up!!  You do know how to use Python dictionaries don't you?

You can find the dictionary _values_ by using the _key_, but you CAN NOT get the key from the 
value.  (For one thing, keys must be unique but values don't need to be.)  You are trying to use 
the values as keys, but the keys here are all strings, there are NO keys that are integers.  Of 
course this is a dictionary key error.

I think someone already told you this previously, but I'll repeat it, perhaps more verbosely, 
but at least in different words....

The items() function gives you a list of key/value pairs as tuples.  (Not exactly, Python 3 give 
you an iterator not an actual list, but anyway...)
You want to sort this list based on the second element of these tuples (the values).  A simple 
call to the sorted() function will sort based on the first element (the keys).  But you can 
alter this function by using the key parameter, and here you can do it with a short lambda 
function.  (And don't confuse the sorted() key parameter with dictionary keys -- two entirely 
different things that just happen to use the same word.)  See if you can follow this short example:

 >>> d = {'Jan':1, 'Feb':2, 'Mar':3, 'Apr':4, 'May':5}     #  A short example dictionary
 >>> list(d.items())                       #  Display unsorted list
[('Jan', 1), ('Apr', 4), ('Mar', 3), ('Feb', 2), ('May', 5)]
 >>> sorted(d.items())                     #  Display list sorted by keys (month names)
[('Apr', 4), ('Feb', 2), ('Jan', 1), ('Mar', 3), ('May', 5)]    #  Not what you want
 >>> keys = sorted(d.items(), key=lambda n: n[1])    #  Sort by values (the 2nd element of items)
 >>> for month in keys:                    #  Print this sorted list
	print('Month {} is {}'.format(month[1], month[0]))

Month 1 is Jan
Month 2 is Feb
Month 3 is Mar
Month 4 is Apr
Month 5 is May

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#47542

Am 10.06.2013 10:04, schrieb Νικόλαος Κούρας:
> months = { 'Ιανουάριος':1, 'Φεβρουάριος':2, 'Μάρτιος':3, 'Απρίλιος':4, 'Μάϊος':5, 'Ιούνιος':6, \
> 	   'Ιούλιος':7, 'Αύγουστος':8, 'Σεπτέμβριος':9, 'Οκτώβριος':10, 'Νοέμβριος':11, 'Δεκέμβριος':12 }
>
> for key in sorted( months.keys() ):
> ================
>
> I'm having trouble ordering a dictionary though.

I can't find a problem here. I tried simple dictionaries containing 
numbers as keys using Python 3.3, and sorting the keys works without any 
problem there. What exactly is the "trouble" you are having? Be a bit 
more precise and describe what you saw and, just in case, also what you 
expected to see.

BTW: You have a line continuation there using a backslash. This isn't 
necessary, since the pair of {} automatically tell Python the target range.


Good luck!

Uli

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