Re: Zipping a dictionary whose values are lists
On 13 April 2012 17:35, Kiuhnm kiuhnm03.4t.yahoo...@mail.python.org wrote: On 4/13/2012 17:58, Alexander Blinne wrote: zip(*[x[1] for x in sorted(d.items(), key=lambda y: y[0])]) Or zip(*[d[k] for k in sorted(d.keys())]) .keys() is superfluous here: zip(*(d[k] for k in sorted(d))) -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
RE: Zipping a dictionary whose values are lists
The below code should work:
zip(*d.values())
when you do *d.values() its going to return tuple of elements, which then
further be can be zipped to
achieve your desired result.
Regards,
Shambhu Rajak
Python Lover
-Original Message-
From: tkp...@gmail.com [mailto:tkp...@gmail.com]
Sent: 12/04/2012 9:58 PM
To: python-list@python.org
Subject: Zipping a dictionary whose values are lists
I using Python 3.2 and have a dictionary
d = {0:[1,2], 1:[1,2,3], 2:[1,2,3,4]}
whose values are lists I would like to zip into a list of tuples. If I
explicitly write:
list(zip([1,2], [1,2,3], [1,2,3,4])
[(1, 1, 1), (2, 2, 2)]
I get exactly what I want. On the other hand, I have tried
list(zip(d))
[(0,), (1,), (2,)]
list(zip(d.values()))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
list(zip(d[i] for i in d))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
list(zip(*d))
Traceback (most recent call last):
File pyshell#48, line 1, in module
list(zip(*d))
TypeError: zip argument #1 must support iteration
and nothing quite works. What am I doing wrong?
Sincerely
Thomas Philips
--
http://mail.python.org/mailman/listinfo/python-list
Re: Zipping a dictionary whose values are lists
Am 12.04.2012 18:38, schrieb Kiuhnm: Almost. Since d.values() = [[1,2], [1,2,3], [1,2,3,4]], you need to use list(zip(*d.values())) which is equivalent to list(zip([1,2], [1,2,3], [1,2,3,4])) Kiuhnm While this accidently works in this case, let me remind you that d.values() does not return the elements of the d in any specific order. (It is a non-random but implementation-specific order, see http://docs.python.org/library/stdtypes.html#dict.items.) Thus if you need the correct order (as suggested by the dict keys) an explicit sorting step is required, for example zip(*[x[1] for x in sorted(d.items(), key=lambda y: y[0])]) Greetings -- http://mail.python.org/mailman/listinfo/python-list
Re: Zipping a dictionary whose values are lists
Alexander Blinne wrote: zip(*[x[1] for x in sorted(d.items(), key=lambda y: y[0])]) Why not zip(*[x[1] for x in sorted(d.items())])? -- http://mail.python.org/mailman/listinfo/python-list
Re: Zipping a dictionary whose values are lists
On 4/13/2012 17:58, Alexander Blinne wrote: Am 12.04.2012 18:38, schrieb Kiuhnm: Almost. Since d.values() = [[1,2], [1,2,3], [1,2,3,4]], you need to use list(zip(*d.values())) which is equivalent to list(zip([1,2], [1,2,3], [1,2,3,4])) Kiuhnm While this accidently works in this case, let me remind you that d.values() does not return the elements of the d in any specific order. The OP said nothing about ordering. The fact that the keys are ordered might be accidental :) (It is a non-random but implementation-specific order, see http://docs.python.org/library/stdtypes.html#dict.items.) Thus if you need the correct order (as suggested by the dict keys) an explicit sorting step is required, for example zip(*[x[1] for x in sorted(d.items(), key=lambda y: y[0])]) Or zip(*[d[k] for k in sorted(d.keys())]) Kiuhnm -- http://mail.python.org/mailman/listinfo/python-list
Re: Zipping a dictionary whose values are lists
zip(*d.values())
On 12 April 2012 20:28, tkp...@gmail.com wrote:
I using Python 3.2 and have a dictionary
d = {0:[1,2], 1:[1,2,3], 2:[1,2,3,4]}
whose values are lists I would like to zip into a list of tuples. If I
explicitly write:
list(zip([1,2], [1,2,3], [1,2,3,4])
[(1, 1, 1), (2, 2, 2)]
I get exactly what I want. On the other hand, I have tried
list(zip(d))
[(0,), (1,), (2,)]
list(zip(d.values()))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
list(zip(d[i] for i in d))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
list(zip(*d))
Traceback (most recent call last):
File pyshell#48, line 1, in module
list(zip(*d))
TypeError: zip argument #1 must support iteration
and nothing quite works. What am I doing wrong?
Sincerely
Thomas Philips
--
http://mail.python.org/mailman/listinfo/python-list
--
С уважением, Аносов Павел
--
http://mail.python.org/mailman/listinfo/python-list
Re: Zipping a dictionary whose values are lists
On 4/12/2012 18:28, tkp...@gmail.com wrote:
I using Python 3.2 and have a dictionary
d = {0:[1,2], 1:[1,2,3], 2:[1,2,3,4]}
whose values are lists I would like to zip into a list of tuples. If I
explicitly write:
list(zip([1,2], [1,2,3], [1,2,3,4])
[(1, 1, 1), (2, 2, 2)]
I get exactly what I want. On the other hand, I have tried
list(zip(d))
[(0,), (1,), (2,)]
list(zip(d.values()))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
Almost. Since d.values() = [[1,2], [1,2,3], [1,2,3,4]], you need to use
list(zip(*d.values()))
which is equivalent to
list(zip([1,2], [1,2,3], [1,2,3,4]))
Kiuhnm
--
http://mail.python.org/mailman/listinfo/python-list
Re: Zipping a dictionary whose values are lists
tkp...@gmail.com wrote:
I using Python 3.2 and have a dictionary
d = {0:[1,2], 1:[1,2,3], 2:[1,2,3,4]}
whose values are lists I would like to zip into a list of tuples. If I
explicitly write:
list(zip([1,2], [1,2,3], [1,2,3,4])
[(1, 1, 1), (2, 2, 2)]
I get exactly what I want. On the other hand, I have tried
list(zip(d))
[(0,), (1,), (2,)]
list(zip(d.values()))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
list(zip(d[i] for i in d))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
list(zip(*d))
Traceback (most recent call last):
File pyshell#48, line 1, in module
list(zip(*d))
TypeError: zip argument #1 must support iteration
and nothing quite works. What am I doing wrong?
You have all the building blocks ;)
d = {0:[1,2], 1:[1,2,3], 2:[1,2,3,4]}
list(zip(*d.values()))
[(1, 1, 1), (2, 2, 2)]
The order of the values is undefined, so you may want to sort the lists by
key first:
list(zip(*[v for k, v in sorted(d.items())]))
[(1, 1, 1), (2, 2, 2)]
Well, I guess it doesn't really matter for that example...
--
http://mail.python.org/mailman/listinfo/python-list
Re: Zipping a dictionary whose values are lists
On Thu, 12 Apr 2012 09:28:03 -0700 (PDT)
tkp...@gmail.com wrote:
I using Python 3.2 and have a dictionary
d = {0:[1,2], 1:[1,2,3], 2:[1,2,3,4]}
whose values are lists I would like to zip into a list of tuples. If
I explicitly write:
list(zip([1,2], [1,2,3], [1,2,3,4])
[(1, 1, 1), (2, 2, 2)]
I get exactly what I want. On the other hand, I have tried
list(zip(d))
[(0,), (1,), (2,)]
list(zip(d.values()))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
list(zip(d[i] for i in d))
[([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
list(zip(*d))
Traceback (most recent call last):
File pyshell#48, line 1, in module
list(zip(*d))
TypeError: zip argument #1 must support iteration
and nothing quite works. What am I doing wrong?
Try this:
list(zip(*d.values()))
d.values() is a list, but zip wants the individual values as separate
arguments.
HTH,
Dan
--
http://mail.python.org/mailman/listinfo/python-list
