_ is just a plain variable name in Python. It is sometimes when a variable is
needed to receive a value that won't be used.
Like in some other interactive systems (Mathematica, etc, but with a
different syntax) _ has a use in the interactive shell, it contains the
last unassigned result:
a = 2
[EMAIL PROTECTED] wrote:
Following Ron Adam solution (and using [] instead of list() in the last
line), this may be a possible solution of the problem, that is often
quite fast:
def psort16(s1, s2):
try:
d = dict(izip(s2, s1))
except TypeError:
_indices =
Ron Adam wrote:
Alex Martelli wrote:
Ron Adam [EMAIL PROTECTED] wrote:
...
Considering the number time I sort keys after getting them, It's the
behavior I would prefer. Maybe a more dependable dict.sortedkeys()
method would be nice. ;-)
sorted(d) is guaranteed to do exactly the same
Kent Johnson wrote:
[a for b,a in sorted(zip(B,A))]
also, [a for _,a in sorted(zip(B,A))]
didn't read refs, tested above python-2.2.3.
--
Sun Yi Ming
you can logout any time you like,
but you can never leave...
--
http://mail.python.org/mailman/listinfo/python-list
Dolmans Sun wrote:
Kent Johnson wrote:
[a for b,a in sorted(zip(B,A))]
also, [a for _,a in sorted(zip(B,A))]
didn't read refs, tested above python-2.2.3.
Is there something here I can't see, or did you just
change a variable name and present that as another
solution?
--
Ron Adam wrote:
Ron Adam wrote:
Alex Martelli wrote:
Ron Adam [EMAIL PROTECTED] wrote:
...
Considering the number time I sort keys after getting them, It's the
behavior I would prefer. Maybe a more dependable dict.sortedkeys()
method would be nice. ;-)
sorted(d) is guaranteed to do
Scott David Daniels wrote:
Ron Adam wrote:
Ron Adam wrote:
This probably should be:
def psort11(s1, s2):
d = dict(zip(s2,s1))
assert len(d) == len(s1)
s1[:] = list(d[v] for v in sorted(d))
You could do this to avoid all of those lookups:
def psort_xx(s1, s2):
Following Ron Adam solution (and using [] instead of list() in the last
line), this may be a possible solution of the problem, that is often
quite fast:
def psort16(s1, s2):
try:
d = dict(izip(s2, s1))
except TypeError:
_indices = range(len(s1))
Magnus Lycka wrote:
Is there something here I can't see, or did you just
change a variable name and present that as another
solution?
Heh, I have use python for a very short time. Base your statements, I test
in python, found `_' is a valid variable name. So I don't know it is a just
a plain
Simon Sun wrote:
Magnus Lycka wrote:
Is there something here I can't see, or did you just
change a variable name and present that as another
solution?
Heh, I have use python for a very short time. Base your statements, I test
in python, found `_' is a valid variable name. So I don't know it
Hello,
I have two lists, one with strings (filenames, actually), and one with a
real-number
rank, like:
A=['hello','there','this','that']
B=[3,4,2,5]
I'd like to sort list A using the values from B, so the result would be in this
example,
A=['this','hello','there','that']
The sort method on
Brian Blais wrote:
Hello,
I have two lists, one with strings (filenames, actually), and one with a
real-number
rank, like:
A=['hello','there','this','that']
B=[3,4,2,5]
I'd like to sort list A using the values from B, so the result would be
in this example,
Brian Blais wrote:
Hello,
I have two lists, one with strings (filenames, actually), and one with a
real-number
rank, like:
A=['hello','there','this','that']
B=[3,4,2,5]
I'd like to sort list A using the values from B, so the result would be
in this example,
Brian Blais [EMAIL PROTECTED] wrote:
Hello,
I have two lists, one with strings (filenames, actually), and one with a
real-number rank, like:
A=['hello','there','this','that']
B=[3,4,2,5]
I'd like to sort list A using the values from B, so the result would be in
this example,
Brian Blais wrote:
Hello,
I have two lists, one with strings (filenames, actually), and one with a
real-number
rank, like:
A=['hello','there','this','that']
B=[3,4,2,5]
I'd like to sort list A using the values from B, so the result would be
in this example,
Steven Bethard wrote:
Here's a solution that makes use of the key= argument to sorted():
A = ['hello','there','this','that']
B = [3,4,2,5]
indices = range(len(A))
indices.sort(key=B.__getitem__)
[A[i] for i in indices]
['this', 'hello', 'there', 'that']
Basically, it sorts the
Your solution Steven Bethard looks very intelligent, here is a small
speed test, because sorting a list according another one is a quite
common operation.
(Not all solutions are really the same, as Alex has shown).
from itertools import izip, imap
from operator import itemgetter
from random
[EMAIL PROTECTED] wrote:
Your solution Steven Bethard looks very intelligent, here is a small
speed test, because sorting a list according another one is a quite
common operation.
(Not all solutions are really the same, as Alex has shown).
Try this one.
def psort10(s1, s2):
d =
Ron Adam [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] wrote:
Your solution Steven Bethard looks very intelligent, here is a small
speed test, because sorting a list according another one is a quite
common operation.
(Not all solutions are really the same, as Alex has shown).
Try this
It's faster on my system because d.keys() is already sorted. But that may not
be the case on other versions of python.
In my version it's a little slower. But what system are you using where
keys is already sorted? IronPython maybe?
Bye,
bearophile
--
Alex Martelli wrote:
Ron Adam [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] wrote:
Your solution Steven Bethard looks very intelligent, here is a small
speed test, because sorting a list according another one is a quite
common operation.
(Not all solutions are really the same, as Alex has
[EMAIL PROTECTED] wrote:
It's faster on my system because d.keys() is already sorted. But that may
not be the case on other versions of python.
In my version it's a little slower. But what system are you using where
keys is already sorted? IronPython maybe?
Bye,
bearophile
Python
Ron Adam wrote:
Python 2.4.1 (#65, Mar 30 2005, 09:13:57) [MSC v.1310 32 bit (Intel)]
on win32
I was a bit surprised by them being sorted. I just happend to try
d.keys() in place of s2, and it sped up. I was expecting it to be a
bit slower.
Purely an implementation accident - based
Ron Adam [EMAIL PROTECTED] wrote:
...
Considering the number time I sort keys after getting them, It's the
behavior I would prefer. Maybe a more dependable dict.sortedkeys()
method would be nice. ;-)
sorted(d) is guaranteed to do exactly the same thing as sorted(d.keys())
AND to be
Alex Martelli wrote:
Ron Adam [EMAIL PROTECTED] wrote:
...
Considering the number time I sort keys after getting them, It's the
behavior I would prefer. Maybe a more dependable dict.sortedkeys()
method would be nice. ;-)
sorted(d) is guaranteed to do exactly the same thing as
Delaney, Timothy (Tim) wrote:
Ron Adam wrote:
Python 2.4.1 (#65, Mar 30 2005, 09:13:57) [MSC v.1310 32 bit (Intel)]
on win32
I was a bit surprised by them being sorted. I just happend to try
d.keys() in place of s2, and it sped up. I was expecting it to be a
bit slower.
Purely an
26 matches
Mail list logo
