sort one list using the values from another list
Brian Blais
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 lists does in-place sorting. Is there a way to do what I want here?
thanks,
Brian Blais
--
-----------------
Kent Johnson
> 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']
Here are two ways:
>>> A=['hello','there','this','that']
>>> B=[3,4,2,5]
>>> zip(*sorted(zip(B,A)))[1]
('this', 'hello', 'there', 'that')
>>> [a for b,a in sorted(zip(B,A))]
['this', 'hello', 'there', 'that']
I prefer the second one, I think it is more readable, will use less
memory (doesn't have to create a new list B in the final step) and it's
even faster on my computer:
D:\Projects\CB>python -m timeit -s
"A=['hello','there','this','that'];B=[3,4,2,5]" "zip(*sorted(zip(B,A)))[1]"
100000 loops, best of 3: 6.29 usec per loop
D:\Projects\CB>python -m timeit -s
"A=['hello','there','this','that'];B=[3,4,2,5]" "[a for b,a in
sorted(zip(B,A))]"
100000 loops, best of 3: 5.53 usec per loop
(I'm bored this morning :-)
There's probably a clever way to do it using the key parameter to sort
but I can't think of it...
>
> The sort method on lists does in-place sorting. Is there a way to do
> what I want here?
The example above does not sort in place, if you want it to be in place use
A[:] = [a for b,a in sorted(zip(B,A))]
Kent
Jeffrey Schwab
> 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 lists does in-place sorting. Is there a way to do
> what I want here?
If A has no duplicate elements, you could create a hash mapping A's
elements to their respective precedences, then provide a sort criterion
that accessed the hash. Alternatively, you could do something like this:
from operator import itemgetter
result = map(itemgetter(0), sorted(zip(A, B), key=itemgetter(1)))
Alex Martelli
> 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 lists does in-place sorting. Is there a way to do what
> I want here?
Sure, many ways, mostly ones based on the "Decorate-Sort-Undecorate"
idiom and its incarnation as the key= optional argument to function
sorted and list method sort. I believe that a more explicit DSU is
ideal (clearer, though maybe a tad slower) in your case:
_aux = zip(B, A)
_aux.sort()
A[:] = [a for b, a in _aux]
Twisting things to use the speedy key= approach seems more "contorted"
and less general in this case, and it also needs an auxiliary structure
(to avoid slowing things down by repeated .index calls) However, things
may be different if you need to consider the possibility that B has some
duplicate items, and deeply care for such a case NOT to result in
comparisons of items of A. The above approach would then have to be
changed, e.g., to:
_aux = zip(B, enumerate(A))
_aux.sort()
A[:] = [a for (b, (i, a)) in _aux]
where I've also added a pair of redundant parentheses to make the
nesting structure of items of _aux more obvious. Of course, each of
these has a variant using sorted instead of sort, and for those you
could use izip from itertools rather than built-in zip, and do
everything within one single statement, etc, etc.
Alex
Steven Bethard
> 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']
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 indices to A -- [0, 1, 2, 3] -- in the order
given by B, and then selects the items from A in the appropriate order.
Duncan Booth
> 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 indices to A -- [0, 1, 2, 3] -- in the order
> given by B, and then selects the items from A in the appropriate order.
>
>
That's impressive. I'm sure like a lot of other people I looked at the
question and thought it ought to be possible to use key to get this result
without a load of zipping and unzipping but just couldn't quite see it.
If you combine your technique with 'sorted', you get the one line version:
>>> [A[i] for i in sorted(range(len(A)), key=B.__getitem__)]
bearoph...@lycos.com
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 import randint, choice, randint, shuffle
from string import lowercase
def timeit(function, *arg1, **arg2):
"""timeit(function, *arg1, **arg2): given a function and its
parameters, calls it
and computes its running time, in seconds. It result is
discarded."""
t1 = clock()
function(*arg1, **arg2)
t2 = clock()
return round(t2-t1, 2)
def psort1(s1, s2):
s1[:] = [a for b,a in sorted(zip(s2, s1))]
def psort2(s1, s2):
aux = zip(s2, enumerate(s1))
aux.sort()
s1[:] = [a for b, (i, a) in aux]
def psort3(s1, s2):
_indices = range(len(s1))
_indices.sort(key=s2.__getitem__)
s1[:] = [s1[i] for i in _indices]
def psort4(s1, s2):
_indices = range(len(s1))
_indices.sort(key=s2.__getitem__)
s1[:] = map(s1.__getitem__, _indices)
def psort5(s1, s2):
s1[:] = zip(*sorted(zip(s2, s1)))[1]
def psort6(s1, s2):
s1[:] = map(itemgetter(0), sorted(zip(s1, s2), key=itemgetter(1)))
def psort7(s1, s2):
s1[:] = [a for b,a in sorted(izip(s2, s1))]
def psort8(s1, s2):
s1[:] = zip(*sorted(izip(s2, s1)))[1]
def psort9(s1, s2):
s1[:] = map(itemgetter(0), sorted(izip(s1, s2), key=itemgetter(1)))
n = 100000
s1 = ["".join(choice(lowercase) for i in xrange(randint(2,8))) for k in
xrange(n)]
s2 = range(n)
shuffle(s2)
for psort in sorts:
s1c = list(s1)
print timeit(psort, s1c, s2), "s"
Timings on my PC, Python 2.4, PIII 500 MHz:
2.87
3.82
1.6
1.56
4.35
2.49
2.75
4.29
2.35
psort4 is my variant of your solution, and it seems the faster one.
Note: one liners are bad, and expecially bad in Python that is designed
to be a readable language, so let's avoid them as much as possible. I
have used one of them to generate s1, but I'll not use them in
production code or in my libraries, etc.
Bye,
bearophile
Ron Adam
> 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 = dict(zip(s2,s1))
s1[:] = (d[n] for n in sorted(d.keys()))
It's faster on my system because d.keys() is already sorted. But that
may not be the case on other versions of python.
Ron
Alex Martelli
If there are duplicates in s2, this solution will silently lose some
items from s1. I would at least include an assert len(s2)==len(d) as
the second statement to get some insurance that this doesn't occur.
Alex
bearoph...@lycos.com
In my version it's a little slower. But what system are you using where
keys is already sorted? IronPython maybe?
Bye,
bearophile
Ron Adam
Good point, and a function that only works part time isn't good. ;-)
Ron Adam
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.
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. ;-)
Like Alex pointed out you need to make sure there aren't duplicates in
the list used for keys.
def psort10(s1, s2):
d = dict(zip(s2,s1))
assert len(d) == len(s1)
s1[:] = (d[n] for n in sorted(d.keys()))
Ron
Delaney, Timothy (Tim)
> 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 entirely on the hash values of
the objects you've used (i.e. small integers). As you add more data to
the dictionary, you will definitely reach the point where you no longer
get keys out in sorted order.
Tim Delaney
Alex Martelli
...
> 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 faster (would be pretty weird if it weren't faster...!).
E.g., ...:
helen:~ alex$ python -mtimeit -s'd=dict(enumerate("tarazoplay"))'
'sorted(d.keys())'
100000 loops, best of 3: 6.82 usec per loop
helen:~ alex$ python -mtimeit -s'd=dict(enumerate("tarazoplay"))'
'sorted(d)'
100000 loops, best of 3: 5.98 usec per loop
Alex
Ron Adam
Yes, it did decrease it. And simplified it as well. ;)
def psort11(s1, s2):
d = dict(zip(s2, s1))
assert len(d) == len(s1)
sorted(d)
s1[:] = d.values()
psort1 0.554 s
psort2 0.727 s
psort3 0.295 s
psort4 0.293 s
psort5 0.831 s
psort6 0.438 s
psort7 0.575 s
psort8 0.845 s
psort9 0.424 s
psort10 0.235 s
psort11 0.206 s
Ron Adam
That's what I thought.
Ron Adam
> Alex Martelli wrote:
>> Ron Adam <r...@ronadam.com> 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 faster (would be pretty weird if it weren't faster...!).
>>
>> E.g., ...:
>>
>> helen:~ alex$ python -mtimeit -s'd=dict(enumerate("tarazoplay"))'
>> 'sorted(d.keys())'
>> 100000 loops, best of 3: 6.82 usec per loop
>>
>> helen:~ alex$ python -mtimeit -s'd=dict(enumerate("tarazoplay"))'
>> 'sorted(d)'
>> 100000 loops, best of 3: 5.98 usec per loop
>>
>>
>> Alex
>
>
> Yes, it did decrease it. And simplified it as well. ;)
>
>
> def psort11(s1, s2):
> d = dict(zip(s2, s1))
> assert len(d) == len(s1)
> sorted(d)
> s1[:] = d.values()
>
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))
Dolmans Sun
> >>> [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...
Magnus Lycka
> 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?
Scott David Daniels
> Ron Adam wrote:
>> Alex Martelli wrote:
>>> Ron Adam <r...@ronadam.com> 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 faster (would be pretty weird if it weren't faster...!).
>>>
>>> E.g., ...:
>>>
>>> helen:~ alex$ python -mtimeit -s'd=dict(enumerate("tarazoplay"))'
>>> 'sorted(d.keys())'
>>> 100000 loops, best of 3: 6.82 usec per loop
>>>
>>> helen:~ alex$ python -mtimeit -s'd=dict(enumerate("tarazoplay"))'
>>> 'sorted(d)'
>>> 100000 loops, best of 3: 5.98 usec per loop
>>>
>>>
>>> Alex
>>
>>
>> Yes, it did decrease it. And simplified it as well. ;)
>>
>> def psort11(s1, s2):
>> d = dict(zip(s2, s1))
>> assert len(d) == len(s1)
>> sorted(d)
>> s1[:] = d.values()
Dictionaries are not ordered, the "sorted" line does nothing except produce
a sorted list of the dictionary's keys which is ignored.
> 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):
pairs = sorted(dict(zip(s2, s1)).iteritems())
assert len(pairs) == len(s1)
s1[:] = [s1_value for s2_value, s1_value in pairs]
--
-Scott David Daniels
scott....@acm.org
Ron Adam
This version takes twice as long on my system, although the result may
vary on other platforms. Dictionary lookups are very fast.
Although replacing zip with izip cut the time in half.
def psort11(s1, s2):
d = dict(izip(s2,s1))
assert len(d) == len(s1)
s1[:] = list(d[v] for v in sorted(d))
Ron
bearoph...@lycos.com
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))
_indices.sort(key=s2.__getitem__)
s1[:] = map(s1.__getitem__, _indices)
else:
if len(d) == len(s1):
s1[:] = [d[v] for v in sorted(d)]
else:
_indices = range(len(s1))
_indices.sort(key=s2.__getitem__)
s1[:] = map(s1.__getitem__, _indices)
Bye,
bearophile
Simon Sun
> 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 variable or like something pattern matching in Haskell.
--
Simon Sun
Kent Johnson
> 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 variable or like something pattern matching in Haskell.
_ 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. So your
example was correct and idiomatic but functionally identical to mine.
Kent
bearoph...@lycos.com
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 * 5
>>> _
Traceback (most recent call last):
File "<interactive input>", line 1, in ?
NameError: name '_' is not defined
>>> 2 * 5
10
>>> _
10
Bye,
bearophile
Ron Adam
Looks good, but I think It would be simpler to just do.
def psort17(s1, s2):
try:
d = dict(izip(s2, s1))
assert len(d) != len(s2)
s1[:] = [d[v] for v in sorted(d)]
except Exception:
_indices = range(len(s1))
_indices.sort(key=s2.__getitem__)
s1[:] = map(s1.__getitem__, _indices)
We don't need to specify which exception. Any problems will just get
raised again on the second try it also fails.
Cheers,
Ron