Hello,
This is a question for the best method (in terms of performance
only) to choose a random element from a list among those that satisfy
a certain property.
This is the setting: I need to pick from a list a random element
that satisfies a given property. All or none of the elements may have
the property. Most of the time, many of the elements will satisfy the
property, and the property is a bit expensive to evaluate. Chance of
having the property are uniform among elements.
A simple approach is:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Returns None if no element has the property
'''
random.shuffle(a_list)
for i in a_list:
if property(i): return i
but that requires to shuffle the list every time.
A second approach, that works if we know that at least one element of
the list has the property, is:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Loops forever if no element has the property
'''
while 1:
i=random.choice(a_list)
if property(i): return i
which is more efficient (on average) if many elements of the list have
the property and less efficient if only few elements of the list has
the property (and goes crazy if no element has the property)
Yet another one:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
'''
b_list=[x for x in a_list if property(x)]
try:
return random.choice(b_list)
finally: return None
but this one checks the property on all the elements, which is no
good.
I don't need strong random numbers, so a simple solution like:
import random
globalRNG=random.Random()
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Works only if len(a_list)+1 is prime: uses Fermat's little theorem
'''
a=globalRNG(1,len(a_list))
ind=a
for i in xrange(len(a_list)):
x=a_list[a1]
if property(x):return x
ind*=a
but this works only if len(a_list)+1 is prime!!! Now this one could be
saved if there were an efficient method to find a prime number given
than a number n but not very much greater...
Any other ideas? Thanks everybody 15 2513
On Jan 4, 7:55 pm, c...@mailinator.com wrote:
Hello,
This is a question for the best method (in terms of performance
only) to choose a random element from a list among those that satisfy
a certain property.
This is the setting: I need to pick from a list a random element
that satisfies a given property. All or none of the elements may have
the property. Most of the time, many of the elements will satisfy the
property, and the property is a bit expensive to evaluate. Chance of
having the property are uniform among elements.
A simple approach is:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Returns None if no element has the property
'''
random.shuffle(a_list)
for i in a_list:
if property(i): return i
but that requires to shuffle the list every time.
A second approach, that works if we know that at least one element of
the list has the property, is:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Loops forever if no element has the property
'''
while 1:
i=random.choice(a_list)
if property(i): return i
which is more efficient (on average) if many elements of the list have
the property and less efficient if only few elements of the list has
the property (and goes crazy if no element has the property)
Yet another one:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
'''
b_list=[x for x in a_list if property(x)]
try:
return random.choice(b_list)
finally: return None
but this one checks the property on all the elements, which is no
good.
I don't need strong random numbers, so a simple solution like:
import random
globalRNG=random.Random()
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Works only if len(a_list)+1 is prime: uses Fermat's little theorem
'''
a=globalRNG(1,len(a_list))
ind=a
for i in xrange(len(a_list)):
x=a_list[a1]
if property(x):return x
ind*=a
but this works only if len(a_list)+1 is prime!!! Now this one could be
saved if there were an efficient method to find a prime number given
than a number n but not very much greater...
Any other ideas? Thanks everybody
Caching might help.
If random_pick is called several times with the same list(s) then
cache the result of
[property(i) for i in a_list] against a_list.
If random_pick is called several times with list(s) whith multiple
instances of list items then cache
property(i) against i for i in a_list .
You could do both.
You might investigate wether property(i) could be implemented using a
faster algorithm, maybe table lookup, or interpolation from initial
table lookup.
 Paddy.
Caching might help.
If random_pick is called several times with the same list(s) then
cache the result of
[property(i) for i in a_list] against a_list.
If random_pick is called several times with list(s) with multiple
instances of list items then cache
property(i) against i for i in a_list .
You could do both.
You might investigate wether property(i) could be implemented using a
faster algorithm, maybe table lookup, or interpolation from initial
table lookup.
 Paddy.
Thanks, Paddy. Those are interesting general comments for the general
problem.
By the way, I noticed two things:
I forgot to write randint in the third approach:
a=globalRNG.randint(1,len(a_list))
The warning "The group you are posting to is a Usenet group. Messages
posted to this group will make your email address visible to anyone on
the Internet." means a person, but not a bot, may see my email
address, so it is safe to use my real address next time...
On Jan 4, 7:55*pm, c...@mailinator.com wrote:
* Hello,
* This is a question for the best method (in terms of performance
only) to choose a random element from a list among those that satisfy
a certain property.
* This is the setting: I need to pick from a list a random element
that satisfies a given property. All or none of the elements may have
the property. Most of the time, many of the elements will satisfy the
property, and the property is a bit expensive to evaluate. Chance of
having the property are uniform among elements.
Here's some code that uses a cached random sorting of the list. It
assumes you'll want more than one random element. It never calls
'prop' on the same element twice, and it's O(n) even when the elements
that pass 'prop' are sparse. I hope this is useful to you!
import random
class RandomPicker(object):
def __init__(self, seq, prop=lambda x:True):
seq = list(seq)
random.shuffle(seq)
# Store with the item whether we've computed prop on it
already.
self.random_list = [(x, False) for x in seq]
self.prop = prop
def pick(self):
for i, (x, p) in enumerate(self.random_list):
if p or self.prop(x):
if not p:
# Record the fact that self.prop passed.
self.random_list[i] = (x, True)
# Chop off the items that prop failed on.
self.random_list = self.random_list[i:]
r = self.random_list
# Instead of shuffling the whole list again, just
insert
# x back in the list randomly. Since the remaining
elements
# are randomly sorted already, this is ok.
n = random.randint(0, len(r)  1)
r[0], r[n] = r[n], r[0]
return x
# Nothing in the list passes the 'prop' test.
return None
# Example: pick 100 odd integers from 0 to 1000.
a = RandomPicker(xrange(1000), lambda x: x % 2 == 1)
print [a.pick() for i in xrange(100)]

Paul Hankin
On Jan 4, 7:55*pm, c...@mailinator.com wrote:
* Hello,
* This is a question for the best method (in terms of performance
only) to choose a random element from a list among those that satisfy
a certain property.
* This is the setting: I need to pick from a list a random element
that satisfies a given property. All or none of the elements may have
the property. Most of the time, many of the elements will satisfy the
property, and the property is a bit expensive to evaluate. Chance of
having the property are uniform among elements.
The generator function below yields an infinite sequence of randomly
picked elements from the list who satisfy the property, or nothing if
the list contains no element satisfying the property. It guarantees
that each time, prop() will either not be called or will be called
just enough to find one single item that satisfies it. The catch is
that you need to have an estimate for the number of items that satisfy
the property in the list.
import random
from itertools import islice, ifilter
def picker(lst, prop, np):
# lst: list of items to pick elements from
# prop: property that picked elements must fulfil
# np: (good estimate of) number of items that
# satisfy the property
random.shuffle(lst)
plst = [] # items we know fulfil prop
for item in ifilter(prop, lst):
# The next random item may be one already yielded
while True:
i = random.randrange(np)
if i >= len(plst): break
yield plst[i]
# Or it may be a new one
plst.append(item)
if len(plst) np: np = len(plst)
yield item
# Now we know all items fulfilling prop
if not plst: return
while True:
yield plst[random.randrange(len(plst))]
def test(picker, n=1000):
res = {}
for val in islice(picker, n):
res[val] = res.get(val, 0) + 1
return res
Example:
>>p = picker(range(20), lambda x: x % 2, 10) test(p)
{1: 113, 3: 96, 5: 87, 7: 91, 9: 109, 11: 91, 13: 106, 15: 101, 17:
109, 19: 97}
>>p = picker(range(20), lambda x: False, 10) p.next()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
I don't know if there is a good idea in there, I'll let you be the
judge :)

Arnaud
Hello, Paul and Arnaud.
While I think about your answers: do you think there is any way to
avoid shuffle?
It may take unnecessary long on a long list most of whose elements
have the property.
On Jan 5, 5:07 pm, c...@mailinator.com wrote:
Hello, Paul and Arnaud.
While I think about your answers: do you think there is any way to
avoid shuffle?
It may take unnecessary long on a long list most of whose elements
have the property.
Umm...
You provide nice answers in the case many elements are picked from the
same list.
Any ideas for the case when the picker is called many times on a
program, but never twice with the same list?
Any other ideas?
How about this:
def random_pick(list, property):
L = len(list)
pos = start = random.randrange(L)
while 1:
x = list[pos]
if property(x): return x
pos = (pos + 1) % L
if pos == start:
raise ValueError, "no such item"
Regards,
Martin
On Jan 5, 5:12*pm, "Martin v. Löwis" <mar...@v.loewis.dewrote:
Any other ideas?
How about this:
def random_pick(list, property):
* * L = len(list)
* * pos = start = random.randrange(L)
* * while 1:
* * * * x = list[pos]
* * * * if property(x): return x
* * * * pos = (pos + 1) % L
* * * * if pos == start:
* * * * * *raise ValueError, "no such item"
This might be acceptable for the OP's use, but it's strongly biased
towards values which follow a long stream of things that fail
property.
print [random_pick(range(100), lambda x:x 90)
for i in range(999)].count(91)
929
If it were uniform, it should be around 111.

Paul Hankin
On Jan 5, 4:14*pm, c...@mailinator.com wrote:
On Jan 5, 5:07 pm, c...@mailinator.com wrote:
Hello, Paul and Arnaud.
While I think about your answers: do you think there is any way to
avoid shuffle?
It may take unnecessary long on a long list most of whose elements
have the property.
Umm...
You provide nice answers in the case many elements are picked from the
same list.
Any ideas for the case when the picker is called many times on a
program, but never twice with the same list?
Here's a pragmatic optimisation for any algorithm: first test some
elements at random in case you get lucky or most of the elements are
good.
Eg, that optimisation applied to the naive shuffle algorithm.
import random
import itertools
def pick_random_fast(seq, prop):
L = len(seq)
stabs = 5
for i in xrange(stabs):
r = random.randrange(L)
if prop(seq[r]): return seq[r]
random.shuffle(seq)
return itertools.ifilter(prop, seq).next()
I've used 5 'stabs' here. Perhaps it should be a function of L, but I
suppose you can tune it for your data.

Paul Hankin
On Sat, 05 Jan 2008 08:14:46 0800, caca wrote:
On Jan 5, 5:07 pm, c...@mailinator.com wrote:
>Hello, Paul and Arnaud. While I think about your answers: do you think there is any way to avoid shuffle? It may take unnecessary long on a long list most of whose elements have the property.
Umm...
You provide nice answers in the case many elements are picked from the
same list.
Any ideas for the case when the picker is called many times on a
program, but never twice with the same list?
ISTM the best thing would be to reimplement the shuffle algorithm, but to
stop shuffling as soon as you get a hit. The drawback is that it's a
destructive operation, but that doesn't sound like it's an issue for you.
Here's something for starters:
def random_element_with_property(x,test_property_func) :
for i in xrange(len(x)1):
j = random.randrange(i+1,len(x))
tmp = x[j]
if test_property_func(tmp):
return tmp
x[j] = x[i]
x[i] = tmp
return None
Then, for example, use it like this:
def odd(i): return i&1
e = random_element_with_property(range(20),odd)
Carl Banks
On Jan 5, 8:14 am, c...@mailinator.com wrote:
On Jan 5, 5:07 pm, c...@mailinator.com wrote:
Hello, Paul and Arnaud.
While I think about your answers: do you think there is any way to
avoid shuffle?
It may take unnecessary long on a long list most of whose elements
have the property.
Umm...
You provide nice answers in the case many elements are picked from the
same list.
Any ideas for the case when the picker is called many times on a
program, but never twice with the same list?
Here's my stab:
from random import randint, seed
from time import time
from sys import stdout
seed(time())
iterations = 0#DEBUG
def pick_random(seq, prop=bool):
temp = list(seq)
global iterations#DEBUG
while temp:
iterations += 1#DEBUG
i = randint(0, len(temp)  1)
if prop(temp[i]): return temp[i]
else: del temp[i]
def generate_list(length):
l = list()
for x in xrange(length): l.append(randint(0,1) * randint(1,1000))
return l
count = 0
for x in xrange(1000):
count += 1
print pick_random(generate_list(1000)),
print
print "AVERAGE ITERATIONS:", float(iterations) / count
The average number of iterations is 1/p where p is the chance of your
property being true. It's independent of list size! Just remove the
DEBUG lines and it's ready for use.
Buck
On Jan 5, 9:12 am, "Martin v. Löwis" <mar...@v.loewis.dewrote:
Any other ideas?
How about this:
def random_pick(list, property):
L = len(list)
pos = start = random.randrange(L)
while 1:
x = list[pos]
if property(x): return x
pos = (pos + 1) % L
if pos == start:
raise ValueError, "no such item"
Regards,
Martin
I thought about this, but in the sequence "00012" (and property =
bool) the 1 will be returned four times as often as the 2. Maybe
that's ok...
On Jan 5, 4:16 am, c...@mailinator.com wrote:
The warning "The group you are posting to is a Usenet group. Messages
posted to this group will make your email address visible to anyone on
the Internet." means a person, but not a bot, may see my email
address, so it is safe to use my real address next time...
Wrong. A bot may be able to read and scan all messages sent through
Usenet.
On Jan 5, 6:37 am, Paddy <paddy3...@googlemail.comwrote:
On Jan 4, 7:55 pm, c...@mailinator.com wrote:
Hello,
This is a question for the best method (in terms of performance
only) to choose a random element from a list among those that satisfy
a certain property.
This is the setting: I need to pick from a list a random element
that satisfies a given property. All or none of the elements may have
the property. Most of the time, many of the elements will satisfy the
property, and the property is a bit expensive to evaluate. Chance of
having the property are uniform among elements.
A simple approach is:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Returns None if no element has the property
'''
random.shuffle(a_list)
for i in a_list:
if property(i): return i
but that requires to shuffle the list every time.
A second approach, that works if we know that at least one element of
the list has the property, is:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Loops forever if no element has the property
'''
while 1:
i=random.choice(a_list)
if property(i): return i
which is more efficient (on average) if many elements of the list have
the property and less efficient if only few elements of the list has
the property (and goes crazy if no element has the property)
Yet another one:
import random
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
'''
b_list=[x for x in a_list if property(x)]
try:
return random.choice(b_list)
finally: return None
but this one checks the property on all the elements, which is no
good.
I don't need strong random numbers, so a simple solution like:
import random
globalRNG=random.Random()
def random_pick(a_list,property):
'''Returns a random element from a list that has the property
Works only if len(a_list)+1 is prime: uses Fermat's little theorem
'''
a=globalRNG(1,len(a_list))
ind=a
for i in xrange(len(a_list)):
x=a_list[a1]
if property(x):return x
ind*=a
but this works only if len(a_list)+1 is prime!!! Now this one could be
saved if there were an efficient method to find a prime number given
than a number n but not very much greater...
Any other ideas? Thanks everybody
Caching might help.
If random_pick is called several times with the same list(s) then
cache the result of
[property(i) for i in a_list] against a_list.
If random_pick is called several times with list(s) whith multiple
instances of list items then cache
property(i) against i for i in a_list .
You could do both.
You might investigate wether property(i) could be implemented using a
faster algorithm, maybe table lookup, or interpolation from initial
table lookup.
 Paddy.
Here's a caching decorator that you could apply to function property: http://aspn.activestate.com/ASPN/Coo.../Recipe/498245
 Paddy.
Just for fun, I profiled my answer versus the final answer...
This mailing list is awesome!
PS:ajaksu, I have to leave now, I hope bukzor's answer was enough to
you (at least for the moment) This discussion thread is closed Replies have been disabled for this discussion. Similar topics
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