Re: Efficient way to sum a product of numbers...
After simplifying my problem, I can say that I want to get the sum of
the product of two culumns:
Say
m= [[ 'a', 1], [ 'b', 2],[ 'a', 3]]
assuming you meant ['c', 3] here...^
r={'a':4, 'b':5, 'c':6}
What I need is the calculation
1*4 + 2*5 + 3*4 = 4 + 10 + 12 = 26
and you mean 3*6 here instead of 3*4, which is 18 instead of
12, making the whole sum 4+10+18=32
Then it sounds like you could do something like
result = sum(v * r[k] for k,v in m)
where m is any arbitrary iterable of tuples. If the keys (the
letters) aren't guaranteed to be in r, then you can use
defaults (in this case 0, but could just as likely be 1
depending on your intent):
result = sum(v * r.get(k,0) for k,v in m)
If the conditions above don't hold, you'll have to introduce me
to your new math. ;-)
-tkc
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Re: Efficient way to sum a product of numbers...
On Aug 31, 6:30 pm, Tim Chase python.l...@tim.thechases.com wrote:
After simplifying my problem, I can say that I want to get the sum of
the product of two culumns:
Say
m= [[ 'a', 1], [ 'b', 2],[ 'a', 3]]
assuming you meant ['c', 3] here... ^ r={'a':4, 'b':5, 'c':6}
What I need is the calculation
1*4 + 2*5 + 3*4 = 4 + 10 + 12 = 26
and you mean 3*6 here instead of 3*4, which is 18 instead of
12, making the whole sum 4+10+18=32
Then it sounds like you could do something like
result = sum(v * r[k] for k,v in m)
where m is any arbitrary iterable of tuples. If the keys (the
letters) aren't guaranteed to be in r, then you can use
defaults (in this case 0, but could just as likely be 1
depending on your intent):
result = sum(v * r.get(k,0) for k,v in m)
If the conditions above don't hold, you'll have to introduce me
to your new math. ;-)
-tkc
Hello Tim,
There is no mistake in my original post, so I really meant [ 'a', 3]
Imagine that m contains time sheets of suppliers
supplier 'a' has worked for you 1 hour
supplier 'b' has worked for you 2 hour
supplier 'a' has worked for you 3 hour
Now
supplier 'a' charges $4 per hour
supplier 'b' charges $5 per hour
supplier 'c' charges $6 per hour
I want to know how much I will be charged this month by my pannel of
suppliers.
1*4 + 2*5 + 3*4 = 4 + 10 + 12 = 26
This is what I am after.
I expect all my suppliers to have handed me in advance the per hour
fee. If at least one hasn't, I must know that the result is undefined.
Hope this helps
Vicente Soler
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Re: Efficient way to sum a product of numbers...
vsoler wrote:
On Aug 31, 6:30 pm, Tim Chase python.l...@tim.thechases.com wrote:
After simplifying my problem, I can say that I want to get the sum of
the product of two culumns:
Say
m= [[ 'a', 1], [ 'b', 2],[ 'a', 3]]
assuming you meant ['c', 3] here...^ r={'a':4, 'b':5, 'c':6}
What I need is the calculation
1*4 + 2*5 + 3*4 = 4 + 10 + 12 = 26
and you mean 3*6 here instead of 3*4, which is 18 instead of
12, making the whole sum 4+10+18=32
Then it sounds like you could do something like
result = sum(v * r[k] for k,v in m)
where m is any arbitrary iterable of tuples. If the keys (the
letters) aren't guaranteed to be in r, then you can use
defaults (in this case 0, but could just as likely be 1
depending on your intent):
result = sum(v * r.get(k,0) for k,v in m)
If the conditions above don't hold, you'll have to introduce me
to your new math. ;-)
There is no mistake in my original post, so I really meant [ 'a', 3]
Ah...that makes more sense of the data. My answer still holds
then. Use the r[k] version instead of the r.get(...) version,
and it will throw an exception if the rate doesn't exist in your
mapping. (a KeyError if you want to catch it)
-tkc
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Re: Efficient way to sum a product of numbers...
vsoler vicente.so...@gmail.com writes:
m= [[ 'a', 1], [ 'b', 2],[ 'a', 3]]
r={'a':4, 'b':5, 'c':6}
What I need is the calculation
1*4 + 2*5 + 3*4 = 4 + 10 + 12 = 26
sum(r[k]*w for k,w in m)
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Re: Efficient way to sum a product of numbers...
On Aug 31, 6:59 pm, Tim Chase python.l...@tim.thechases.com wrote:
vsoler wrote:
On Aug 31, 6:30 pm, Tim Chase python.l...@tim.thechases.com wrote:
After simplifying my problem, I can say that I want to get the sum of
the product of two culumns:
Say
m= [[ 'a', 1], [ 'b', 2],[ 'a', 3]]
assuming you meant ['c', 3] here... ^ r={'a':4, 'b':5, 'c':6}
What I need is the calculation
1*4 + 2*5 + 3*4 = 4 + 10 + 12 = 26
and you mean 3*6 here instead of 3*4, which is 18 instead of
12, making the whole sum 4+10+18=32
Then it sounds like you could do something like
result = sum(v * r[k] for k,v in m)
where m is any arbitrary iterable of tuples. If the keys (the
letters) aren't guaranteed to be in r, then you can use
defaults (in this case 0, but could just as likely be 1
depending on your intent):
result = sum(v * r.get(k,0) for k,v in m)
If the conditions above don't hold, you'll have to introduce me
to your new math. ;-)
There is no mistake in my original post, so I really meant [ 'a', 3]
Ah...that makes more sense of the data. My answer still holds
then. Use the r[k] version instead of the r.get(...) version,
and it will throw an exception if the rate doesn't exist in your
mapping. (a KeyError if you want to catch it)
-tkc
It works!!!
Thank you
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Re: Efficient way to sum a product of numbers...
31-08-2009 o 18:19:28 vsoler vicente.so...@gmail.com wrote:
Say
m= [[ 'a', 1], [ 'b', 2],[ 'a', 3]]
r={'a':4, 'b':5, 'c':6}
What I need is the calculation
1*4 + 2*5 + 3*4 = 4 + 10 + 12 = 26
That is, for each row list in variable 'm' look for its first element
in variable 'r' and multiply the value found by the second element in
row 'm'. After that, sum all the products.
What's an efficient way to do it? I have thousands of these
calculations to make on a big data file.
31-08-2009 o 18:30:27 Tim Chase python.l...@tim.thechases.com wrote:
result = sum(v * r[k] for k,v in m)
You can also check if this isn't more efficient:
from itertools import starmap
from operator import mul
result = sum(starmap(mul, ((r[name], hour) for name, hour in m)))
Or, if you had m in form of two lists:
names = ['a', 'b', 'a']
hours = [1, 2, 3]
...then you could do:
from itertools import imap as map # - remove if you use Py3.x
from operator import mul
result = sum(map(mul, map(r.__getitem__, names), hours))
Cheers,
*j
PS. I've done a quick test on my computer (Pentium 4, 2.4Ghz, Linux):
setup = from itertools import starmap, imap ; from operator import
mul; import random, string; names =
[rndom.choice(string.ascii_letters) for x in xrange(1)]; hours =
[random.randint(1, 12) for x in xrange(1000)]; m = zip(names, hours);
workers = set(names); r = dict(zip(workers, (random.randint(1, 10) for
x in xrange(en(workers)
tests = (
... 'sum(v * r[k] for k,v in m)',
... 'sum(starmap(mul, ((r[name], hour) for name, hour in m)))',
... 'sum(imap(mul, imap(r.__getitem__, names), hours))',
... )
for t in tests:
... print t
... timeit.repeat(t, setup, number=1000)
... print
...
sum(v * r[k] for k,v in m)
[6.2493009567260742, 6.1892399787902832, 6.2634339332580566]
sum(starmap(mul, ((r[name], hour) for name, hour in m)))
[9.3293819427490234, 10.280816078186035, 9.2766909599304199]
sum(imap(mul, imap(r.__getitem__, names), hours))
[5.7341709136962891, 5.5898380279541016, 5.7318859100341797]
--
Jan Kaliszewski (zuo) z...@chopin.edu.pl
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Re: Efficient way to sum a product of numbers...
31-08-2009 o 22:28:56 Jan Kaliszewski z...@chopin.edu.pl wrote: setup = from itertools import starmap, imap ; from operator import mul; import random, string; names = [rndom.choice(string. ascii_letters) for x in xrange(1)]; hours = [random.randint( 1, 12) for x in xrange(1000)]; m = zip(names, hours); workers = set(names); r = dict(zip(workers, (random.randint(1, 10) for x i n xrange(en(workers) Erratum -- should be: setup = ( ... 'from itertools import starmap, imap;' ... 'from operator import mul;' ... 'import random, string; names' ... ' = [random.choice(string.ascii_letters)' ... 'for x in xrange(1)];' ... 'hours = [random.randint(1, 12)' ... for x in xrange(1)];' ... 'm = zip(names, hours);' ... 'workers = set(names);' ... 'r = dict(zip(workers, (random.randint(1, 10)' ... ' for x in xrange(len(workers)' ... ) -- Jan Kaliszewski (zuo) z...@chopin.edu.pl -- http://mail.python.org/mailman/listinfo/python-list
