Re: [Tutor] Calculating and returning possible combinations of elements from a given set
2010/7/28 Dave Angel da...@ieee.org
ZUXOXUS wrote:
snip
My doubt now is whether I can change the way python show the combinations.
I mean, here's what python actually does:
for prod in itertools.product('abc', repeat=3):
print(prod)
('a', 'a', 'a')
('a', 'a', 'b')
('a', 'a', 'c')
('a', 'b', 'a')
('a', 'b', 'b')
('a', 'b', 'c')
[...] etc.
what if I want the combinations listed in a... well, in a list, kind of
like
this:
('aaa', 'aab', aac', 'aba', 'abb', 'abc' [...]etc.)
can I do that?
I have checked how the function works (see below), perhaps I have to just
change couple of lines of the code and voilá, the result displayed as I
want... But unfortunately I'm too newbie for this, or this is too
hardcore:
def product(*args, **kwds):
# product('ABCD', 'xy') -- Ax Ay Bx By Cx Cy Dx Dy
# product(range(2), repeat=3) -- 000 001 010 011 100 101 110 111
pools = map(tuple, args) * kwds.get('repeat', 1)
result = [[]]
for pool in pools:
result = [x+[y] for x in result for y in pool]
for prod in result:
yield tuple(prod)
Any ideas will be very much appreciated.
snip, double-included history
Well itertools.product() already returns an iterator that's equivalent to a
list of tuples. You can print that list simply by doing something like:
print list(itertools.product('abc', repeat=3))
So your question is how you can transform such a list into a list of
strings instead.
so try each of the following.
for prod in itertools.product('abc', repeat=3):
print .join(prod)
print [.join(prod) for prod in itertools.product('abc', repeat=3)]
DaveA
***
Hi DaveA, the second option returns exactly the result I wanted:
print([.join(prod) for prod in itertools.product('abc', repeat=3)])
['aaa', 'aab', 'aac', 'aba', 'abb', 'abc', 'aca', 'acb', 'acc', 'baa',
'bab', 'bac', 'bba', 'bbb', 'bbc', 'bca', 'bcb', 'bcc', 'caa', 'cab', 'cac',
'cba', 'cbb', 'cbc', 'cca', 'ccb', 'ccc']
Thank you very much.
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Re: [Tutor] Calculating and returning possible combinations of elements from a given set
2010/7/28 Dave Angel da...@ieee.org
ZUXOXUS wrote:
Oh, I think i got it:
for prod in itertools.product('ABC', 'ABC'):
print(prod)
('A', 'A')
('A', 'B')
('A', 'C')
('B', 'A')
('B', 'B')
('B', 'C')
('C', 'A')
('C', 'B')
('C', 'C')
Thank you very much!!
2010/7/28 ZUXOXUS zuxo...@gmail.com
You're top-posting, which loses all the context. In this forum, put your
comments after whatever lines you're quoting.
Your latest version gets the product of two. But if you want an arbitrary
number, instead of repeating the iterable ('ABC' in your case), you can use
a repeat count. That's presumably what you were trying to do in your
earlier incantation. But you forgot the 'repeat' keyword:
for prod in itertools.product('ABC', repeat=4):
will give you all the four-tuples.
DaveA
Hey
Sorry for the top-posting, I forgot this rule
Thanks for the reminder, Dave Angel, and for the 'repeat' keyword!
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Re: [Tutor] Calculating and returning possible combinations of elements from a given set
ZUXOXUS wrote:
2010/7/28 Dave Angel da...@ieee.org
snip
Your latest version gets the product of two. But if you want an arbitrary
number, instead of repeating the iterable ('ABC' in your case), you can use
a repeat count. That's presumably what you were trying to do in your
earlier incantation. But you forgot the 'repeat' keyword:
for prod in itertools.product('ABC', repeat=4):
will give you all the four-tuples.
DaveA
snip
Thanks for the reminder, Dave Angel, and for the 'repeat' keyword!
Since you're new to Python, it's probably useful to expound a little
bit, on how you could have figured it out from the help documentation.
itertools.product(/*iterables/[, /repeat/])
Cartesian product of input iterables.
Equivalent to nested for-loops in a generator expression. For
example, product(A, B) returns the same as ((x,y) for x in A for y
in B).
The nested loops cycle like an odometer with the rightmost element
advancing on every iteration. This pattern creates a lexicographic
ordering so that if the input’s iterables are sorted, the product
tuples are emitted in sorted order.
To compute the product of an iterable with itself, specify the
number of repetitions with the optional /repeat/ keyword argument.
For example, product(A, repeat=4) means the same as product(A, A, A, A).
Now that example at the end is exactly what you need here. But let's
look at the first line.
See the *iterables in the formal parameter list. The leading * means you
can put 1, 2, or as many iterables as you like, and they'll each be
treated as an independent dimension in the cartesian product. So when
you put the arguments,
...product(ABC, 2)
the 2 is treated as an iterable, which it isn't. Thus your error. When
there are an arbitrary number of such arguments, followed by another
optional parameter, there's no way the compiler could guess in which
sense you wanted the 2 to be used. So you have to use that parameter's
name as a keyword in your call. If you have a function declared as
def product(*iterables, repeat):
.
then you can call it with
count = 4
product(list1, list2, repeat=count)
and by specifying the 'repeat' keyword, the system knows to stop copying
your arguments into the iterables collection.
Actually, I suspect that if you specify a repeat count, you can only
supply one iterable, but I'm really talking about the language here.
DaveA
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Re: [Tutor] Calculating and returning possible combinations of elements from a given set
2010/7/28 Dave Angel da...@ieee.org
snip
Your latest version gets the product of two. But if you want an arbitrary
number, instead of repeating the iterable ('ABC' in your case), you can
use
a repeat count. That's presumably what you were trying to do in your
earlier incantation. But you forgot the 'repeat' keyword:
for prod in itertools.product('ABC', repeat=4):
will give you all the four-tuples.
DaveA
snip
Thanks for the reminder, Dave Angel, and for the 'repeat' keyword!
Since you're new to Python, it's probably useful to expound a little bit, on
how you could have figured it out from the help documentation.
itertools.product(/*iterables/[, /repeat/])
Cartesian product of input iterables.
Equivalent to nested for-loops in a generator expression. For
example, product(A, B) returns the same as ((x,y) for x in A for y
in B).
The nested loops cycle like an odometer with the rightmost element
advancing on every iteration. This pattern creates a lexicographic
ordering so that if the input’s iterables are sorted, the product
tuples are emitted in sorted order.
To compute the product of an iterable with itself, specify the
number of repetitions with the optional /repeat/ keyword argument.
For example, product(A, repeat=4) means the same as product(A, A, A, A).
Now that example at the end is exactly what you need here. But let's look at
the first line.
See the *iterables in the formal parameter list. The leading * means you can
put 1, 2, or as many iterables as you like, and they'll each be treated as
an independent dimension in the cartesian product. So when you put the
arguments,
...product(ABC, 2)
the 2 is treated as an iterable, which it isn't. Thus your error. When there
are an arbitrary number of such arguments, followed by another optional
parameter, there's no way the compiler could guess in which sense you wanted
the 2 to be used. So you have to use that parameter's name as a keyword in
your call. If you have a function declared as
def product(*iterables, repeat):
.
then you can call it with
count = 4
product(list1, list2, repeat=count)
and by specifying the 'repeat' keyword, the system knows to stop copying
your arguments into the iterables collection.
Actually, I suspect that if you specify a repeat count, you can only supply
one iterable, but I'm really talking about the language here.
DaveA
Wow, Thank you DaveA, that was very useful.
However, as it usually happens, answers trigger new questions.
My doubt now is whether I can change the way python show the combinations.
I mean, here's what python actually does:
for prod in itertools.product('abc', repeat=3):
print(prod)
('a', 'a', 'a')
('a', 'a', 'b')
('a', 'a', 'c')
('a', 'b', 'a')
('a', 'b', 'b')
('a', 'b', 'c')
[...] etc.
what if I want the combinations listed in a... well, in a list, kind of like
this:
('aaa', 'aab', aac', 'aba', 'abb', 'abc' [...]etc.)
can I do that?
I have checked how the function works (see below), perhaps I have to just
change couple of lines of the code and voilá, the result displayed as I
want... But unfortunately I'm too newbie for this, or this is too hardcore:
def product(*args, **kwds):
# product('ABCD', 'xy') -- Ax Ay Bx By Cx Cy Dx Dy
# product(range(2), repeat=3) -- 000 001 010 011 100 101 110 111
pools = map(tuple, args) * kwds.get('repeat', 1)
result = [[]]
for pool in pools:
result = [x+[y] for x in result for y in pool]
for prod in result:
yield tuple(prod)
Any ideas will be very much appreciated.
2010/7/28 Dave Angel da...@ieee.org
ZUXOXUS wrote:
2010/7/28 Dave Angel da...@ieee.org
snip
Your latest version gets the product of two. But if you want an
arbitrary
number, instead of repeating the iterable ('ABC' in your case), you can
use
a repeat count. That's presumably what you were trying to do in your
earlier incantation. But you forgot the 'repeat' keyword:
for prod in itertools.product('ABC', repeat=4):
will give you all the four-tuples.
DaveA
snip
Thanks for the reminder, Dave Angel, and for the 'repeat' keyword!
Since you're new to Python, it's probably useful to expound a little bit,
on how you could have figured it out from the help documentation.
itertools.product(/*iterables/[, /repeat/])
Cartesian product of input iterables.
Equivalent to nested for-loops in a generator expression. For
example, product(A, B) returns the same as ((x,y) for x in A for y
in B).
The nested loops cycle like an odometer with the rightmost element
advancing on every iteration. This pattern creates a lexicographic
ordering so that if the input’s iterables are sorted, the product
tuples are emitted in sorted order.
To compute the product of an iterable with itself, specify the
number of repetitions with the optional /repeat/ keyword argument.
For example, product(A, repeat=4) means the same as product(A, A,
Re: [Tutor] Calculating and returning possible combinations of elements from a given set
ZUXOXUS wrote:
snip
My doubt now is whether I can change the way python show the combinations.
I mean, here's what python actually does:
for prod in itertools.product('abc', repeat=3):
print(prod)
('a', 'a', 'a')
('a', 'a', 'b')
('a', 'a', 'c')
('a', 'b', 'a')
('a', 'b', 'b')
('a', 'b', 'c')
[...] etc.
what if I want the combinations listed in a... well, in a list, kind of like
this:
('aaa', 'aab', aac', 'aba', 'abb', 'abc' [...]etc.)
can I do that?
I have checked how the function works (see below), perhaps I have to just
change couple of lines of the code and voilá, the result displayed as I
want... But unfortunately I'm too newbie for this, or this is too hardcore:
def product(*args, **kwds):
# product('ABCD', 'xy') -- Ax Ay Bx By Cx Cy Dx Dy
# product(range(2), repeat=3) -- 000 001 010 011 100 101 110 111
pools = map(tuple, args) * kwds.get('repeat', 1)
result = [[]]
for pool in pools:
result = [x+[y] for x in result for y in pool]
for prod in result:
yield tuple(prod)
Any ideas will be very much appreciated.
snip, double-included history
Well itertools.product() already returns an iterator that's equivalent
to a list of tuples. You can print that list simply by doing something
like:
print list(itertools.product('abc', repeat=3))
So your question is how you can transform such a list into a list of
strings instead.
so try each of the following.
for prod in itertools.product('abc', repeat=3):
print .join(prod)
print [.join(prod) for prod in itertools.product('abc', repeat=3)]
DaveA
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Re: [Tutor] Calculating and returning possible combinations of elements from a given set
On 07/28/2010 01:20 AM, ZUXOXUS wrote: Hi all pythoners I've got a probably easy to answer question. Say I've got a collections of strings, e.g.: 'man', 'bat', 'super', 'ultra'. They are in a list, or in a sequence or whatever, say a bag of words And now I want to know how many couples I can do with them, and I want the program to show me the actual couples: 'manman', 'manbat', 'mansuper', 'manultra', 'batbat', 'batman', 'batsuper', etc. But hey, why building up new words from just two strings? I also want to know the possible combinations of three words, four words, and perhaps, why not, five words. So, is it easy to do? Sorry, I'm new in programing, and am probably far from being a math-master I'm clueless, I think probably the code have some FOR I IN SEQUENCE... but then what? I don't know how to say: take every element and paste it to another one from the bag, and with another one, and with another one,... If it's too complex, I dont need the whole code recipe, just need some clues, or perhaps a useful link Thank you very much in advance! Take a look in the itertools module http://docs.python.org/library/itertools.html Check the section *Combinatoric generators: (website doesn't have an anchor link for that, search around a bit) Nick * ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Calculating and returning possible combinations of elements from a given set
On 27/07/2010 23:20, ZUXOXUS wrote: Hi all pythoners I've got a probably easy to answer question. Say I've got a collections of strings, e.g.: 'man', 'bat', 'super', 'ultra'. They are in a list, or in a sequence or whatever, say a bag of words And now I want to know how many couples I can do with them, and I want the program to show me the actual couples: 'manman', 'manbat', 'mansuper', 'manultra', 'batbat', 'batman', 'batsuper', etc. But hey, why building up new words from just two strings? I also want to know the possible combinations of three words, four words, and perhaps, why not, five words. So, is it easy to do? Sorry, I'm new in programing, and am probably far from being a math-master I'm clueless, I think probably the code have some FOR I IN SEQUENCE... but then what? I don't know how to say: take every element and paste it to another one from the bag, and with another one, and with another one,... If it's too complex, I dont need the whole code recipe, just need some clues, or perhaps a useful link Thank you very much in advance! ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor The lazy way. http://docs.python.org/library/itertools.html Look for combinations(). HTH. Mark Lawrence. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Calculating and returning possible combinations of elements from a given set
On Tue, 2010-07-27 at 23:31 +0100, Mark Lawrence wrote: On 27/07/2010 23:20, ZUXOXUS wrote: Hi all pythoners I've got a probably easy to answer question. Say I've got a collections of strings, e.g.: 'man', 'bat', 'super', 'ultra'. They are in a list, or in a sequence or whatever, say a bag of words And now I want to know how many couples I can do with them, and I want the program to show me the actual couples: 'manman', 'manbat', 'mansuper', 'manultra', 'batbat', 'batman', 'batsuper', etc. But hey, why building up new words from just two strings? I also want to know the possible combinations of three words, four words, and perhaps, why not, five words. So, is it easy to do? Sorry, I'm new in programing, and am probably far from being a math-master I'm clueless, I think probably the code have some FOR I IN SEQUENCE... but then what? I don't know how to say: take every element and paste it to another one from the bag, and with another one, and with another one,... If it's too complex, I dont need the whole code recipe, just need some clues, or perhaps a useful link Thank you very much in advance! ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor The lazy way. http://docs.python.org/library/itertools.html Look for combinations(). From the examples listed by the OP (see: manbat, batman) I actually believe he is looking for the cartesian product. The syntax should therefore be (DISCLAIMER: I am going by memory without having the possibility of testing right now): import itertools for prod in itertools.product(my_list_of_words, 2): print prod (The two as second parameter tells that you are wishing to have a bi-dimensional product, with both axis having the same list of words. See docs for more info) ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Calculating and returning possible combinations of elements from a given set
Sharp thanks, but:
I try to reproduce the example from the table, but:
import itertools
combinations('ABC', 2)
Traceback (most recent call last):
File pyshell#27, line 1, in module
combinations('ABC', 2)
NameError: name 'combinations' is not defined
If im not mistaken, it should return AB, AC, BA, etc.
I'm using Python 3.1
2010/7/28 Mark Lawrence breamore...@yahoo.co.uk
On 27/07/2010 23:20, ZUXOXUS wrote:
Hi all pythoners
I've got a probably easy to answer question.
Say I've got a collections of strings, e.g.: 'man', 'bat', 'super',
'ultra'.
They are in a list, or in a sequence or whatever, say a bag of words
And now I want to know how many couples I can do with them, and I want the
program to show me the actual couples: 'manman', 'manbat', 'mansuper',
'manultra', 'batbat', 'batman', 'batsuper', etc.
But hey, why building up new words from just two strings? I also want to
know the possible combinations of three words, four words, and perhaps,
why
not, five words.
So, is it easy to do?
Sorry, I'm new in programing, and am probably far from being a math-master
I'm clueless, I think probably the code have some FOR I IN SEQUENCE... but
then what? I don't know how to say: take every element and paste it to
another one from the bag, and with another one, and with another one,...
If it's too complex, I dont need the whole code recipe, just need some
clues, or perhaps a useful link
Thank you very much in advance!
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The lazy way.
http://docs.python.org/library/itertools.html
Look for combinations().
HTH.
Mark Lawrence.
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Re: [Tutor] Calculating and returning possible combinations of elements from a given set
Mac,
this is what I get:
for prod in itertools.product('ABC', 2):
print(prod)
Traceback (most recent call last):
File pyshell#34, line 1, in module
for prod in itertools.product('ABC', 2):
TypeError: 'int' object is not iterable
hmm, what might be that 'int' object? 2?
2010/7/28 ZUXOXUS zuxo...@gmail.com
Sharp thanks, but:
I try to reproduce the example from the table, but:
import itertools
combinations('ABC', 2)
Traceback (most recent call last):
File pyshell#27, line 1, in module
combinations('ABC', 2)
NameError: name 'combinations' is not defined
If im not mistaken, it should return AB, AC, BA, etc.
I'm using Python 3.1
2010/7/28 Mark Lawrence breamore...@yahoo.co.uk
On 27/07/2010 23:20, ZUXOXUS wrote:
Hi all pythoners
I've got a probably easy to answer question.
Say I've got a collections of strings, e.g.: 'man', 'bat', 'super',
'ultra'.
They are in a list, or in a sequence or whatever, say a bag of words
And now I want to know how many couples I can do with them, and I want
the
program to show me the actual couples: 'manman', 'manbat', 'mansuper',
'manultra', 'batbat', 'batman', 'batsuper', etc.
But hey, why building up new words from just two strings? I also want to
know the possible combinations of three words, four words, and perhaps,
why
not, five words.
So, is it easy to do?
Sorry, I'm new in programing, and am probably far from being a
math-master
I'm clueless, I think probably the code have some FOR I IN SEQUENCE...
but
then what? I don't know how to say: take every element and paste it to
another one from the bag, and with another one, and with another one,...
If it's too complex, I dont need the whole code recipe, just need some
clues, or perhaps a useful link
Thank you very much in advance!
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The lazy way.
http://docs.python.org/library/itertools.html
Look for combinations().
HTH.
Mark Lawrence.
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Re: [Tutor] Calculating and returning possible combinations of elements from a given set
Oh, I think i got it:
for prod in itertools.product('ABC', 'ABC'):
print(prod)
('A', 'A')
('A', 'B')
('A', 'C')
('B', 'A')
('B', 'B')
('B', 'C')
('C', 'A')
('C', 'B')
('C', 'C')
Thank you very much!!
2010/7/28 ZUXOXUS zuxo...@gmail.com
Mac,
this is what I get:
for prod in itertools.product('ABC', 2):
print(prod)
Traceback (most recent call last):
File pyshell#34, line 1, in module
for prod in itertools.product('ABC', 2):
TypeError: 'int' object is not iterable
hmm, what might be that 'int' object? 2?
2010/7/28 ZUXOXUS zuxo...@gmail.com
Sharp thanks, but:
I try to reproduce the example from the table, but:
import itertools
combinations('ABC', 2)
Traceback (most recent call last):
File pyshell#27, line 1, in module
combinations('ABC', 2)
NameError: name 'combinations' is not defined
If im not mistaken, it should return AB, AC, BA, etc.
I'm using Python 3.1
2010/7/28 Mark Lawrence breamore...@yahoo.co.uk
On 27/07/2010 23:20, ZUXOXUS wrote:
Hi all pythoners
I've got a probably easy to answer question.
Say I've got a collections of strings, e.g.: 'man', 'bat', 'super',
'ultra'.
They are in a list, or in a sequence or whatever, say a bag of words
And now I want to know how many couples I can do with them, and I want
the
program to show me the actual couples: 'manman', 'manbat', 'mansuper',
'manultra', 'batbat', 'batman', 'batsuper', etc.
But hey, why building up new words from just two strings? I also want to
know the possible combinations of three words, four words, and perhaps,
why
not, five words.
So, is it easy to do?
Sorry, I'm new in programing, and am probably far from being a
math-master
I'm clueless, I think probably the code have some FOR I IN SEQUENCE...
but
then what? I don't know how to say: take every element and paste it to
another one from the bag, and with another one, and with another one,...
If it's too complex, I dont need the whole code recipe, just need some
clues, or perhaps a useful link
Thank you very much in advance!
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The lazy way.
http://docs.python.org/library/itertools.html
Look for combinations().
HTH.
Mark Lawrence.
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Re: [Tutor] Calculating and returning possible combinations of elements from a given set
ZUXOXUS wrote:
Oh, I think i got it:
for prod in itertools.product('ABC', 'ABC'):
print(prod)
('A', 'A')
('A', 'B')
('A', 'C')
('B', 'A')
('B', 'B')
('B', 'C')
('C', 'A')
('C', 'B')
('C', 'C')
Thank you very much!!
2010/7/28 ZUXOXUS zuxo...@gmail.com
You're top-posting, which loses all the context. In this forum, put your
comments after whatever lines you're quoting.
Your latest version gets the product of two. But if you want an
arbitrary number, instead of repeating the iterable ('ABC' in your
case), you can use a repeat count. That's presumably what you were
trying to do in your earlier incantation. But you forgot the 'repeat'
keyword:
for prod in itertools.product('ABC', repeat=4):
will give you all the four-tuples.
DaveA
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