Re: List problem
On 2012-12-02, Thomas Bach thb...@students.uni-mainz.de wrote: On Sun, Dec 02, 2012 at 04:16:01PM +0100, Lutz Horn wrote: len([x for x in l if x[1] == 'VBD']) Another way is sum(1 for x in l if x[1] == 'VBD') which saves the list creation. To also index them: vbdix = [i for i, a in emumerate(l) if a[1] == 'VBD'] vbdno = len(indices) -- Neil Cerutti -- http://mail.python.org/mailman/listinfo/python-list
Re: List problem
In 8c0a3ea9-2560-47eb-a9c7-3770e41fe...@googlegroups.com
subhabangal...@gmail.com writes:
Dear Group,
I have a list of the following pattern,
[('', ''), ('Eastern', 'NNP'), ('Army', 'NNP'), ('Commander', 'NNP'), (=
'Lt', 'NNP'), ('Gen', 'NNP'), ('Dalbir', 'NNP'), ('Singh', 'NNP'), ('Suhag'=
, 'NNP'), ('briefed', 'VBD'), ('the', 'DT'), ('Army', 'NNP'), ('chief', 'NN=
'), ('on', 'IN'), ('the', 'DT'), ('operational', 'JJ'), ('preparedness', 'N=
N'), ('and', 'CC'), ('the', 'DT'), ('security', 'NN'), ('scenario', 'NN'), =
('in', 'IN'), ('the', 'DT'), ('eastern', 'NN'), ('region', 'NN'), (',', ','=
), ('', ''), ('defence', 'NN'), ('spokesperson', 'NN'), ('Group', 'NNP'=
), ('Capt', 'NNP'), ('T', 'NNP'), ('K', 'NNP'), ('Singha', 'NNP'), ('said',=
'VBD'), ('here', 'RB')]
Now, as we see it has multiple VBD elements.
I want to recognize,count and index them all.
That depends on exactly what you mean by 'reorganize' and 'index'. But
here's a start:
items = [('', ''), ('Eastern', 'NNP'), ('Army', 'NNP'),
('Commander', 'NNP'), ('Lt', 'NNP'), ('Gen', 'NNP'),
('Dalbir', 'NNP'), ('Singh', 'NNP'), ('Suhag' , 'NNP'),
('briefed', 'VBD'), ('the', 'DT'), ('Army', 'NNP'),
('chief', 'NN'), ('on', 'IN'), ('the', 'DT'),
('operational', 'JJ'), ('preparedness', 'NN'), ('and', 'CC'),
('the', 'DT'), ('security', 'NN'), ('scenario', 'NN'),
('in', 'IN'), ('the', 'DT'), ('eastern', 'NN'), ('region', 'NN'),
(',', ','), ('', ''), ('defence', 'NN'),
('spokesperson', 'NN'), ('Group', 'NNP'), ('Capt', 'NNP'),
('T', 'NNP'), ('K', 'NNP'), ('Singha', 'NNP'), ('said', 'VBD'),
('here', 'RB')]
vbds = [item[0] for item in items if item[1] == 'VBD']
print vbds
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Re: List problem
Him
Am 02.12.2012 um 16:03 schrieb subhabangal...@gmail.com:
I have a list of the following pattern,
[('', ''), ('Eastern', 'NNP'), ('Army', 'NNP'), ('Commander', 'NNP'),
('Lt', 'NNP'), ('Gen', 'NNP'), ('Dalbir', 'NNP'), ('Singh', 'NNP'), ('Suhag',
'NNP'), ('briefed', 'VBD'), ('the', 'DT'), ('Army', 'NNP'), ('chief', 'NN'),
('on', 'IN'), ('the', 'DT'), ('operational', 'JJ'), ('preparedness', 'NN'),
('and', 'CC'), ('the', 'DT'), ('security', 'NN'), ('scenario', 'NN'), ('in',
'IN'), ('the', 'DT'), ('eastern', 'NN'), ('region', 'NN'), (',', ','), ('',
''), ('defence', 'NN'), ('spokesperson', 'NN'), ('Group', 'NNP'), ('Capt',
'NNP'), ('T', 'NNP'), ('K', 'NNP'), ('Singha', 'NNP'), ('said', 'VBD'),
('here', 'RB')]
Now, as we see it has multiple VBD elements.
I want to recognize,count and index them all.
len([x for x in l if x[1] == 'VBD'])
Lutz
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Re: List problem
On Sun, Dec 02, 2012 at 04:16:01PM +0100, Lutz Horn wrote: len([x for x in l if x[1] == 'VBD']) Another way is sum(1 for x in l if x[1] == 'VBD') which saves the list creation. Regards, Thomas. -- http://mail.python.org/mailman/listinfo/python-list
Re: List problem
On Sunday, December 2, 2012 9:29:22 PM UTC+5:30, Thomas Bach wrote: On Sun, Dec 02, 2012 at 04:16:01PM +0100, Lutz Horn wrote: len([x for x in l if x[1] == 'VBD']) Another way is sum(1 for x in l if x[1] == 'VBD') which saves the list creation. Regards, Thomas. Thanks. After I posted I got a solution as, [x for x, y in enumerate(chunk_word) if /VB in y] but you are smarter. Thanks. Regards, Subhabrata. -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On Sunday, September 23, 2012 2:31:48 PM UTC-7, jimbo1qaz wrote: I have a nested list. Whenever I make a copy of the list, changes in one affect the other, even when I use list(orig) or even copy the sublists one by one. I have to manually copy each cell over for it to work. Link to broken code: http://jimbopy.pastebay.net/1090401 No, actually that's the OK code. http://jimbopy.pastebay.net/1090494 is the broken one. -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On 23 September 2012 22:31, jimbo1qaz jimmyli1...@gmail.com wrote: I have a nested list. Whenever I make a copy of the list, changes in one affect the other, even when I use list(orig) or even copy the sublists one by one. I have to manually copy each cell over for it to work. Link to broken code: http://jimbopy.pastebay.net/1090401 There are many things wrong with that code but I can't tell what you're referring to. Can you paste the code into your post (rather than just a link to it)? Can you also explain what you want it to do and at what point it does the wrong thing? Oscar -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On Mon, Sep 24, 2012 at 7:44 AM, jimbo1qaz jimmyli1...@gmail.com wrote: On Sunday, September 23, 2012 2:31:48 PM UTC-7, jimbo1qaz wrote: I have a nested list. Whenever I make a copy of the list, changes in one affect the other, even when I use list(orig) or even copy the sublists one by one. I have to manually copy each cell over for it to work. Link to broken code: http://jimbopy.pastebay.net/1090401 No, actually that's the OK code. http://jimbopy.pastebay.net/1090494 is the broken one. The first thing I'd change about that code is the whole thing of using try/exec/except to suppress IndexError. Definitely not good code. I'm not wholly certain, but I think you might run into weird issues with negative OOBounds indices (since Python treats a negative list index as counting from the far end). This is nothing to do with your originally requested issue, which I can't see the cause of in your script there. But when you assign a list, you just get another reference to the same list. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On 09/23/2012 05:44 PM, jimbo1qaz wrote: On Sunday, September 23, 2012 2:31:48 PM UTC-7, jimbo1qaz wrote: I have a nested list. Whenever I make a copy of the list, changes in one affect the other, even when I use list(orig) or even copy the sublists one by one. I have to manually copy each cell over for it to work. Link to broken code: http://jimbopy.pastebay.net/1090401 No, actually that's the OK code. http://jimbopy.pastebay.net/1090494 is the broken one. I also would prefer an inline posting of the code, but if it's too big to post here, it's probably too big for me to debug here. The usual reason for such a symptom is a nested list, where you have multiple references to the same inner list inside the outer. When you change one of those, you change all of them. alist = [1, 2, 3] blist = [alist, alist, alist] # or blist = alist * 3 print blist alist.append(49) print blist davea@think:~/temppython$ python jimbo.py [[1, 2, 3], [1, 2, 3], [1, 2, 3]] [[1, 2, 3, 49], [1, 2, 3, 49], [1, 2, 3, 49]] Solution to this is to make sure that only copies of alist get into blist. One way is blist = [alist[:], alist[:], alist[:]] More generally, you can get into this type of trouble whenever you have non-immutable objects inside the list. Understand, this is NOT a flaw in the language. It's perfectly reasonable to be able to do so, in fact essential in many cases, when you want it to be the SAME item. -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On Sun, 23 Sep 2012 14:31:48 -0700, jimbo1qaz wrote:
I have a nested list. Whenever I make a copy of the list, changes in one
affect the other,
Then you aren't making a copy.
py first_list = [1, 2, 3]
py second_list = first_list # THIS IS NOT A COPY
py second_list.append()
py print first_list
[1, 2, 3, ]
even when I use list(orig)
Nonsense. Either you are confused, or there is something you aren't
telling us. Calling list *does* make a copy:
py first_list = [1, 2, 3]
py second_list = list(first_list)
py second_list.append()
py print first_list
[1, 2, 3]
What aren't you telling us? My guess is that there are TWO lists
involved, and you're only copying ONE:
py a = [1, 2, 3]
py a.append([ham, spam, cheese])
py print a
[1, 2, 3, ['ham', 'spam', 'cheese']]
py b = list(a) # make a copy of a, but not the contents of a
py b.append(99) # change b
py b[-1].append(tomato)
py print a
[1, 2, 3, ['ham', 'spam', 'cheese', 'tomato']]
Notice that b is a copy of a: changing b does not change a. But the
embedded list within a is *not* copied, so whether you append to that
list via a or b is irrelevant, both see the same change because there is
only one inner list in two places.
It might be more obvious if you give the shared sublist a name:
c = ['ham', 'spam', 'tomato']
a = [1, 2, 3, c]
b = [1, 2, 3, c] # a copy of a, but not a copy of c
Now it should be obvious that any changes to c will show up in both a and
b, regardless of how you change it. All three of these will have the
exact same effect:
a[-1].append('eggs')
b[-1].append('eggs')
c.append('eggs')
The way to copy lists, and all their sublists, and their sublists, and so
on all the way down, is with the copy module:
import copy
b = copy.deepcopy(a)
but if you are doing this a lot, (1) your code will be slow, and (2) you
can probably redesign your code to avoid so many copies.
By the way, instead of dumping lots of irrelevant code on us, please take
the time to narrow your problem down to the smallest possible piece of
code. We're volunteers here, and you are not paying us to wade through
your code trying to determine where your problem lies. That is up to you:
you narrow down to the actual problem, then ask for help.
Please read http://sscce.org/ for more details of how, and why, you
should do this.
Thank you.
--
Steven
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Re: List Problem
On Sunday, September 23, 2012 2:31:48 PM UTC-7, jimbo1qaz wrote: I have a nested list. Whenever I make a copy of the list, changes in one affect the other, even when I use list(orig) or even copy the sublists one by one. I have to manually copy each cell over for it to work. Link to broken code: http://jimbopy.pastebay.net/1090401 OK, deepcopy fixed it! And I fixed the catch indexerror thing too. -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On 9/23/2012 3:44 PM, jimbo1qaz wrote: On Sunday, September 23, 2012 2:31:48 PM UTC-7, jimbo1qaz wrote: I have a nested list. Whenever I make a copy of the list, changes in one affect the other, even when I use list(orig) or even copy the sublists one by one. I have to manually copy each cell over for it to work. Link to broken code: http://jimbopy.pastebay.net/1090401 No, actually that's the OK code. http://jimbopy.pastebay.net/1090494 is the broken one. I've not been following this thread fully, but why not just use x=list(y) to copy the list? The issue is that when you assign i=[1,2,3] and then j = i, j is just a reference to i, which is why you change either and the other changes. -- Take care, Ty http://tds-solutions.net The aspen project: a barebones light-weight mud engine: http://code.google.com/p/aspenmud He that will not reason is a bigot; he that cannot reason is a fool; he that dares not reason is a slave. -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On Mon, Sep 24, 2012 at 10:56 AM, Littlefield, Tyler ty...@tysdomain.com wrote: I've not been following this thread fully, but why not just use x=list(y) to copy the list? The issue is that when you assign i=[1,2,3] and then j = i, j is just a reference to i, which is why you change either and the other changes. The problem is with lists as elements of that list, so the key is deepcopy. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem...
On Tue, 28 Sep 2010 20:11:51 +0100, Rog wrote: On Tue, 28 Sep 2010 11:59:08 -0700, geremy condra wrote: On Tue, Sep 28, 2010 at 11:44 AM, Rog r...@pynguins.com wrote: Hi all, Have been grappling with a list problem for hours... a = [2, 3, 4, 5,.] b = [4, 8, 2, 6,.] Basicly I am trying to place a[0], b[0] in a seperate list IF a[2] and b[2] is present. I have tried sets, zip etc with no success. I am tackling Euler projects with Python 3.1, with minimal knowledge, and having to tackle the language as I progress. Enjoyable frustration :) I'm not clear on what your actual problem is, could you restate it? It sounds like you want to copy the ith element out of a and b into some other list- call it c- when the (i+2)th element meets some condition. What's the condition? Geremy Condra The condition is that the i-th element is inverted, but not equal. eg 4,2 - 2,4 , 34,5 - 5,34 etc. Hope that is clearer. Clear as mud. Perhaps you should given an example. Given input a = [2, 3, 4, 5, 6, 7] b = [4, 8, 2, 6, 10, 42] what output are you expecting, and how would you work it out by hand? -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem...
On 29 sep, 14:17, Steven D'Aprano st...@remove-this- cybersource.com.au wrote: On Tue, 28 Sep 2010 20:11:51 +0100, Rog wrote: On Tue, 28 Sep 2010 11:59:08 -0700, geremy condra wrote: On Tue, Sep 28, 2010 at 11:44 AM, Rog r...@pynguins.com wrote: Hi all, Have been grappling with a list problem for hours... a = [2, 3, 4, 5,.] b = [4, 8, 2, 6,.] Basicly I am trying to place a[0], b[0] in a seperate list IF a[2] and b[2] is present. I have tried sets, zip etc with no success. I am tackling Euler projects with Python 3.1, with minimal knowledge, and having to tackle the language as I progress. Enjoyable frustration :) I'm not clear on what your actual problem is, could you restate it? It sounds like you want to copy the ith element out of a and b into some other list- call it c- when the (i+2)th element meets some condition. What's the condition? Geremy Condra The condition is that the i-th element is inverted, but not equal. eg 4,2 - 2,4 , 34,5 - 5,34 etc. Hope that is clearer. Clear as mud. Perhaps you should given an example. Given input a = [2, 3, 4, 5, 6, 7] b = [4, 8, 2, 6, 10, 42] what output are you expecting, AFAICT, the OP expects [2, 4] in this case, but it's not clear what he'd expect for let's say: a = [2, 3, 21, 4, 5, 6, 7] b = [4, 8, 22, 2, 6, 10, 42] (the 'reversed' pair is at i+3, not i+2) or a = [0, 2, 3, 4, 5, 6, 7] b = [3, 4, 8, 2, 6, 10, 42] (the first pair is at pos 1, not 0) or a = [2, 3, 4, 8, 6, 7] b = [4, 8, 2, 3, 10, 42] (there's a second 'non-reversed/reversed' match at positions resp. 1 3) -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem...
On Wed, 29 Sep 2010 05:52:32 -0700, bruno.desthuilli...@gmail.com wrote: On 29 sep, 14:17, Steven D'Aprano st...@remove-this- cybersource.com.au wrote: On Tue, 28 Sep 2010 20:11:51 +0100, Rog wrote: On Tue, 28 Sep 2010 11:59:08 -0700, geremy condra wrote: On Tue, Sep 28, 2010 at 11:44 AM, Rog r...@pynguins.com wrote: Hi all, Have been grappling with a list problem for hours... a = [2, 3, 4, 5,.] b = [4, 8, 2, 6,.] Basicly I am trying to place a[0], b[0] in a seperate list IF a[2] and b[2] is present. I have tried sets, zip etc with no success. I am tackling Euler projects with Python 3.1, with minimal knowledge, and having to tackle the language as I progress. Enjoyable frustration :) I'm not clear on what your actual problem is, could you restate it? It sounds like you want to copy the ith element out of a and b into some other list- call it c- when the (i+2)th element meets some condition. What's the condition? Geremy Condra The condition is that the i-th element is inverted, but not equal. eg 4,2 - 2,4 , 34,5 - 5,34 etc. Hope that is clearer. Clear as mud. Perhaps you should given an example. Given input a = [2, 3, 4, 5, 6, 7] b = [4, 8, 2, 6, 10, 42] what output are you expecting, AFAICT, the OP expects [2, 4] in this case, but it's not clear what he'd expect for let's say: a = [2, 3, 21, 4, 5, 6, 7] b = [4, 8, 22, 2, 6, 10, 42] (the 'reversed' pair is at i+3, not i+2) or a = [0, 2, 3, 4, 5, 6, 7] b = [3, 4, 8, 2, 6, 10, 42] (the first pair is at pos 1, not 0) or a = [2, 3, 4, 8, 6, 7] b = [4, 8, 2, 3, 10, 42] (there's a second 'non-reversed/reversed' match at positions resp. 1 3) It is true that I would have needed any 'non-reversed/reversed' pairs, these would have been the amicable pairs I was looking for. The method to recognise them eluded me. Eyes are not good enough :) I have now joined Python-List and will follow the correct route. I have used a method suggested that has made the problem redundant, though it will bug me from now on. g = [] def divsum(n): return sum(i for i in range(1, n) if not n % i) for x in range(1, 2, -1): c = divsum(x) v = divsum(c) if v == x and v != c: g.append(divsum(x)) else: continue print(sum(g)) -- Rog http://www.rog.pynguins.com -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem...
On Thu, Sep 30, 2010 at 3:20 AM, Rog r...@pynguins.com wrote: On Wed, 29 Sep 2010 05:52:32 -0700, bruno.desthuilli...@gmail.com wrote: On 29 sep, 14:17, Steven D'Aprano st...@remove-this- cybersource.com.au wrote: On Tue, 28 Sep 2010 20:11:51 +0100, Rog wrote: On Tue, 28 Sep 2010 11:59:08 -0700, geremy condra wrote: On Tue, Sep 28, 2010 at 11:44 AM, Rog r...@pynguins.com wrote: Hi all, Have been grappling with a list problem for hours... a = [2, 3, 4, 5,.] b = [4, 8, 2, 6,.] Basicly I am trying to place a[0], b[0] in a seperate list IF a[2] and b[2] is present. I have tried sets, zip etc with no success. I am tackling Euler projects with Python 3.1, with minimal knowledge, and having to tackle the language as I progress. Enjoyable frustration :) I'm not clear on what your actual problem is, could you restate it? It sounds like you want to copy the ith element out of a and b into some other list- call it c- when the (i+2)th element meets some condition. What's the condition? Geremy Condra The condition is that the i-th element is inverted, but not equal. eg 4,2 - 2,4 , 34,5 - 5,34 etc. Hope that is clearer. Clear as mud. Perhaps you should given an example. Given input a = [2, 3, 4, 5, 6, 7] b = [4, 8, 2, 6, 10, 42] what output are you expecting, AFAICT, the OP expects [2, 4] in this case, but it's not clear what he'd expect for let's say: a = [2, 3, 21, 4, 5, 6, 7] b = [4, 8, 22, 2, 6, 10, 42] (the 'reversed' pair is at i+3, not i+2) or a = [0, 2, 3, 4, 5, 6, 7] b = [3, 4, 8, 2, 6, 10, 42] (the first pair is at pos 1, not 0) or a = [2, 3, 4, 8, 6, 7] b = [4, 8, 2, 3, 10, 42] (there's a second 'non-reversed/reversed' match at positions resp. 1 3) It is true that I would have needed any 'non-reversed/reversed' pairs, these would have been the amicable pairs I was looking for. The method to recognise them eluded me. Eyes are not good enough :) I have now joined Python-List and will follow the correct route. I have used a method suggested that has made the problem redundant, though it will bug me from now on. g = [] def divsum(n): return sum(i for i in range(1, n) if not n % i) You can optimize divsum. Remember that factors exist in pairs except when numbers are squares. Like say 12 have factors = 1, 12, 2, 6, 3, 4 so you can run the loop till sqrt(n) for n since floor(sqrt(n)) = 3 where n = 12 now factors are , (1, n/1), (2, n/2), (3, n/2). for a square number say n = 9, the factors are (1, 3, 12) If you run the loop till sqrt(n) for n since floor(sqrt(n)) = 3 where n = 9 we get the factors as (1, 9), (3, 3) One of the factor will be redundant, so you need to take care of that. for x in range(1, 2, -1): c = divsum(x) v = divsum(c) Can be done as, c, v = divsum(x), divsum(divsum(x)) if v == x and v != c: g.append(divsum(x)) No need for else part, what is the use ? else: continue print(sum(g)) You could pack this up as, sum(i for i in range(1, 0, -2) if divsum(divsum(i)) == i and divsum(i) != i) 31626 -- Rog http://www.rog.pynguins.com -- http://mail.python.org/mailman/listinfo/python-list -- ~l0nwlf -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem...
On Wed, Sep 29, 2010 at 12:14 AM, Rog r...@pynguins.com wrote: Hi all, Have been grappling with a list problem for hours... a = [2, 3, 4, 5,.] b = [4, 8, 2, 6,.] Basicly I am trying to place a[0], b[0] in a seperate list IF a[2] and b[2] is present. You are not exactly clear with your problem statement. Care to elaborate it more. I have tried sets, zip etc with no success. I am tackling Euler projects with Python 3.1, with minimal knowledge, and having to tackle the language as I progress. Enjoyable frustration :) -- Rog http://www.rog.pynguins.com -- http://mail.python.org/mailman/listinfo/python-list -- ~l0nwlf -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem...
On Tue, Sep 28, 2010 at 11:44 AM, Rog r...@pynguins.com wrote: Hi all, Have been grappling with a list problem for hours... a = [2, 3, 4, 5,.] b = [4, 8, 2, 6,.] Basicly I am trying to place a[0], b[0] in a seperate list IF a[2] and b[2] is present. I have tried sets, zip etc with no success. I am tackling Euler projects with Python 3.1, with minimal knowledge, and having to tackle the language as I progress. Enjoyable frustration :) I'm not clear on what your actual problem is, could you restate it? It sounds like you want to copy the ith element out of a and b into some other list- call it c- when the (i+2)th element meets some condition. What's the condition? Geremy Condra -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem...
On Tue, 28 Sep 2010 11:59:08 -0700, geremy condra wrote: On Tue, Sep 28, 2010 at 11:44 AM, Rog r...@pynguins.com wrote: Hi all, Have been grappling with a list problem for hours... a = [2, 3, 4, 5,.] b = [4, 8, 2, 6,.] Basicly I am trying to place a[0], b[0] in a seperate list IF a[2] and b[2] is present. I have tried sets, zip etc with no success. I am tackling Euler projects with Python 3.1, with minimal knowledge, and having to tackle the language as I progress. Enjoyable frustration :) I'm not clear on what your actual problem is, could you restate it? It sounds like you want to copy the ith element out of a and b into some other list- call it c- when the (i+2)th element meets some condition. What's the condition? Geremy Condra The condition is that the i-th element is inverted, but not equal. eg 4,2 - 2,4 , 34,5 - 5,34 etc. Hope that is clearer. Thanks for the response. -- Rog http://www.rog.pynguins.com -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem...
On Wed, Sep 29, 2010 at 1:15 AM, rog r...@pynguins.com wrote: Shashwat Anand wrote: On Wed, Sep 29, 2010 at 12:14 AM, Rog r...@pynguins.com mailto: r...@pynguins.com wrote: Hi all, Have been grappling with a list problem for hours... a = [2, 3, 4, 5,.] b = [4, 8, 2, 6,.] Basicly I am trying to place a[0], b[0] in a seperate list IF a[2] and b[2] is present. You are not exactly clear with your problem statement. Care to elaborate it more. I have tried sets, zip etc with no success. I am tackling Euler projects with Python 3.1, with minimal knowledge, and having to tackle the language as I progress. Enjoyable frustration :) -- Rog http://www.rog.pynguins.com -- http://mail.python.org/mailman/listinfo/python-list -- ~l0nwlf Let d(/n/) be defined as the sum of proper divisors of /n/ (numbers less than /n/ which divide evenly into /n/). If d(/a/) = /b/ and d(/b/) = /a/, where /a/ ≠ /b/, then /a/ and /b/ are an amicable pair and each of /a/ and /b/ are called amicable numbers. For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220. Evaluate the sum of all the amicable numbers under 1. from Project Euler. I have all answers contained in two lists but need to extract the amicable pairs. eg a = [200, 4, 284,.] b = [284, 9, 200, ..] Hope that helps. Thanks for the link. I am not sure how and what you managed, but what I understand you need to calculate sum of divisors. A quick unoptimized solution can be, def divsum(n):return sum(i for i in range(1, n) if not n % i)... divsum(220)284 divsum(284)220 Now all is left is looping and little maths, since upper bound is not high, brute force is Ok. spoiler http://codepad.org/g6PRyoiV /spoiler # For reference Also I guess you missed 'Reply All', please take care of it next time. Rog -- ~l0nwlf -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On Wed, Dec 10, 2008 at 3:40 PM, dongzhi [EMAIL PROTECTED] wrote: If I execute part[1], I have got 'a'. If I execute part[2], I have got ' '. But, if I execute part[1::2], I have got ['a', '', '']. I don't know why. Please tell me why. Perhaps you meant: part[1:2] pydoc list This will tell you there are 3 arguments to __getslice__ i, j, y Without doing any further reading I presume from my experience that this works much like range and xrange and that the 3rd parameter is the step size. Consider this: x = [9, 1, 2, 3, 4] x [9, 1, 2, 3, 4] x[1:2] [1] x[1::2] [1, 3] x[1::1] [1, 2, 3, 4] x[1::2] [1, 3] x[1::3] [1, 4] cheers James -- -- -- Problems are solved by method -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On Dec 10, 2:00 pm, James Mills [EMAIL PROTECTED] wrote: On Wed, Dec 10, 2008 at 3:40 PM, dongzhi [EMAIL PROTECTED] wrote: If I execute part[1], I have got 'a'. If I execute part[2], I have got ' '. But, if I execute part[1::2], I have got ['a', '', '']. I don't know why. Please tell me why. Perhaps you meant: part[1:2] pydoc list This will tell you there are 3 arguments to __getslice__ i, j, y Without doing any further reading I presume from my experience that this works much like range and xrange and that the 3rd parameter is the step size. Consider this: x = [9, 1, 2, 3, 4] x [9, 1, 2, 3, 4] x[1:2] [1] x[1::2] [1, 3] x[1::1] [1, 2, 3, 4] x[1::2] [1, 3] x[1::3] [1, 4] cheers James -- -- -- Problems are solved by method Dear James Mills, Thanks a lot. Now, I know the reason. Best Regards, Liu Ming -- http://mail.python.org/mailman/listinfo/python-list
Re: List Problem
On Tue, 09 Dec 2008 21:40:08 -0800, dongzhi wrote:
I have one problem for List. Like that:
format='just a little test'
part = format.split('')
print part
the result is : ['just ', 'a', ' ', '', 'little', '', ' test']
the list part have 7 element.
If I execute part[1], I have got 'a'. If I execute part[2], I have got
' '. But, if I execute part[1::2], I have got ['a', '', '']. I don't
know why. Please tell me why.
You're slicing your list with the arguments start at 1, stop at the end,
using a step size of 2. It's basically the same as ``part[1], part[1+2],
part[1+2+2], ...``. Perhaps you wanted to do ``part[1:3]`` (meaning
start at 1, stop before 3).
See the Python Reference for details.
http://docs.python.org/library/stdtypes.html#sequence-types-str-unicode-
list-tuple-buffer-xrange
http://docs.python.org/reference/expressions.html#id8
HTH,
--
Robert Stargaming Lehmann
--
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Re: List problem
Alan Bromborsky wrote: I wish to create a list of empty lists and then put something in one of the empty lists. Below is what I tried, but instead of appending 1 to a[2] it was appended to all the sub-lists in a. What am I doing wrong? a = 6*[[]] a [[], [], [], [], [], []] a[2].append(1) a [[1], [1], [1], [1], [1], [1]] What you've done is equivalent to x = [] a = [x, x, x, x, x, x] del x An idiom for what you want is a = [[] for y in xrange (6)] which will populate a with 6 distinct empty lists. Cheers, Mel. -- http://mail.python.org/mailman/listinfo/python-list
Re: List problem
[EMAIL PROTECTED] wrote: For example i write the following code in the Python command line; list = ['One,Two,Three,Four'] Then enter this command, which will then return the following; ['One,Two,Three,Four'] This is already wrong. Assignments do not return anything. Georg -- http://mail.python.org/mailman/listinfo/python-list
Re: List problem
[EMAIL PROTECTED] wrote: For example i write the following code in the Python command line; list = ['One,Two,Three,Four'] Then enter this command, which will then return the following; ['One,Two,Three,Four'] Now the problem, reading through the Python tutorial's, it describe's that list's can sliced, concatenated and so on etc So i write the following list[1:] + ['Five'] Then the error returned is that 'str' and 'list' could not be concatenated which is where it baffles me. if you got that error, you clearly didn't type what you say you typed: list = ['One,Two,Three,Four'] list ['One,Two,Three,Four'] list[1:] + ['Five'] ['Five'] (note that the first list contains a single string; if you want to put multiple strings in a list, you need to be more careful with where you put the quotes.) /F -- http://mail.python.org/mailman/listinfo/python-list
Re: List problem
[EMAIL PROTECTED] wrote: For example i write the following code in the Python command line; list = ['One,Two,Three,Four'] Then enter this command, which will then return the following; ['One,Two,Three,Four'] Now the problem, reading through the Python tutorial's, it describe's that list's can sliced, concatenated and so on etc So i write the following list[1:] + ['Five'] Then the error returned is that 'str' and 'list' could not be concatenated which is where it baffles me. I'm understanding that this mean's that the string 'Five' could not be joined onto the list, as i havn't defined it in the array. Viewing the Python tutrial they have the following, and working code; a[:2] + ['bacon', 2*2] ['spam', 'eggs', 'bacon', 4] How can they have the string 'bacon' and have it returned with no error? Your error message doesn't match your command. Now if you typed: list[0]+['Five'] that gives you the error you showed. What you meant to type was: l = ['One','Two','Three','Four'] This is a list of four strings, what you entered was a list of one string. NOTE never call a variable 'list' as it will mask the built-in list method (same goes for str, tuple, int, float, etc). -Larry Bates -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
placid: This may be a solution: l1 = ['acXXX1', 'XXX2', 'wXXX3', 'kXXX5'] l2 = [ 'bXXX1', 'xXXX2', 'efXXX3', 'yXXX6', 'zZZZ9'] import re findnum = re.compile(r[0-9]+$) s1 = set(int(findnum.search(el).group()) for el in l1) s2 = set(int(findnum.search(el).group()) for el in l2) nmax = max(max(s1), max(s2)) # XXXnmax is surely unavailable missing = set(range(1, nmax)) - s1 - s2 print [XXX%d % i for i in sorted(missing)] # Output: ['XXX4', 'XXX7', 'XXX8'] If you need more speed you can replace some of those sets (like the range one) with fors. Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
placid wrote: Hi all, I have two lists that contain strings in the form string + number for example list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX5'] the second list contains strings that are identical to the first list, so lets say the second list contains the following list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX6'] and now what ive been trying to do is find the first string that is available, i.e a string that is in neither of the two lists so the following code should only print XXX4 then return. for i in xrange(1,10): numpart = str(1) + str(%04i %i) str = XXX + numpart for list1_elm in list1: if list1_elm == str: break else: for list2_elm in list2: if list2_elm == str: break else: print str return Cheer Just a thought I would probably use sets and see if the value that you are looking for using the union of the two lists. (Yes it won't scale to really big lists) for instance if you do the following set1 = set(list1) set2 = set(list2) you can then do a single check if XXX%d % i not in set1.union(set2): # or set1 | set2 # do something Rgds Tim -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
placid wrote: But there may be other characters before XXX (which XXX is constant). A better example would be, that string s is like a file name and the characters before it are the absolute path, where the strings in the first list can have a different absolute path then the second list entries. But the filenames are always exact. So you need to split the entries bases on \\ (windows machine) and match on this ? Cheers If you're actually working with filenames and paths then you should use os.path.basename() to get just the filename parts of the paths. test = set(map(os.path.basename, list1)) test |= set(map(os.path.basename, list2)) (Note: I *was* being stupid last night, the + operator doesn't work for sets. You want to use | ) Peace, ~Simon -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
placid wrote: Hi all, I have two lists that contain strings in the form string + number for example list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX5'] the second list contains strings that are identical to the first list, so lets say the second list contains the following list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX6'] and now what ive been trying to do is find the first string that is available, i.e a string that is in neither of the two lists so the following code should only print XXX4 then return. for i in xrange(1,10): numpart = str(1) + str(%04i %i) str = XXX + numpart for list1_elm in list1: if list1_elm == str: break else: for list2_elm in list2: if list2_elm == str: break else: print str return Cheer I don't know how close the following is to what you want ( or how efficient etc...). If both lists are the same up to a certain point, then the first function should do, if not, try the second function. Gerard from itertools import izip, dropwhile def get_first_missing1( seq1, seq2 ): i = int( seq1[0][-1] ) for x1, x2 in izip( seq1, seq2 ): if int(x1[-1]) != i and int(x2[-1]) != i: return x1[:-1] + str(i) i += 1 return -1 def get_first_missing2( seq1, seq2 ): i = int( seq1[0][-1] ) j = int( seq2[0][-1] ) if j i: seq1, seq2 = seq2, seq1 i, j = j, i return get_first_missing1( list(dropwhile(lambda s: int(s[-1]) j, seq1)), seq2 ) L1 = [ 'XXX1', 'XXX2', 'XXX3', 'XXX5'] L2 = [ 'YYY1', 'YYY2', 'YYY3', 'YYY6'] print get_first_missing1(L1, L2) print get_first_missing2(L1, L2) 'XXX4' 'XXX4' L1 = [ 'XXX1', 'XXX2', 'XXX3', 'XXX5'] L2 = [ 'YYY2', 'YYY3', 'YYY5', 'YYY6'] print get_first_missing1(L1, L2) print get_first_missing2(L1, L2) 'XXX4' 'XXX4' -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
Gerard Flanagan wrote: placid wrote: Hi all, I have two lists that contain strings in the form string + number for example list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX5'] the second list contains strings that are identical to the first list, so lets say the second list contains the following list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX6'] and now what ive been trying to do is find the first string that is available, i.e a string that is in neither of the two lists so the following code should only print XXX4 then return. for i in xrange(1,10): numpart = str(1) + str(%04i %i) str = XXX + numpart for list1_elm in list1: if list1_elm == str: break else: for list2_elm in list2: if list2_elm == str: break else: print str return Cheer I don't know how close the following is to what you want ( or how efficient etc...). If both lists are the same up to a certain point, then the first function should do, if not, try the second function. Gerard from itertools import izip, dropwhile def get_first_missing1( seq1, seq2 ): i = int( seq1[0][-1] ) for x1, x2 in izip( seq1, seq2 ): if int(x1[-1]) != i and int(x2[-1]) != i: return x1[:-1] + str(i) i += 1 return -1 def get_first_missing2( seq1, seq2 ): i = int( seq1[0][-1] ) j = int( seq2[0][-1] ) if j i: seq1, seq2 = seq2, seq1 i, j = j, i return get_first_missing1( list(dropwhile(lambda s: int(s[-1]) j, seq1)), seq2 ) L1 = [ 'XXX1', 'XXX2', 'XXX3', 'XXX5'] L2 = [ 'YYY1', 'YYY2', 'YYY3', 'YYY6'] print get_first_missing1(L1, L2) print get_first_missing2(L1, L2) 'XXX4' 'XXX4' ehm...a bit limited in what it will handle, now that I look at it! like more than ten items in a list - '11'[-1] == '1'...no time to test further, sorry:( Gerard -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
placid [EMAIL PROTECTED] writes:
list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX5']
the second list contains strings that are identical to the first list,
so lets say the second list contains the following
list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX6']
I think you meant list2 for the second one. So:
import re
list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX5']
list2 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX6']
def num(s):
# get the number out of one of the strings
# (throw exception if no number is there)
digits = re.search('\d+$', s)
return int(digits.group(0))
def get_lowest_unused(list1, list2):
prev = 0
for n in sorted(set(map(num,list1+list2))):
if n != prev+1:
return prev+1
prev = n
print get_lowest_unused(list1, list2)
You could do all this with iterators and save a little memory, but
that's more confusing.
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Re: list problem
Thank you all for the replies, i now have a better solution. Cheers -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
placid wrote: Hi all, I have two lists that contain strings in the form string + number for example list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX5'] the second list contains strings that are identical to the first list, so lets say the second list contains the following list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX6'] and now what ive been trying to do is find the first string that is available, i.e a string that is in neither of the two lists so the following code should only print XXX4 then return. for i in xrange(1,10): numpart = str(1) + str(%04i %i) str = XXX + numpart for list1_elm in list1: if list1_elm == str: break else: for list2_elm in list2: if list2_elm == str: break else: print str return Cheer Well first off, don't use 'str' for a variable name. Second, %04i % i creates a string, don't call str() on it. Third, str(1) will always be 1 so just add that to your format string already 1%04i % i (And if the XXX part is also constant then add that too: XXX1%04i % i) Finally, you can say: for i in xrange(1,10): s = XXX1%04i % i if s not in list1 and s not in list2: print s HTH, ~Simon -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
Simon Forman wrote: Finally, you can say: for i in xrange(1,10): s = XXX1%04i % i if s not in list1 and s not in list2: print s HTH, ~Simon D'oh! Forgot to break. for i in xrange(1,10): s = XXX1%04i % i if s not in list1 and s not in list2: print s break Peace, ~Simon -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
Simon Forman wrote: placid wrote: Hi all, I have two lists that contain strings in the form string + number for example list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX5'] the second list contains strings that are identical to the first list, so lets say the second list contains the following list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX6'] and now what ive been trying to do is find the first string that is available, i.e a string that is in neither of the two lists so the following code should only print XXX4 then return. for i in xrange(1,10): numpart = str(1) + str(%04i %i) str = XXX + numpart for list1_elm in list1: if list1_elm == str: break else: for list2_elm in list2: if list2_elm == str: break else: print str return Cheer Well first off, don't use 'str' for a variable name. Second, %04i % i creates a string, don't call str() on it. Third, str(1) will always be 1 so just add that to your format string already 1%04i % i thanks for the tips (And if the XXX part is also constant then add that too: XXX1%04i % i) Finally, you can say: for i in xrange(1,10): s = XXX1%04i % i if s not in list1 and s not in list2: print s But there may be other characters before XXX (which XXX is constant). A better example would be, that string s is like a file name and the characters before it are the absolute path, where the strings in the first list can have a different absolute path then the second list entries. But the filenames are always exact. So you need to split the entries bases on \\ (windows machine) and match on this ? Cheers -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem
placid wrote: Simon Forman wrote: placid wrote: Hi all, I have two lists that contain strings in the form string + number for example list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX5'] the second list contains strings that are identical to the first list, so lets say the second list contains the following list1 = [ ' XXX1', 'XXX2', 'XXX3', 'XXX6'] and now what ive been trying to do is find the first string that is available, i.e a string that is in neither of the two lists so the following code should only print XXX4 then return. for i in xrange(1,10): numpart = str(1) + str(%04i %i) str = XXX + numpart for list1_elm in list1: if list1_elm == str: break else: for list2_elm in list2: if list2_elm == str: break else: print str return Cheer Well first off, don't use 'str' for a variable name. Second, %04i % i creates a string, don't call str() on it. Third, str(1) will always be 1 so just add that to your format string already 1%04i % i thanks for the tips (And if the XXX part is also constant then add that too: XXX1%04i % i) Finally, you can say: for i in xrange(1,10): s = XXX1%04i % i if s not in list1 and s not in list2: print s But there may be other characters before XXX (which XXX is constant). A better example would be, that string s is like a file name and the characters before it are the absolute path, where the strings in the first list can have a different absolute path then the second list entries. But the filenames are always exact. So you need to split the entries bases on \\ (windows machine) and match on this ? Cheers hmm, a slightly different problem than your OP. Yeah, I would build a new list (or set) from the contents of BOTH lists with the prefixes stripped off and test your target string against that. You might also be able to do something with the endswith() method of strings. test = set(n[3:] for n in list1) + set(n[3:] for n in list2) if s not in test: print s It's late though, so I may be being stupid. ;-) Peace, ~Simon -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem 4 newbie
Hi, if u check the id's of a and b lists and also its elements, you will obeserve that the id's of a and b have changed but id's of their elements have not changed. If you make a deep copy of the list a and then make your changes in that list, it shud work. this can be done using the copy module. hope that helps, vaibhav id(a) -1208622388 id(b) -1208622324 for el in a: ... print id(el) ... -1208622836 -1208622708 -1208622772 -1208622420 for el in b: ... print id(el) ... -1208622836 -1208622708 -1208622772 -1208622420 import copy c = copy.deepcopy(a) id(c) -1208464564 for el in c: ... print id(el) ... -1208465172 -1208464276 -1208464180 -1208463988 -- http://mail.python.org/mailman/listinfo/python-list
Re: list problem 4 newbie
manstey wrote:
for index in reversed(range(0,len(a)-1)):
if '75' in b[index][1]:
b[index][1].remove('75')
b[index][1].append('99')
What on earth is all that messing around in the for loop intended to do? If
you want a range from len(a)-2 to 0 inclusive then just do it in range
directly (and did you really mean not to look at the last element?), if you
actually just wanted to iterate through b in reverse, then just iterate
through b in reverse:
b = copy.deepcopy(a)
for element in reversed(b):
if '75' in element[1]:
element[1].remove('75')
element[1].append('99')
--
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Re: list problem 4 newbie
Thanks very much. Deepcopy works fine, as does reversed(b). I thought I
needed the index number but I didn't.
Duncan Booth wrote:
manstey wrote:
for index in reversed(range(0,len(a)-1)):
if '75' in b[index][1]:
b[index][1].remove('75')
b[index][1].append('99')
What on earth is all that messing around in the for loop intended to do? If
you want a range from len(a)-2 to 0 inclusive then just do it in range
directly (and did you really mean not to look at the last element?), if you
actually just wanted to iterate through b in reverse, then just iterate
through b in reverse:
b = copy.deepcopy(a)
for element in reversed(b):
if '75' in element[1]:
element[1].remove('75')
element[1].append('99')
--
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