instance attribute "a" defined outside __init__
Hi, When running below code, I get error saying: instance attribute "a" defined outside __init__ class Foo: var = 9 def add(self, a, b): self.a = a self.b = b print a+b def __init__(self): print 10 bar = Foo() # create object bar.add(4, 7) # call method by object.method print bar.var # access class variable why the output prints: 10 11 9 why not serially? 9 11 10 -- https://mail.python.org/mailman/listinfo/python-list
Re: Sorting a dictionary
This one I think I know... Try with: for k in sorted(word_count) : print k,"=",word_count[k] You need to do the sorting before iterating over the keys... Jaime On Tue, May 12, 2009 at 1:54 PM, Ronn Ross wrote: > I'm attempting to sort for the results of a dictionary. I would like to > short by the 'key' in ascending order. I have already made several attempts > using: sorted() and .sort(). > Here is my loop: > for key,value in word_count.items(): > print sorted(key), "=", value > > Thanks > > -- > http://mail.python.org/mailman/listinfo/python-list > > -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: putting date strings in order
If you simply want to generate an ordered list of months, start with
it in order:
dates = ["x_jan",...,"x_dec"]
and if the desired starting month is
start = 6 # i.e. x_jun
dates = dates[start - 1:] + dates[:start - 1]
If you have to sort the list itself, I would use an intermediate
dictionary to hold the positions, as in:
months = {"x_jan" : 1, ..., "x_dec" : 12}
You can then sort the list with :
dates.sort(None,lambda x : months[x])
and then do the
dates = dates[start - 1:] + dates[:start - 1]
or alternatively you can get the sorting directly as you want it with:
dates.sort(None,lambda x : (months[x] - start + 1) % 12)
Jaime
On Tue, May 12, 2009 at 1:58 PM, noydb wrote:
> On May 11, 11:30 pm, Paul Rubin wrote:
>> noydb writes:
>> > Anyone have any good ideas? I was curious to see what people came up
>> > with.
>>
>> Is this a homework assignment? Some hints:
>>
>> 1) figure out how to compare two month names for chronological order,
>> leaving out the issue of the starting month not being january.
>> 2) figure out how to adjust for the starting month. The exact
>> semantics of the "%" operator might help do this concisely.
>
> Ha! No, this is not a homework assignment. I just find myself to be
> not the most eloquent and efficient scripter and wanted to see how
> others would approach it.
>
> I'm not sure how I follow your suggestion. I have not worked with the
> %. Can you provide a snippet of your idea in code form?
>
> I thought about assigning a number string (like 'x_1') to any string
> containing 'jan' -- so x_jan would become x_1, and so on. Then I
> could loop through with a counter on the position of the number (which
> is something i will need to do, comparing one month to the next
> chronological month, then that next month to its next month, and so
> on). And as for the starting postion, the user could declare, ie, aug
> the start month. aug is position 8. therefore subtract 7 from each
> value, thus aug becomes 1 but then I guess early months would have
> to be add 5, such that july would become 12. Ugh, seems sloppy to me.
>
> Something like that seems poor to me. Anybody have a bteer
> idea, existing code???
> --
> http://mail.python.org/mailman/listinfo/python-list
>
--
(\__/)
( O.o)
( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus
planes de dominación mundial.
--
http://mail.python.org/mailman/listinfo/python-list
Re: putting date strings in order
On Tue, May 12, 2009 at 5:02 PM, MRAB wrote: > John Machin wrote: >> >> MRAB mrabarnett.plus.com> writes: >>> >>> Sort the list, passing a function as the 'key' argument. The function >>> should return an integer for the month, eg 0 for 'jan', 1 for 'feb'. If >>> you want to have a different start month then add >> >> and if you don't like what that produces, try subtract :-) >> > Oops! > >>> the appropriate >>> integer for that month (eg 0 for 'jan', 1 for 'feb') and then modulo 12 >>> to make it wrap around (there are only 12 months in a year), returning >>> the result. >> > Actually, subtract the start month, add 12, and then modulo 12. Both on my Linux and my Windows pythons, modulos of negative numbers are properly taken, returning always the correct positive number between 0 and 11. I seem to recall, from my distant past, that Perl took pride on this being a language feature. Anyone knows if that is not the case with python, and so not adding 12 before taking the modulo could result in wrong results in some implementations? > -- > http://mail.python.org/mailman/listinfo/python-list > -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Representing a Tree in Python
Dijkstra's algorithm computes shortest paths between a node and _ALL_ other nodes in the graph. It is usually stopped once computing the shortest path to the target node is done, but that's simply for efficiency, not a limitation of the algorithm. So you should be able to tweak the code you are using so that it provides you with all you are looking for. I'd be surprised if graphine (which, by the way, looks great, CTO) or any other graph package didn't implement it, so switching to that may be the most efficient thing to do. On the other hand, if you want to post your code and links to the Dijkstra code you are using it may be possible to help you with the tweaking... Jaime On Wed, May 13, 2009 at 8:31 AM, godshorse wrote: > On May 13, 11:54 am, CTO wrote: >> On May 13, 12:10 am, godshorse wrote: >> >> > Hello, >> >> > I want to find out the shortest path tree from a root to several nodes >> > in a graph data structure. I found a Dijkstra code from internet that >> > finds shortest path between only two nodes. How can i extend it to a >> > tree?. And what is the best way to represent a tree in Python?. >> >> > Thank you, >> >> Well, I'm biased, but I like http://graphine.org>. >> As an example, to build a five node tree: >> >> >>> from graph.base import Graph >> >>> g = Graph() >> >>> for i in range(5): >> >> ... g.add_node(i) >> ... >> >> >>> g.add_edge(0, 1) >> >>> g.add_edge(0, 2) >> >>> g.add_edge(1, 3) >> >>> g.add_edge(1, 4) >> >> And to find the shortest path between, say, node 0 and node 4: >> >> >>> start = g[0] >> >>> end = g[4] >> >>> distance, edges = g.get_shortest_paths(start)[end] >> >>> distance >> 2 >> >>> edges >> >> [Edge(name=(0,1)), Edge(name=(1,4))] >> >> Let me know what you think if you decide to use it- I'm looking for >> feedback. >> >> Geremy Condra > > Thanks very much for your reply Geremy. That site was interesting. > > Actually the Graph building part is already completed now. I used a > dictionary for that and it works fine. for Dijkstra shortest path > problem your suggestion can be used. > > But let me clear the my problem again. I have a graph. and I want to > find 'shortest path tree' from a root node to several nodes. as a > example if we have a graph of 5 nodes from 1 to 5, I need to build the > shortest path tree from node 1 to nodes 2,3,5. So my question is > instead of keeping separate lists for each destination node's shortest > path. How can I represent and store them in a tree structure using > python. Then I can easily find out what are the common nodes in the > path to each destination. > > Thanks once again. > -- > http://mail.python.org/mailman/listinfo/python-list > -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Just wondering
map is creating a new list of 10,000,000 items, not modifying the values inside list a. That's probably where your time difference comes from... Jaime On Fri, May 15, 2009 at 1:56 PM, Gediminas Kregzde wrote: > Hello, > > I'm Vilnius college II degree student and last semester our teacher > introduced us to python > I've used to program with Delphi, so I very fast adopted to python > > Now I'm developing cross platform program and use huge amounts of > data. Program is needed to run as fast as it coud. I've read all tips > about geting performance, but got 1 bug: map function is slower than > for loop for about 5 times, when using huge amounts of data. > It is needed to perform some operations, not to return data. > > I'm adding sample code: > from time import time > > def doit(i): > pass > > def main(): > a = [0] * 1000 > t = time() > map(doit, a) > print "map time: " + str(time() - t) > > def main2(): > t = time() > a = [0] * 1000 > for i in a: > pass > print "loop time: " + str(time() - t) > > main() # takes approximately 5x times longer than main2() > main2() > > I'm wondering were is catch? > > I'm using python 2.6 on windows xp sp2 machine > > P.S. Sorry for my broken English > -- > http://mail.python.org/mailman/listinfo/python-list > -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Just wondering
> (1) building another throwaway list and > (2) function call overhead for calling doit() > > You can avoid (1) by using filter() instead of map() Are you sure of this? My python returns, when asked for help(filter) : Help on built-in function filter in module __builtin__: filter(...) filter(function or None, sequence) -> list, tuple, or string Return those items of sequence for which function(item) is true. If function is None, return the items that are true. If sequence is a tuple or string, return the same type, else return a list. I'm afraid there is no builtin function to do an inplace modification of a sequence... -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Representing a Tree in Python
If you run Dijkstra without a third argument, i.e. without an end
node, the it will compute the shortest paths from your start node to
all nodes in the tree. So by doing something like:
start_node = 1
end_nodes = [2,3,5]
D, P = Dijkstra(G, start_node)
You will then have in D a dictionary with distances to all nodes, and
in P a dictionary of preceding nodes along the shortest path for each
node. You could tweak the Dijkstra function so that it stops when the
minimum distances to all elements of end_nodes have been calculated,
without having to solve for the whole graph...
Apart from that, how you want to store the shortest path information
is up to you, but if you stick to the "graph as a dictionary of
dictionaries" you can construct the subgraph that contains only the
nodes in the shortest paths between the start and the end nodes with
code like this one:
shortest_tree = {}
current_layer = end_nodes
while current_layer != [start_node] :
new_layer = set()
for node in current_layer :
new_node = P[node]
if new_node in shortest_tree :
if node not in shortest_tree[new_node] :
shortest_tree[new_node][node] = G[new_node][node]
else :
shodtest_tree[new_node] = {}
shortest_tree[new_node][node] = G[new_node][node]
new_layer.add(new_node)
current_layer = [j for j in new_layer]
I haven't tested the code, and there may be more efficient ways of
achieving the same, though...
Jaime
On Wed, May 13, 2009 at 10:49 AM, godshorse wrote:
> On May 13, 3:19 pm, Jaime Fernandez del Rio
> wrote:
>> Dijkstra's algorithm computes shortest paths between a node and _ALL_
>> other nodes in the graph. It is usually stopped once computing the
>> shortest path to the target node is done, but that's simply for
>> efficiency, not a limitation of the algorithm. So you should be able
>> to tweak the code you are using so that it provides you with all you
>> are looking for. I'd be surprised if graphine (which, by the way,
>> looks great, CTO) or any other graph package didn't implement it, so
>> switching to that may be the most efficient thing to do.
>>
>> On the other hand, if you want to post your code and links to the
>> Dijkstra code you are using it may be possible to help you with the
>> tweaking...
>>
>> Jaime
>>
>>
>>
>> On Wed, May 13, 2009 at 8:31 AM, godshorse wrote:
>> > On May 13, 11:54 am, CTO wrote:
>> >> On May 13, 12:10 am, godshorse wrote:
>>
>> >> > Hello,
>>
>> >> > I want to find out the shortest path tree from a root to several nodes
>> >> > in a graph data structure. I found a Dijkstra code from internet that
>> >> > finds shortest path between only two nodes. How can i extend it to a
>> >> > tree?. And what is the best way to represent a tree in Python?.
>>
>> >> > Thank you,
>>
>> >> Well, I'm biased, but I like http://graphine.org>.
>> >> As an example, to build a five node tree:
>>
>> >> >>> from graph.base import Graph
>> >> >>> g = Graph()
>> >> >>> for i in range(5):
>>
>> >> ... g.add_node(i)
>> >> ...
>>
>> >> >>> g.add_edge(0, 1)
>> >> >>> g.add_edge(0, 2)
>> >> >>> g.add_edge(1, 3)
>> >> >>> g.add_edge(1, 4)
>>
>> >> And to find the shortest path between, say, node 0 and node 4:
>>
>> >> >>> start = g[0]
>> >> >>> end = g[4]
>> >> >>> distance, edges = g.get_shortest_paths(start)[end]
>> >> >>> distance
>> >> 2
>> >> >>> edges
>>
>> >> [Edge(name=(0,1)), Edge(name=(1,4))]
>>
>> >> Let me know what you think if you decide to use it- I'm looking for
>> >> feedback.
>>
>> >> Geremy Condra
>>
>> > Thanks very much for your reply Geremy. That site was interesting.
>>
>> > Actually the Graph building part is already completed now. I used a
>> > dictionary for that and it works fine. for Dijkstra shortest path
>> > problem your suggestion can be used.
>>
>> > But let me clear the my problem again. I have a graph. and I want to
>> > find 'shortest path tree' from a root node to several nodes. as a
>> > example if we have a graph of 5 nodes from 1 to 5, I need to build the
>> > shortest path tree from node 1 to nodes 2,3,5. So my question is
>> > instead of keeping separate lists for each desti
Re: Regarding sort()
Hi Dhananjay, Sort has several optional arguments, the function's signature is as follows: s.sort([cmp[, key[, reverse]]]) If you store your data as a list of lists, to sort by the third column you could do something like: data.sort(None, lambda x : x[2]) For more complex sortings, as the one you propose, you need to define a compare function, that determines if an object preceeds another or not. In your case, something like: def list_cmp( a, b) : if a[2] > b[2] : return 1 elif a[2] < b[2] : return -1 else : # a[2] == b[2] if a[5] > b[5] : return 1 elif a[5] < b[5] : return -1 else : # a[5] == b[5] return 0 data.sort(list_cmp) should do the trick for you... Jaime On Mon, May 25, 2009 at 9:29 AM, Dhananjay wrote: > Hello All, > > I have data set as follows: > > 24 GLU 3 47 LYS 6 3.909233 1 > 42 PRO 5 785 VAL 74 4.145114 1 > 54 LYS 6 785 VAL 74 4.305017 1 > 55 LYS 6 785 VAL 74 4.291098 1 > 56 LYS 7 785 VAL 74 3.968647 1 > 58 LYS 7 772 MET 73 4.385121 1 > 58 LYS 7 778 MET 73 4.422980 1 > 58 LYS 7 779 MET 73 3.954990 1 > 58 LYS 7 785 VAL 74 3.420554 1 > 59 LYS 7 763 GLN 72 4.431955 1 > 59 LYS 7 767 GLN 72 3.844037 1 > 59 LYS 7 785 VAL 74 3.725048 1 > > > > > I want to sort the data on the basis of 3rd column first and latter want to > sort the sorted data (in first step) on the basis of 6th column. > > I tried sort() function but could not get the way how to use it. > > I am new to programming, please tell me how can I sort. > > Thanking you in advance > > regards > > > > > -- > -- > Dhananjay C Joshi > Project Assistant > Lab of Structural Biology, > CDFD, Bldg.7, Gruhakalpa > 5-4-399/B, Nampally > Hyderabad- 51, India > Tel: +91-40-24749404 > Fax: +91-40-24749448 > -- > > -- > http://mail.python.org/mailman/listinfo/python-list > > -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Regarding sort()
On Mon, May 25, 2009 at 10:15 AM, Chris Rebert wrote: > Erm, using a compare function rather than a key function slows down > sorting /significantly/. In fact, the `cmp` parameter to list.sort() > has been *removed* in Python 3.0 because of this. Makes a lot of sense, as you only have to run the slow python function N times, rather than doing it N log N times... So I guess I have also learned today that the internal cmp can handle lists and tuples properly, i.e (1,7) > (1,5), definitely a much neater way of doing things: the reasons are starting to pile to fare 2.6 goodbye and move on to 3.0... > > Also, top-posting is evil. Please avoid it in the future. Must be the mad scientist wanting to take over the world in me. ;-) Note taken... Jaime > > Cheers, > Chris > -- > http://blog.rebertia.com > -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
exec statement with/without prior compile...
These weekend I've been tearing down to pieces Michele Simionato's
decorator module, that builds signature preserving decorators. At the
heart of it all there is a dynamically generated function, which works
something similar to this...
...
src = """def function(a,b,c) :\nreturn _caller_(a,b,c)\n"""
evaldict = {'_caller_' : _caller_}
code = compile(src, '', 'single')
exec code in evaldict
new_func = evaldict[function]
...
I have found fooling around with this code, that the compile step can
be completely avoided and go for a single:
exec src in evaldict
Now, I'm sure there is a good reason for that additional step, but I
haven't been able to find what the difference between both approaches
is.
And since I'm asking, could something similar, i.e. defining a new
function and get a handle to it, be achieved with eval?
Thanks in advance,
Jaime Fernández
http://numericalrecipes.blogspot.com
--
(\__/)
( O.o)
( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus
planes de dominación mundial.
--
http://mail.python.org/mailman/listinfo/python-list
Re: How do I sample randomly based on some probability(wightage)?
On Tue, May 26, 2009 at 8:42 PM, Sumitava Mukherjee wrote: > On May 26, 11:39 pm, Sumitava Mukherjee wrote: > > [Oh, I forgot to mention. I am looking for sampling without replacement.] That means that, you'll have to code something that updates the sums of probabilities after each extraction... Would there be a smart way of doing this, of just adding the whole updated list again is the best, or at least good enough? -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: List comprehension and string conversion with formatting
On Tue, Jun 9, 2009 at 5:38 PM,
stephen_b wrote:
> I'd like to convert a list of floats to a list of strings constrained
> to one .1f format. These don't work. Is there a better way?
>
> [".1f" % i for i in l]
> or
> [(".1f" % i) for i in l]
There's a missing %, this does work...
["%.1f" % i for i in l]
Jaime
--
(\__/)
( O.o)
( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus
planes de dominación mundial.
--
http://mail.python.org/mailman/listinfo/python-list
Re: persistent composites
On Sun, Jun 14, 2009 at 4:27 PM, Aaron Brady wrote: > > Before I go and flesh out the entire interfaces for the provided > types, does anyone have a use for it? A real-world application of persistent data structures can be found here: http://stevekrenzel.com/persistent-list Jaime -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Measuring Fractal Dimension ?
On Wed, Jun 17, 2009 at 4:50 AM, Lawrence D'Oliveiro wrote: > In message <[email protected]>, wrote: > >> Lawrence D'Oliveiro writes: >> >>> I don't think any countable set, even a countably-infinite set, can have >>> a fractal dimension. It's got to be uncountably infinite, and therefore >>> uncomputable. >> >> I think the idea is you assume uniform continuity of the set (as >> expressed by a parametrized curve). That should let you approximate >> the fractal dimension. > > Fractals are, by definition, not uniform in that sense. I had my doubts on this statement being true, so I've gone to my copy of Gerald Edgar's "Measure, Topology and Fractal Geometry" and Proposition 2.4.10 on page 69 states: "The sequence (gk), in the dragon construction of the Koch curve converges uniformly." And uniform continuity is a very well defined concept, so there really shouldn't be an interpretation issue here either. Would not stick my head out for it, but I am pretty sure that a continuous sequence of curves that converges to a continuous curve, will do so uniformly. Jaime -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Measuring Fractal Dimension ?
On Wed, Jun 17, 2009 at 1:52 PM, Mark Dickinson wrote: > Maybe James is thinking of the standard theorem > that says that if a sequence of continuous functions > on an interval converges uniformly then its limit > is continuous? Jaime was simply plain wrong... The example that always comes to mind when figuring out uniform convergence (or lack of it), is the step function , i.e. f(x)= 0 if x in [0,1), x(x)=1 if x >= 1, being approximated by the sequence f_n(x) = x**n if x in [0,1), f_n(x) = 1 if x>=1, where uniform convergence is broken mostly due to the limiting function not being continuous. I simply was too quick with my extrapolations, and have realized I have a lot of work to do for my "real and functional analysis" exam coming in three weeks... Jaime P.S. The snowflake curve, on the other hand, is uniformly continuous, right? -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with dictionaries and multidimensial lists
On Tue, Jun 23, 2009 at 4:45 PM, Cameron
Pulsford wrote:
> Hey all, I have a dictionary that looks like this (small example version)
> {(1, 2): 0} named a
>
> so I can do a[1,2] which returns 0. What I also have is a list of
> coordinates into a 2 dimensional array that might look like this b =
> [[1,2]]. Is there anyway I can call a[b[0]] and have it return 0?
a[tuple(b[0])] should do it...
Jaime
--
(\__/)
( O.o)
( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus
planes de dominación mundial.
--
http://mail.python.org/mailman/listinfo/python-list
Re: installing 2.6 on vista64
I have installed the 32 bit 2.6 and the 64 bit 3.1 on my machine running Vista 64 without any issues. Which of course has nothing to do with your problem On Thu, Jul 23, 2009 at 7:06 PM, DwBear75 wrote: > I just downloaded and attempted to install python 2.6.2. The > installer proceeds to do its work then dies, leaving an entry in the > eventlog: > > Windows Installer installed the product. Product Name: Python 2.6.2. > Product Version: 2.6.2150. Product Language: 1033. Installation > success or error status: 1602. > > Googling for this I wasn't able to narrow the results down to > something usable. Anyone know of issues and how to fix them installing > on vista 64 (yes, I have 8 gigs of ram) > -- > http://mail.python.org/mailman/listinfo/python-list > -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial. -- http://mail.python.org/mailman/listinfo/python-list
