@Nasser Or even clearer x_Integer?Positive or x_Integer?NonNegative.

@Verbeia sure. good point.

@Mr.Wizard maybe when I get a chance. have been quite busy.

speaking of which, have to run. Bye all.

@rm-rf What are the relation between Fig. 1,2 and3 and how to generate it?

if anyone like to test my latest demo, see if it is ok, here is the notebook: 12000.org/tmp/082213 if you spot any problems, something not clear, etc... thanks.

@hwlau See the a = DiskMatrix[... message starred on the right. I just ran it on all the images in ExampleData

Fig1 is the original images, Fig2 is at an intermediate time step and Fig3 is after a long time...

Actually, the DiskMatrix + GradientFilter + subtraction + normalization should be amenable to analytical alternative solution, but sadly patterns arising from that are unlikely to be particularly interesting...

Hi - just here to try and gauge how many people here find it hard to work with Notebooks in a source control system - with all the Cells, RowBox's. Do many people here just use txt file extension?

@JeremyThompson My personal hunch (not based on actual educative foot-shooting) is that version control, in the sense it's used with typical source code, is unusable with Mma notebooks. Only meaningful usage scenario is to use it as a some sort of snapshotting backup service.

use my method of source control: v1.nb, v2.nb, v3.nb and so on

@JeremyThompson there are some Q&As with regard to version control around. No problem with .m packages, at any rate.

e.g.:

hey everyone

@Sosi Alo

hi @Rojo! how are you?

@Sosi Good. Just woke up. You?

hehe exactly the same

i mean, i woke up a bit ago, but have done nothing yet

where are you from?

@Sosi Argentina, you?

Portugal here

Lisbon?

I have family there

...so yeah, this was not such a useful morning (it's already 1pm)

I'm from very close to lisbon!

have you ever visited them?

Sim, duas vezes acho. Mas a primeira era muito pequeno. A ultima vez foi já ha7 anos mais ou menos

wow!! Muito bem!! :D

ahahah

:)

Yo tanbien hablo un poqito de espanol, pero no se escribir :P

Jaja, muy bien también

@Rojo other than Alt + ., or Alt + ,, is there a more direct way of just quitting the kernel by force?

@Sosi Those alts don't quit th ekernel, they just try to interrupt the evaluation

To force quit the kernel, go to the Evaluation menu->Quit kernel

or check this question

@Rojo yep, that was exactly it. I was having the same problem as the OP

btw, do you know where can I find MenuSetup.tr? the only one that I find in the Mma installation folder is SystemMenuSetup.ml

@Rojo thanks, it was there

No problem

Does anyone have any idea how can I find the central point in a ContourPlot? (Imagining the contour defines a circle, I would be looking for the centre)

@Sosi This is not obvious for non-convex regions.

Even then you need to define what you mean by "central point".

@halirutan I wonder if this is just a stupid question, given that I have no obvious mathematic way of defining it

@Sosi Almost no question is stupid, but you need to consider that the general case is not so easy to answer.

@halirutan yeah.. :\

@Sosi Take for instance this here

ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}, Contours -> 1]

which center do you want here?

@halirutan in my own problem I don't have multiple regions, just a single one (I used Cos in my example just because I couldn't come up with anything else)

but I recon that this is probably very hard to answer without explicit code

@Sosi Then what you can try easily is the center of masses if you view your region as solid object

@halirutan googling! Thanks!

I get the idea

this

@Sosi Taking your example:

data = Table[ Boole[Sin[x] + Sin[y] < 0], {y, -4.7, 1.6, 0.05}, {x, -4.7, 1.6, 0.05}];
Image[data]
Mean[Position[data, 1]] // N

This gives you the center of masses in pixel coordinates.

@Sosi The same.

@halirutan (also, could you post that as answer? )

Oh, just with Mean? Is that so simple?

wow

hm

@Sosi What I would do is taking the image and use ComponentMeasurements

@Sosi So if you have data as above then

img = Image[data];
ComponentMeasurements[MorphologicalComponents[img], "Centroid"]

@halirutan Oh nice! I see!

The two other objects are the small points to the left and at the bottom which are recognised as separate objects.

wow this is cool

yeah

@Sosi The good thing here is that you have all the fancy measurements which are provided

      	"Centroid"	center of mass coordinates
      	"Medoid"	coordinate of the closest element to the centroid
      	"MeanCentroidDistance"	mean distance of all elements from the centroid
      	"MaxCentroidDistance"	maximum distance of all elements from the centroid
      	"MinCentroidDistance"	minimum distance of all elements from the centroid

And the other 3 millions

oh my...

eheh this is cool!

I didn't know about ComponentMeasurements, but I was suspecting something like it could show up :P

Can you post this code you just gave me as answer to my question? This way I'll give you credit and you get some points!

please

@Sosi Or you integrate like the formula for a continuos objects suggest

ip = ListInterpolation[data];
1/NIntegrate[ip[x, y], {x, 1, 127}, {y, 1, 127}, MaxRecursion -> 0]*
 NIntegrate[ip[x, y]*{x, y}, {x, 1, 127}, {y, 1, 127}, MaxRecursion -> 0]

this is the formula from Wiki

@halirutan thank you so much!

@Sosi And yes, you could extract the polygonized region from a ConourPlot too and calculated the center on basis of the polygons, but if you have a function anyway, I would stick to that.

Lets wait what the other users suggest as answer ;-)

Going to dinner, see you all later!

@halirutan have a nice one! Oh, and someone suggested something similar to what you said above

@Sosi Yes, Simons suggestion is also possible. You see, there are many ways. Maybe I fill in some alternatives later.

@halirutan thank you so much for all the help!

@halirutan enjoy your dinner

Interesting news for v9 users re: MapAt, assuming this isn't already disseminated:

For v9 users: Using MapAt with Span

@YvesKlett I asked SE and they said that the process is the same as before... so please post a meta question inviting applications and feel free to list yourself as a volunteer in the answers :)

@Mr.Wizard GREEEEEEEEEEAT

Yeah, I pinged you just to celebrate

Sorry

@Rojo lol -- that's OK.

Okay, out again. Bye!