How to group numbers of a vector

Hi,

I have the following vector:

V = [1 2 3 5 7 9 11 13 14 15 16 17 20 22 24 26]

I would like to group the numbers according to their difference (step value) from each other so they look like:

Group 1 = [1 2 3]
Group 2 = [5 7 9 11 13]
Group 3 = [14 15 16 17]
Group 4 = [20 22 24 26]

Could anyone help me to sort out this?

Thank you!

[Image Analyst]

Try diff() and findgroups(), but please explain why 3 is a member of the first group, but NOT the second group. And why is 13 a member of group 2 but not also of group 3? When a number can belong to two groups, what is your special rule for determining which group it should go into?

[BAM]

Yes, the vector that I provided as an example its tricky, numbers 3 and 13 could go into either group, but the actual vectors that I work with consist only odd or even numbers, which makes the discrimination easier.

[Stephen23]

"...but the actual vectors that I work with consist only odd or even numbers, which makes the discrimination easier."

No it doesn't. It does not explain which group 3 belongs to.

You need to tell us the algorithm, not just which groups are "correct" or "incorrect".

[Stephen23]

BAM's "Answer" moved here and formatted properly:

I tested the code and in some cases gives an error (does not group correctly). I am going to provide the vector with which I have experienced the error. Group 1 and 5 are incorrect as it is seen 339 should go in group 2 and 473 should be in a separate group alone. If you find a solution for that I will be very thankful.

V = [315   339   341   343   345   347   349   351   353   355   357   359   361   363   365   371   373   375   377   379   381   387   389   391   393   395   397   399   401   403   405   407   409   411   413   415   417   419   421   423   425   427   429   431   433   435   437   439   441   443   445   447   449   473   727   729   731   733   735   737   739   745   747]

Answer:

group{1} =

     315   339
  group{2} =

     341   343   345   347   349   351   353   355   357   359   361   363   365
  group{3} =

     371   373   375   377   379   381
  group{4} =

    Columns 1 through 22

     387   389   391   393   395   397   399   401   403   405   407   409   411   413   415   417   419   421   423   425   427   429

    Columns 23 through 32

     431   433   435   437   439   441   443   445   447   449
  group{5} =

     473   727   729   731   733   735   737   739
  group{6} =

     745   747

[Stephen23]

" Group 1 and 5 are incorrect as it is seen 339 should go in group 2 ..."

It is not enough to simply write that some groups "... are incorrect", because you have not told us anywhere what the algorithm is that you use to decide this. Image Analyst already asked you to clarify this. You stated that "...but the actual vectors that I work with consist only odd or even numbers, which makes the discrimination easier", but this does not explain 3 in your original example, and the number 339 in your comment above, both of which could fit into either group perfectly. You need to tell us exactly how to decide which group they fit into.

[BAM]

I was asked for a specific rule, so if the difference between the first and second value is <=2 and then between the second and third is <=2, these 3 values of the vector will be in the same group, but if the difference between third and fourth is >2 then the fourth value is discriminated from the first group. If the fourth and fifth values have a difference <=2 they will form the second group and so on.

I hope this time is clearer for you asking me for the rule. Thank you!!!

Answer 1

[Stephen23]

One solution based on diff and regexp:

>> V = [1 2 3 5 7 9 11 13 14 15 16 17 20 22 24 26];
>> X = [0,1,~diff(V,2)];
>> [B,E] = regexp(char('0'+X),'0+1*');
>> C = arrayfun(@(b,e){V(b:e)},B,E);
>> C{:}
ans =
    1    2    3
ans =
    5    7    9   11   13
ans =
   14   15   16   17
ans =
   20   22   24   26

[Bruno Luong]

nice

Answer 2

[Bruno Luong]

OK, try this:

    b = [true, false, diff(V,2)~=0];
    b = b & [true, ~b(1:end-1)];
    b = find(b);
    lgt = [b(2:end), length(V)+1] - b;
    G = mat2cell(V,1,lgt);
    G{:}

[BAM]

No change, still the same incorrect results.

[BAM]

Another vector with error:

V = [343 345 347 349 361 363 365 367 387 389 391 393 395 397 399 401 403 405 407 409 411 413 415 417 419 421 423 425 427 429 431 433 437 439 441 443 445 447 715 717 725 727 729 731 733 735 737 739 741 743 745 747 749 751 755 761 763 765 767 769 771 773 775 777]

In this vector 715 and 717 should go into another separate group and 755 as well.

Results:

G{1} = 343 345 347 349

G{2} = 361 363 365 367

G{3} = 387 389 391 393 395 397 399 401 403 405 407 409 411 413 415 417 419 421 423 425 427 429 431 433

G{4} = 437 439 441 443 445 447

G{5} = 715 717 725 727 729 731 733 735 737 739 741 743 745 747 749 751

G{6} = 755 761 763 765 767 769 771 773 775 777

Answer 3

[Bruno Luong]

Can you try this one, I put the flag so that you can select if the common point that goes to left or right group, if you have other specific rule you need to specify it.

V = [315   339   341   343   345   347   349   351   ...
    353   355   357   359   361   363   365   371   ...
    373   375   377   379   381   387   389   391   ...
    393   395   397   399   401   403   405   407   ...
    409   411   413   415   417   419   421   423   ...
    425   427   429   431   433   435   437   439   ...
    441   443   445   447   449   473   727   729   ...
    731   733   735   737   739   745   747]

GoLeftFlag = false; % choice of the common point goes to the left or right group
n = length(V);
d = diff([true, diff(V,2)==0, true]);
s = [1, find(d==1)];
e = [find(d==-1)+1, n];
collision = [0 e(1:end-1)]>=s;
if GoLeftFlag
    s(collision) = s(collision) + 1;
else
    collision(1) = [];
    e(collision) = e(collision) - 1;
end
i2 = arrayfun(@(s,e) min(bsxfun(@plus,s:2:e,[0;1]),e), ...
              e(1:end-1)+1, s(2:end)-1, 'unif', 0);
c = cell(1,2*size(s,2)-1);
c(1:2:end) = num2cell([s; e],1);
c(2:2:end-1) = i2;
i = cat(2,c{:});
lgt = diff(i,1,1)+1;
G = mat2cell(V,1,lgt);
G{:}

[BAM]

I have tried the code, solves the issue that I had previously, but now messes up the vector which previously was correct. Now the problem is with this vector: V = 346 348 350 352 362 The results that I get: Group{1} = 346 348 350 Group{2} = 352 362; 352 must be in group one.

[Bruno Luong]

We don't see why it does mess up, because you haven't told the rule why 352 should be in group 1, no more logical than it goes to group 2, since in both cases groups have linear data, which is the only rule that is clear so far.

If you reverse the flag

GoLeftFlag = true;

352 will go to group 1.

As long as you don't spell out the rule, no code can follow your empirical direction.

[BAM]

Hi Bruno,

Could you please try developing a code for grouping the numbers according to the rule I am providing as you requested before?

The rule is to group numbers in a same group which have difference between each other no more than 2 (<=2). If the difference is more than 2 (>2) send the next number in another group along with the numbers having a difference <=2 which are following the first number send into separated group. I am giving you an example with the last vector I mention earlier.

V = [346 348 350 352 362], numbers having a difference <=2 are going in same group and those >2 into another one. In this case 346, 348, 350, 352 have difference <=2 between each other and 362 have difference > 2 from 352, thus 362 is sent to a separate group. If there is more numbers in the vector as following ...362, 364, 366] their difference between each other is <=2 and that's why they stay in the same group 362, 364,366. I hope this explanation is clear enough.

Also, is it possible to find automatically the minimum and maximum values of each individual group? Diffirent vectors will have different numbers of groups depending on the size and the numbers they consist.

I am new to MATLAB and my knowlidge is very limited in development of codes.

If you can help me to sort out the problem I will be wery thankful.

Thank you in advance.

[Bruno Luong]

The "rule" of <=2 you just state doesn't seem to bring anything new to me, at least in the example you provide.

My old code works well, since it split at the place where the difference changes, regardless it's 2 to 10 or something else.

>> V = [346 348 350 352 362 364 366],

V =

   346   348   350   352   362   364   366

>> GoLeftFlag = false; % choice of the common point goes to the left or right group
n = length(V);
d = diff([true, diff(V,2)==0, true]);
s = [1, find(d==1)];
e = [find(d==-1)+1, n];
collision = [0 e(1:end-1)]>=s;
if GoLeftFlag
    s(collision) = s(collision) + 1;
else
    collision(1) = [];
    e(collision) = e(collision) - 1;
end
i2 = arrayfun(@(s,e) min(bsxfun(@plus,s:2:e,[0;1]),e), ...
              e(1:end-1)+1, s(2:end)-1, 'unif', 0);
c = cell(1,2*size(s,2)-1);
c(1:2:end) = num2cell([s; e],1);
c(2:2:end-1) = i2;
i = cat(2,c{:});
lgt = diff(i,1,1)+1;
G = mat2cell(V,1,lgt);
G{:}

Results:

ans =

   346   348   350   352


ans =

   362   364   366

[BAM]

Yes, it works with this vector.

Could you do a final try with this one:

V = [ 266 270 273 274 276 280 313 314 315 439 455 456 457 471 472 473 474 475 476 863] ?

Also, is it possible to find the minimum and maximum value of each group automatically?

Thank you for your effort once again!

[Bruno Luong]

Sorry I don't try perpetually again and again.

What we need is you spell out the consistent rule in a formal and clear way to avoid undecidability in case of colision.

[BAM]

Group{1} = 266 270 (270 should be separated)

Group{2} = 273 274 (perfect)

Group{3} = 276 280 (280 should be separated)

Group{4} = 313 314 315 (perfect)

Group{5} = 439 (perfect)

Group{6} = 455 456 457 (perfect)

Group{7} = 471 472 473 474 475 (perfect)

Group{8} = 476 863 (863 should be separated)

This is the result that I get, as I said numbers having a diffirence within 2 are going into same group, numbers having a difference greater than 2 (>2) go into another group.

Sorry, for the poor explanation of the rule, but I don't know any other way how I can explain it.

Thank you so much indeed for your willingness to help.

[Bruno Luong]

OK I've modified the code to make group for only step <= 2. It looks ugly

V = [ 266   270   273   274   276   280   313   314   315   439   455   456  ...
    457   471   472   473  474   475   476   863]

GoLeftFlag = false; % choice of the common point goes to the left or right group
n = length(V);
d = diff([true, diff(V,2)==0, true]); % & [true, c] & [c, true]);
s = [1, find(d==1)];
e = [find(d==-1)+1, n];
collision = [0 e(1:end-1)]>=s;
if GoLeftFlag
    s(collision) = s(collision) + 1;
else
    collision(1) = [];
    e(collision) = e(collision) - 1;
end
i2 = arrayfun(@(s,e) min(bsxfun(@plus,s:2:e,[0;1]),e), ...
              e(1:end-1)+1, s(2:end)-1, 'unif', 0);
c = cell(1,2*size(s,2)-1);
c(1:2:end) = num2cell([s; e],1);
c(2:2:end-1) = i2;
i = cat(2,c{:})';
% Make group only if step is <= 2
dv = diff(V(i),1,2)./diff(i,1,2);
stepbig = dv(:)>2;
s = [n,1];
up = accumarray(i(stepbig,1),1,s);
down = -accumarray(i(stepbig(1:end-1),2)+1,1,s);
g = accumarray(i(~stepbig,1),1,s);
lgt = diff(find([cumsum(up+down)+g; true]));
G = mat2cell(V,1,lgt);
G{:}

The result is

G{1} = 266
G{2} = 270
G{3} = 273   274
G{4} = 276
G{5} = 280
G{6} = 313   314   315
G{7} = 439
G{8} = 455   456   457
G{9} = 471   472   473   474   475
G{10} = 476
G{11} = 863

[BAM]

This is it!

You are a STAR Bruno!!! Your work is much appreciated. Thank you!

Just one last thing. I tried to find the minimum and maximum elements of these groups automatically, but I get errors, obviously I did wrong.

When I specify the group number it works, but then this is becoming a manual way of finding the min and max values.

For instance, I specified Group 9 and it works.

MIN = min(G{9})

MAX = max(G{9})

DIFF = MAX-MIN

MIN = 471

MAX = 475

DIFF = 4

Any ideas how the minimum and maximum elements of each group could be found automatically?