How to display averages in new array?

Hello, I want to average each column in 'B' and arrange the averages in an array so that each row of the new array will display the set of averages relating to the final stock price B(n+1,i).

Thats where I am having difficulty.Thank you for your efforts

n=4
i=1:16
B =
  Columns 1 through 11

    100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  
     90.0000   90.0000   90.0000   90.0000  110.0000   90.0000   90.0000   90.0000  110.0000  110.0000  110.0000   
     81.0000   81.0000   81.0000   99.0000   99.0000   81.0000   99.0000   99.0000   99.0000   99.0000  121.0000   
     72.9000   72.9000   89.1000   89.1000   89.1000   89.1000   89.1000  108.9000   89.1000  108.9000  108.9000  
     65.6100   80.1900   80.1900   80.1900   80.1900   98.0100   98.0100   98.0100   98.0100   98.0100   98.0100  

    Columns 12 through 16

    100.0000  100.0000  100.0000  100.0000  100.000   
    110.0000  110.0000  110.0000  110.0000  110.000
    121.0000  121.0000  121.0000  121.0000  121.000
    108.9000  133.1000  133.1000  108.9000  133.100
    119.7900  119.7900  119.7900  119.7900  146.410

Each column in B is a price trajectory of an asset, the asset moves up or down at each row, but some paths end up at the same point after n+1 time steps.

so I have multiple price paths ending at the same point.

I want to average the paths and group them together such that I will have a set of averages relating to one final stock price.

for example my new array A will look like

so A will look something like this

A=

   81.9020         0         0         0         0         0
   84.8100   88.0500   91.6500   95.6500         0         0
   91.6200   95.2200   99.1820   99.2200  103.1800  107.5800
  103.5300  107.5300  111.9300  116.7700         0         0
  122.1000         0         0         0         0         0

..... so each row of A displays the set of averages for each price path ending at the same point. row 1 of A has the averages for final stock price 65.61 row 2 of A has the averages for final stock price 80.19 for 2 of A has the averages for final stock price 98.01

[David Fletcher]

To be honest, I'm having trouble understanding what this final matrix will look like. If n=4, and i=1:16 does this mean that you want the entire fifth row of B (your stated B(n+1,i)) to be merged with mean(B) giving a 2x16 matrix. If not, some sort of example to clarify what you are trying to achieve might be helpful.

[michaelsmith]

Yeah sure,

each column of B represents a price path of an asset.

The final entry in each column of the matrix B, (B(n+1,i)), represents a final stock price of the asset after n time steps.

I want to create a new array displaying the mean of the price paths so that each row displays the set of averages for each end point.

so for instance new array A is size n+1 x 1:end

A=

mean(B(:,1);

mean(B(:,2),mean(B(:,3)),mean(B(:,4),mean(B(:,5);

mean(B(:,6),mean(B(:,7)),mean(B(:,8),mean(B(:,9),mean(B(:,11),..

mean(B(:,12),mean(B(:,13),....

.....
so each row of A displays the set of averages for each price path ending at the same point.
row 1 of A has the averages for 65.61
row 2 of A has the averages for 80.19
for 2 of A has the averages for 98.01

I hope this clears things up Many thanks

[Image Analyst]

My code below (and in your duplicate question) computes the average of everything from row 1 to the current row. Is that what you want? Because I'm not seeing how the average of any values there equals 65.61. For example, all elements in the first column of B are more than 65.61 so how can the mean be that?

Answer 1

[Image Analyst]

OK, I changed it from a cell array to a double array like you wanted:

B=[100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.000;
  90.0000   90.0000   90.0000   90.0000  110.0000   90.0000   90.0000   90.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.000;
  81.0000   81.0000   81.0000   99.0000   99.0000   81.0000   99.0000   99.0000   99.0000   99.0000  121.0000  121.0000  121.0000  121.0000  121.0000  121.000;
  72.9000   72.9000   89.1000   89.1000   89.1000   89.1000   89.1000  108.9000   89.1000  108.9000  108.9000  108.9000  133.1000  133.1000  108.9000  133.100;
  65.6100   80.1900   80.1900   80.1900   80.1900   98.0100   98.0100   98.0100   98.0100   98.0100   98.0100  119.7900  119.7900  119.7900  119.7900  146.410]
% Extract the last row.
lastRow = B(end, :)
[rows, columns] = size(B)
% Get unique prices in the last row, for example [65.61, 80.19, 98.01]
pricesToSearchFor = unique(lastRow)
for k = 1 : length(pricesToSearchFor)
  thisPrice = pricesToSearchFor(k);
  fprintf('Searching all columns for price $%.2f.\n', thisPrice);
  % Search every column for this unique price.
  columnCounter = 1;
  for column = 1 : columns
    thisColumn = B(:, column);
    % Find out if this price occurs EXACTLY in the column
    lastIndexInColumn = find(thisColumn == thisPrice, 1, 'last');
    if ~isempty(lastIndexInColumn)
      fprintf('Found %.2f at row %d of column %d of B.\n', ...
        thisPrice, lastIndexInColumn, column);
      % If we found the price in this column, add it to a new column of A
      A(k, columnCounter) = mean(thisColumn(1:lastIndexInColumn));
      % Increment the column in A
      columnCounter = columnCounter + 1;
    end
  end
end
A

What it shows is:

A =
       81.902            0            0            0            0            0
       84.818       88.058       91.658       95.658            0            0
       91.622       95.222       99.182       99.222       103.18       107.58
       111.94       116.78       116.78       111.94            0            0
        122.1            0            0            0            0            0

Is this what you want?

Answer 2

[David Fletcher]

B=[100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.000;
 90.0000   90.0000   90.0000   90.0000  110.0000   90.0000   90.0000   90.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.000;
 81.0000   81.0000   81.0000   99.0000   99.0000   81.0000   99.0000   99.0000   99.0000   99.0000  121.0000  121.0000  121.0000  121.0000  121.0000  121.000; 
 72.9000   72.9000   89.1000   89.1000   89.1000   89.1000   89.1000  108.9000   89.1000  108.9000  108.9000  108.9000  133.1000  133.1000  108.9000  133.100;
 65.6100   80.1900   80.1900   80.1900   80.1900   98.0100   98.0100   98.0100   98.0100   98.0100   98.0100  119.7900  119.7900  119.7900  119.7900  146.410]

result=zeros(5,16)

for iter=1:5
    result(iter,:)=mean(B(1:iter,:))
end

I think I understand (maybe)

[David Fletcher]

Just seen the above comment, I'm afraid I'm back to be completely in the dark again. Thank Christ I never became a banker.

[michaelsmith]

Thanks David, ill link you to my comment; https://uk.mathworks.com/matlabcentral/answers/389060-how-to-display-averages-in-new-array#comment_546643

I think you are almost there, I want to group the averages for each final price point so that I can access them easily, like what I have linked

[David Fletcher]

OK, I'll have another look in bit. Just got to go and mop my fevered brow with a copy of the Financial Times

[michaelsmith]

Each column in B is a price trajectory of an asset, the asset moves up or down at each row, but some paths end up at the same point after n+1 time steps.

so I have multiple price paths ending at the same point.

I want to average the paths and group them together such that I will have a set of averages relating to one final stock price.

[Image Analyst]

Did you see my second answer I just posted? The answer is a cell array because some columns of B may not have that exact price in it.

Answer 3

[Image Analyst]

It's just as confusing as your prior, duplicate question. Do you perhaps mean

B =[...
  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000
  90.0000   90.0000   90.0000   90.0000  110.0000   90.0000   90.0000   90.0000  110.0000  110.0000  110.0000
  81.0000   81.0000   81.0000   99.0000   99.0000   81.0000   99.0000   99.0000   99.0000   99.0000  121.0000
  72.9000   72.9000   89.1000   89.1000   89.1000   89.1000   89.1000  108.9000   89.1000  108.9000  108.9000
  65.6100   80.1900   80.1900   80.1900   80.1900   98.0100   98.0100   98.0100   98.0100   98.0100   98.0100  ]

counts = cumsum(ones(size(B)))
cB = cumsum(B)
meanPrices = cB ./ counts

[michaelsmith]

Hi Image Analyst,

thanks for your help.

This comment I added may clear things up of what Im tying to achieve.

https://uk.mathworks.com/matlabcentral/answers/389060-how-to-display-averages-in-new-array#comment_546636

[Image Analyst]

Not really. As you can see, you're saying A(1,1) is the mean of B(:, 1). And the first column of B is [100, 90, 81, 72, 65], the mean of which is 81.6. However you say it should be 65.61. Why?

[michaelsmith]

No no no, sorry,

Im trying to say the row of A(1,:) will contain all the averages of the price paths ending in 65.61.

and the row A(2,:) will contain all the averages of price paths ending in 80.19.

row A(3,:) will contain all the averages of price paths ending in 98.01

[michaelsmith]

so A will look something like this

A=

   81.9020         0         0         0         0         0
   84.8100   88.0500   91.6500   95.6500         0         0
   91.6200   95.2200   99.1820   99.2200  103.1800  107.5800
  103.5300  107.5300  111.9300  116.7700         0         0
  122.1000         0         0         0         0         0

Answer 4

[Image Analyst]

See if this does what you want:

B=[100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.000;
  90.0000   90.0000   90.0000   90.0000  110.0000   90.0000   90.0000   90.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.000;
  81.0000   81.0000   81.0000   99.0000   99.0000   81.0000   99.0000   99.0000   99.0000   99.0000  121.0000  121.0000  121.0000  121.0000  121.0000  121.000;
  72.9000   72.9000   89.1000   89.1000   89.1000   89.1000   89.1000  108.9000   89.1000  108.9000  108.9000  108.9000  133.1000  133.1000  108.9000  133.100;
  65.6100   80.1900   80.1900   80.1900   80.1900   98.0100   98.0100   98.0100   98.0100   98.0100   98.0100  119.7900  119.7900  119.7900  119.7900  146.410]
% Extract the last row.
lastRow = B(end, :)
[rows, columns] = size(B)
% Get unique prices in the last row, for example [65.61, 80.19, 98.01]
pricesToSearchFor = unique(lastRow)
for k = 1 : length(pricesToSearchFor)
  thisPrice = pricesToSearchFor(k);
  fprintf('Searching all columns for price $%.2f.\n', thisPrice);
  % Search every column for this unique price.
  columnCounter = 1;
  for column = 1 : columns
    thisColumn = B(:, column);
    % Find out if this price occurs EXACTLY in the column
    lastIndexInColumn = find(thisColumn == thisPrice, 1, 'last');
    if ~isempty(lastIndexInColumn)
      fprintf('Found %.2f at row %d of column %d of B.\n', ...
        thisPrice, lastIndexInColumn, column);
      % If we found the price in this column, add it to a new column of A
      A{k, columnCounter} = mean(thisColumn(1:lastIndexInColumn));
      % Increment the column in A
      columnCounter = columnCounter + 1;
    end
  end
end
celldisp(A)

Answer 5

[Stephen23]

This is easy in just a few lines.

B = [...
100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.000;
 90.0000   90.0000   90.0000   90.0000  110.0000   90.0000   90.0000   90.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.0000  110.000;
 81.0000   81.0000   81.0000   99.0000   99.0000   81.0000   99.0000   99.0000   99.0000   99.0000  121.0000  121.0000  121.0000  121.0000  121.0000  121.000; 
 72.9000   72.9000   89.1000   89.1000   89.1000   89.1000   89.1000  108.9000   89.1000  108.9000  108.9000  108.9000  133.1000  133.1000  108.9000  133.100;
 65.6100   80.1900   80.1900   80.1900   80.1900   98.0100   98.0100   98.0100   98.0100   98.0100   98.0100  119.7900  119.7900  119.7900  119.7900  146.410];

Method one: accumarray:

M = mean(B,1);
[U,~,R] = unique(B(end,:));
C = cell2mat(arrayfun(@(r)1:nnz(R==r),unique(R),'uni',0));
A = accumarray([R(:),C(:)],M(:),[],[],0)

giving:

A =
     81.902            0            0            0            0            0
     84.818       88.058       91.658       95.658            0            0
     91.622       95.222       99.182       99.222       103.18       107.58
     111.94       116.78       116.78       111.94            0            0
      122.1            0            0            0            0            0

Method two: sub2ind:

[U,~,R] = unique(B(end,:));
C = cell2mat(arrayfun(@(r)1:nnz(R==r),unique(R),'uni',0));
A = zeros(max(R),max(C));
A(sub2ind(size(A),R,C)) = mean(B,1)

giving:

A =
     81.902            0            0            0            0            0
     84.818       88.058       91.658       95.658            0            0
     91.622       95.222       99.182       99.222       103.18       107.58
     111.94       116.78       116.78       111.94            0            0
      122.1            0            0            0            0            0

It seems that the question shows incorrect values for the fourth row.

[Image Analyst]

Yeah, it seems like it. Glad that your answer corroborated my answer, so it seems like we're right.