I have the following 606 x 274 table:
see here
Goal:
For every date calculate lower and upper 20% percentiles and, based on the outcome, create 2 new variables, e.g. 'L' for "lower" and 'U' for "upper", which contain the ticker names as seen in the header of the table.
Step by step:
% Replace NaNs with 'empty' for the percentile calculation (error: input to be cell array)
T(cellfun(#isnan,T)) = {[]}
% Change date format
T.Date=[datetime(T.Date, 'InputFormat', 'eee dd-MMM-yyyy')];
% Take row by row
for row=1:606
% If Value is in upper 20% percentile create new variable 'U' that contains the according ticker names.
% If Value is in lower 20% percentile create new variable 'L' that contains the according ticker names.
end;
So far, experimenting with 'prctile' only yielded a numeric outcome, for a single column. Example:
Y = prctile(T.A2AIM,20,2);
Thanks for your help and ideas!
Generally speaking, if you have an array of numbers:
a = [4 2 1 8 -2];
percentiles can be computed by first sorting the array and then attempting to access the index supplied in the percentile. So prctile(a,20)'s functionality could in principle be replaced by
b = sort(a);
ind = round(length(b)*20/100);
if ind==0
ind = 1;
end
b = b(ind);
% b = -2
However, prctile does a bit more of fancy magic interpolating the input vector to get a value that is less affected by array size. However, you can use the idea above to find the percentile splitting columns. If you chose to do it like I said above, what you want to do to get the headers that correspond to the 20% and 80% percentiles is to loop through the rows, remove the NaNs, get the indeces of the sort on the remaining values and get the particular index of the 20% or 80% percentile. Regrettably, I have an old version of Matlab that does not support tables so I couldn't verify if the header names are returned correctly, but the idea should be clear.
L = cell(size(T,1),1);
U = cell(size(T,1),1);
for row=1:size(T,1)
row_values = T{row,:};
row_values = row_values(2:end); % Remove date column
non_nan_indeces = find(~isnan(row_values));
if not(isempty(non_nan_indeces))
[row_values,sorted_indeces] = sort(row_values(non_nan_indeces));
% The +1 is because we removed the date column
L_ind = non_nan_indeces(sorted_indeces(1:round(0.2*length(row_values))))+1;
U_ind = non_nan_indeces(sorted_indeces(round(0.8*length(row_values)):end))+1;
% I am unsure about this part
L{row} = T.Properties.VariableNames(L_ind);
U{row} = T.Properties.VariableNames(U_ind);
else
L{row} = nan;
U{row} = nan;
end
end;
If you want to use matlab's prctile, you would have to find the returned value's index doing something like this:
L = cell(size(T,1),1);
U = cell(size(T,1),1);
for row=1:size(T,1)
row_values = T{row,:};
row_values = row_values(2:end); % Remove date column
non_nan_indeces = find(~isnan(row_values));
if not(isempty(non_nan_indeces))
[row_values,sorted_indeces] = sort(row_values(non_nan_indeces));
L_val = prctile(row_values(non_nan_indeces),20);
U_val = prctile(row_values(non_nan_indeces),80);
% The +1 is because we removed the date column
L_ind = non_nan_indeces(sorted_indeces(find(row_values<=L_val)))+1;
U_ind = non_nan_indeces(sorted_indeces(find(row_values>=U_val)))+1;
% I am unsure about this part
L{row} = T.Properties.VariableNames(L_ind);
U{row} = T.Properties.VariableNames(U_ind);
else
L{row} = nan;
U{row} = nan;
end
end;
Related
I have some binary matrix. I want to remove all first ones from each column, but keep one if this value is alone in column. I have some code, which produces correct result, but it looks ugly- I should iterate through all columns.
Could You give me a piece of advice how to improve my code?
Non-vectorised code:
% Dummy matrix for SE
M = 10^3;
N = 10^2;
ExampleMatrix = (rand(M,N)>0.9);
ExampleMatrix1=ExampleMatrix;
% Iterate columns
for iColumn = 1:size(ExampleMatrix,2)
idx = find(ExampleMatrix(:,iColumn)); % all nonzeroes elements
if numel(idx) > 1
% remove all ones except first
ExampleMatrix(idx(1),iColumn) = 0;
end
end
I think this does what you want:
ind_col = find(sum(ExampleMatrix, 1)>1); % index of relevant columns
[~, ind_row] = max(ExampleMatrix(:,ind_col), [], 1); % index of first max of each column
ExampleMatrix(ind_row + (ind_col-1)*size(ExampleMatrix,1)) = 0; % linear indexing
The code uses:
the fact that the second output of max gives the index of the first maximum value. In this case max is applied along the first dimension, to find the first maximum of each column;
linear indexing.
I have the following code:
L_sum = zeros(height(ABC),1);
for i = 1:height(ABC)
L_sum(i) = sum(ABC{i, ABC.L(i,4:281)});
end
Here my table:
Problem: My sum function sums the entire row values (col. 4-281) per date whereas I only want those cells to be added whose headers are in the cell array of ABC.L, for any given date.
X = ABC.L{1, 1}; gives (excerpt):
Red arrow: what sum function is referencing (L of same date).
Green arrow: what I am trying to reference now (L of previous date).
Thanks for your help
In general, in matlab you dont need to use for loops to do simple operations like selective sums.
Example:
Data=...
[1 2 3;
4 5 6;
7 8 9;
7 7 7];
NofRows=size(Data,1);
RowsToSum=3:NofRows;
ColToSum=[1,3];
% sum second dimension 2d array
Result=sum(Data(RowsToSum,ColToSum), 2)
% table mode
DataTable=array2table(Data);
Result2=sum(DataTable{RowsToSum,ColToSum}, 2)
To do that you need to first extract the columns you want to sum, and then sum them:
% some arbitrary data:
ABC = table;
ABC.L{1,1} = {'aa','cc'};
ABC.L{2,1} = {'aa','b'};
ABC.L{3,1} = {'aa','d'};
ABC.L{4,1} = {'b','d'};
ABC{1:4,2:5} = magic(4);
ABC.Properties.VariableNames(2:5) = {'aa','b','cc','d'}
% summing the correct columns:
L_sum = zeros(height(ABC),1);
col_names = ABC.Properties.VariableNames; % just to make things shorter
for k = 1:height(ABC)
% the following 'cellfun' compares each column to the values in ABC.L{k},
% and returns a cell array of the result for each of them, then
% 'cell2mat' converts it to logical array, and 'any' combines the
% results for all elements in ABC.L{k} to one logical vector:
col_to_sum = any(cell2mat(...
cellfun(#(x) strcmp(col_names,x),ABC.L{k},...
'UniformOutput', false).'),1);
% then a logical indexing is used to define the columns for summation:
L_sum(k) = sum(ABC{k,col_to_sum});
end
I have a matrix in MATLAB of 50572x4 doubles. The last column has datenum format dates, increasing values from 7.3025e+05 to 7.3139e+05. The question is:
How can I split this matrix into sub-matrices, each that cover intervals of 30 days?
If I'm not being clear enough… the difference between the first element in the 4th column and the last element in the 4th column is 7.3139e5 − 7.3025e5 = 1.1376e3, or 1137.6. I would like to partition this into 30 day segments, and get a bunch of matrices that have a range of 30 for the 4th columns. I'm not quite sure how to go about doing this...I'm quite new to MATLAB, but the dataset I'm working with has only this representation, necessitating such an action.
Note that a unit interval between datenum timestamps represents 1 day, so your data, in fact, covers a time period of 1137.6 days). The straightforward approach is to compare each timestamps with the edges in order to determine which 30-day interval it belongs to:
t = A(:, end) - min(A:, end); %// Normalize timestamps to start from 0
idx = sum(bsxfun(#lt, t, 30:30:max(t))); %// Starting indices of intervals
rows = diff([0, idx, numel(t)]); %// Number of rows in each interval
where A is your data matrix, where the last column is assumed to contain the timestamps. rows stores the number of rows of the corresponding 30-day intervals. Finally, you can employ cell arrays to split the original data matrix:
C = mat2cell(A, rows, size(A, 2)); %// Split matrix into intervals
C = C(~cellfun('isempty', C)); %// Remove empty matrices
Hope it helps!
Well, all you need is to find the edge times and the matrix indexes in between them. So, if your numbers are at datenum format, one unit is the same as one day, which means that we can jump from 30 and 30 units until we get as close as we can to the end, as follows:
startTime = originalMatrix(1,4);
endTime = originalMatrix(end,4);
edgeTimes = startTime:30:endTime;
% And then loop though the edges checking for samples that complete a cycle:
nEdges = numel(edgeTimes);
totalMeasures = size(originalMatrix,1);
subMatrixes = cell(1,nEdges);
prevEdgeIdx = 0;
for curEdgeIdx = 1:nEdges
nearIdx=getNearestIdx(originalMatrix(:,4),edgeTimes(curEdgeIdx));
if originalMatrix(nearIdx,4)>edgeTimes(curEdgeIdx)
nearIdx = nearIdx-1;
end
if nearIdx>0 && nearIdx<=totalMeasures
subMatrix{curEdgeIdx} = originalMatrix(prevEdgeIdx+1:curEdgeIdx,:);
prevEdgeIdx=curEdgeIdx;
else
error('For some reason the edge was not inbound.');
end
end
% Now we check for the remaining days after the edges which does not complete a 30 day cycle:
if curEdgeIdx<totalMeasures
subMatrix{end+1} = originalMatrix(curEdgeIdx+1:end,:);
end
The function getNearestIdx was discussed here and it gives you the nearest point from the input values without checking all possible points.
function vIdx = getNearestIdx(values,point)
if isempty(values) || ~numel(values)
vIdx = [];
return
end
vIdx = 1+round((point-values(1))*(numel(values)-1)...
/(values(end)-values(1)));
if vIdx < 1, vIdx = []; end
if vIdx > numel(values), vIdx = []; end
end
Note: This is pseudocode and may contain errors. Please try to adjust it into your problem.
I am looking for a 'good' way to find a matrix (pattern) in a larger matrix (arbitrary number of dimensions).
Example:
total = rand(3,4,5);
sub = total(2:3,1:3,3:4);
Now I want this to happen:
loc = matrixFind(total, sub)
In this case loc should become [2 1 3].
For now I am just interested in finding one single point (if it exists) and am not worried about rounding issues. It can be assumed that sub 'fits' in total.
Here is how I could do it for 3 dimensions, however it just feels like there is a better way:
total = rand(3,4,5);
sub = total(2:3,1:3,3:4);
loc = [];
for x = 1:size(total,1)-size(sub,1)+1
for y = 1:size(total,2)-size(sub,2)+1
for z = 1:size(total,3)-size(sub,3)+1
block = total(x:x+size(sub,1)-1,y:y+size(sub,2)-1,z:z+size(sub,3)-1);
if isequal(sub,block)
loc = [x y z]
end
end
end
end
I hope to find a workable solution for an arbitrary number of dimensions.
Here is low-performance, but (supposedly) arbitrary dimensional function. It uses find to create a list of (linear) indices of potential matching positions in total and then just checks if the appropriately sized subblock of total matches sub.
function loc = matrixFind(total, sub)
%matrixFind find position of array in another array
% initialize result
loc = [];
% pre-check: do all elements of sub exist in total?
elements_in_both = intersect(sub(:), total(:));
if numel(elements_in_both) < numel(unique(sub))
% if not, return nothing
return
end
% select a pivot element
% Improvement: use least common element in total for less iterations
pivot_element = sub(1);
% determine linear index of all occurences of pivot_elemnent in total
starting_positions = find(total == pivot_element);
% prepare cell arrays for variable length subscript vectors
[subscripts, subscript_ranges] = deal(cell([1, ndims(total)]));
for k = 1:length(starting_positions)
% fill subscript vector for starting position
[subscripts{:}] = ind2sub(size(total), starting_positions(k));
% add offsets according to size of sub per dimension
for m = 1:length(subscripts)
subscript_ranges{m} = subscripts{m}:subscripts{m} + size(sub, m) - 1;
end
% is subblock of total equal to sub
if isequal(total(subscript_ranges{:}), sub)
loc = [loc; cell2mat(subscripts)]; %#ok<AGROW>
end
end
end
This is based on doing all possible shifts of the original matrix total and comparing the upper-leftmost-etc sub-matrix of the shifted total with the sought pattern subs. Shifts are generated using strings, and are applied using circshift.
Most of the work is done vectorized. Only one level of loops is used.
The function finds all matchings, not just the first. For example:
>> total = ones(3,4,5,6);
>> sub = ones(3,3,5,6);
>> matrixFind(total, sub)
ans =
1 1 1 1
1 2 1 1
Here is the function:
function sol = matrixFind(total, sub)
nd = ndims(total);
sizt = size(total).';
max_sizt = max(sizt);
sizs = [ size(sub) ones(1,nd-ndims(sub)) ].'; % in case there are
% trailing singletons
if any(sizs>sizt)
error('Incorrect dimensions')
end
allowed_shift = (sizt-sizs);
max_allowed_shift = max(allowed_shift);
if max_allowed_shift>0
shifts = dec2base(0:(max_allowed_shift+1)^nd-1,max_allowed_shift+1).'-'0';
filter = all(bsxfun(#le,shifts,allowed_shift));
shifts = shifts(:,filter); % possible shifts of matrix "total", along
% all dimensions
else
shifts = zeros(nd,1);
end
for dim = 1:nd
d{dim} = 1:sizt(dim); % vectors with subindices per dimension
end
g = cell(1,nd);
[g{:}] = ndgrid(d{:}); % grid of subindices per dimension
gc = cat(nd+1,g{:}); % concatenated grid
accept = repmat(permute(sizs,[2:nd+1 1]), [sizt; 1]); % acceptable values
% of subindices in order to compare with matrix "sub"
ind_filter = find(all(gc<=accept,nd+1));
sol = [];
for shift = shifts
total_shifted = circshift(total,-shift);
if all(total_shifted(ind_filter)==sub(:))
sol = [ sol; shift.'+1 ];
end
end
For an arbitrary number of dimensions, you might try convn.
C = convn(total,reshape(sub(end:-1:1),size(sub)),'valid'); % flip dimensions of sub to be correlation
[~,indmax] = max(C(:));
% thanks to Eitan T for the next line
cc = cell(1,ndims(total)); [cc{:}] = ind2sub(size(C),indmax); subs = [cc{:}]
Thanks to Eitan T for the suggestion to use comma-separated lists for a generalized ind2sub.
Finally, you should test the result with isequal because this is not a normalized cross correlation, meaning that larger numbers in a local subregion will inflate the correlation value potentially giving false positives. If your total matrix is very inhomogeneous with regions of large values, you might need to search other maxima in C.
so I have a matrix Data in this format:
Data = [Date Time Price]
Now what I want to do is plot the Price against the Time, but my data is very large and has lines where there are multiple Prices for the same Date/Time, e.g. 1st, 2nd lines
29 733575.459548611 40.0500000000000
29 733575.459548611 40.0600000000000
29 733575.459548612 40.1200000000000
29 733575.45954862 40.0500000000000
I want to take an average of the prices with the same Date/Time and get rid of any extra lines. My goal is to do linear intrapolation on the values which is why I must have only one Time to one Price value.
How can I do this? I did this (this reduces the matrix so that it only takes the first line for the lines with repeated date/times) but I don't know how to take the average
function [ C ] = test( DN )
[Qrows, cols] = size(DN);
C = DN(1,:);
for i = 1:(Qrows-1)
if DN(i,2) == DN(i+1,2)
%n = 1;
%while DN(i,2) == DN(i+n,2) && i+n<Qrows
% n = n + 1;
%end
% somehow take average;
else
C = [C;DN(i+1,:)];
end
end
[C,ia,ic] = unique(A,'rows') also returns index vectors ia and ic
such that C = A(ia,:) and A = C(ic,:)
If you use as input A only the columns you do not want to average over (here: date & time), ic with one value for every row where rows you want to combine have the same value.
Getting from there to the means you want is for MATLAB beginners probably more intuitive with a for loop: Use logical indexing, e.g. DN(ic==n,3) you get a vector of all values you want to average (where n is the index of the date-time-row it belongs to). This you need to do for all different date-time-combinations.
A more vector-oriented way would be to use accumarray, which leads to a solution of your problem in two lines:
[DateAndTime,~,idx] = unique(DN(:,1:2),'rows');
Price = accumarray(idx,DN(:,3),[],#mean);
I'm not quite sure how you want the result to look like, but [DataAndTime Price] gives you the three-row format of the input again.
Note that if your input contains something like:
1 0.1 23
1 0.2 47
1 0.1 42
1 0.1 23
then the result of applying unique(...,'rows') to the input before the above lines will give a different result for 1 0.1 than using the above directly, as the latter would calculate the mean of 23, 23 and 42, while in the former case one 23 would be eliminates as duplicate before and the differing row with 42 would have a greater weight in the average.
Try the following:
[Qrows, cols] = size(DN);
% C is your result matrix
C = DN;
% this will give you the indexes where DN(i,:)==DN(i+1)
i = find(diff(DN(:,2)==0);
% replace C(i,:) with the average
C(i,:) = (DN(i,:)+DN(i+1,:))/2;
% delete the C(i+1,:) rows
C(i,:) = [];
Hope this works.
This should work if the repeated time values come in pairs (the average is calculated between i and i+1). Should you have time repeats of 3 or more then try to rethink how to change these steps.
Something like this would work, but I did not run the code so I can't promise there's no bugs.
newX = unique(DN(:,2));
newY = zeros(1,length(newX));
for ix = 1:length(newX)
allOcurrences = find(DN(:,2)==DN(i,2));
% If there's duplicates, take their mean
if numel(allOcurrences)>1
newY(ix) = mean(DN(allOcurrences,3));
else
% If not, use the only Y value
newY(ix) = DN(ix,3);
end
end