Implementation of correlation matrix in MATLAB - matlab

I want to implement the following formula in MATLAB, where u_i^(k) means the i,k element. However, I get different results from the ones I compute by hand... I believe that something is wrong with my MATLAB code. For instance,
I should get:
L_ii =
0.1022 0 0
0 0.1657 0
0 0 2.7321
U_ij =
0.7514 0.3104 0.5823
-0.6513 0.4901 0.5793
-0.1055 -0.8145 0.5704
1,1=1-(0.1022*(+0.7514)^2+0.1657*(+0.3104)^2+2.7321*(+0.5823)^2)=-0.000049
2,2=1-(0.1022*(-0.6513)^2+0.1657*(+0.4901)^2+2.7321*(+0.5793)^2)=-0.000015
3,3=1-(0.1022*(-0.1055)^2+0.1657*(-0.8145)^2+2.7321*(+0.5704)^2)=+0.000030
Any ideas??? Please, help me fix Epsilon first (it might not need to move on Rho. Let's fix Epsilon first...)
EDIT: Here is a sample code:
E_squared_ii = ONES_j - diag(L_ii)' * (U_ij'.^ 2)
And here is the wrong result I get at the moment:
E_squared_ii =
1.0e-15 *
0.444089209850063 0.333066907387547 -0.222044604925031

If I use your values and code, I get the expected result:
>> L_ii
L_ii =
0.1022 0 0
0 0.1657 0
0 0 2.7321
>> U_ij
U_ij =
0.7514 0.3104 0.5823
-0.6513 0.4901 0.5793
-0.1055 -0.8145 0.5704
>> ONES_j
ONES_j =
1 1 1
>> E_squared_ii = ONES_j - diag(L_ii)' * (U_ij'.^ 2)
E_squared_ii =
1.0e-04 *
-0.4935 -0.1451 0.2985
Presumably this means that something isn't the value you think it is...

Related

How to normalize Integer variable to positive negative or zero with bit operations

7 -> 1
0 -> 0
-7 -> -1
I've have code:
(x == 0 ? 0 : x / abs(x)) + 1
but is it possible to avoid division and make it faster?
How about
(x == 0 ? 0 : (x < 0 ? -1 : 1))
The idea was to use bit operations to avoid branching code or value conversion.
Haven't found how to do it with bit operations but apple already add this function
https://developer.apple.com/documentation/swift/int/2886673-signum
signum()
Returns -1 if this value is negative and 1 if it’s positive; otherwise, 0.
so simple) raw test shows ~x100 faster implementation

Compute the Frequency of bigrams in Matlab

I am trying to compute and plot the distribution of bigrams frequencies
First I did generate all possible bigrams which gives 1296 bigrams
then i extract the bigrams from a given file and save them in words1
my question is how to compute the frequency of these 1296 bigrams for the file a.txt?
if there are some bigrams did not appear at all in the file, then their frequencies should be zero
a.txt is any text file
clear
clc
%************create bigrams 1296 ***************************************
chars ='1234567890abcdefghijklmonpqrstuvwxyz';
chars1 ='1234567890abcdefghijklmonpqrstuvwxyz';
bigram='';
for i=1:36
for j=1:36
bigram = sprintf('%s%s%s',bigram,chars(i),chars1(j));
end
end
temp1 = regexp(bigram, sprintf('\\w{1,%d}', 1), 'match');
temp2 = cellfun(#(x,y) [x '' y],temp1(1:end-1)', temp1(2:end)','un',0);
bigrams = temp2;
bigrams = unique(bigrams);
bigrams = rot90(bigrams);
bigram = char(bigrams(1:end));
all_bigrams_len = length(bigrams);
clear temp temp1 temp2 i j chars1 chars;
%****** 1. Cleaning Data ******************************
collection = fileread('e:\a.txt');
collection = regexprep(collection,'<.*?>','');
collection = lower(collection);
collection = regexprep(collection,'\W','');
collection = strtrim(regexprep(collection,'\s*',''));
%*******************************************************
temp = regexp(collection, sprintf('\\w{1,%d}', 1), 'match');
temp2 = cellfun(#(x,y) [x '' y],temp(1:end-1)', temp(2:end)','un',0);
words1 = rot90(temp2);
%*******************************************************
words1_len = length(words1);
vocab1 = unique(words1);
vocab_len1 = length(vocab1);
[vocab1,void1,index1] = unique(words1);
frequencies1 = hist(index1,vocab_len1);
I. Character counting problem for a string
bsxfun based solution for counting characters -
counts = sum(bsxfun(#eq,[string1-0]',65:90))
Output -
counts =
2 0 0 0 0 2 0 1 0 0 ....
If you would like to get a tabulate output of counts against each letter -
out = [cellstr(['A':'Z']') num2cell(counts)']
Output -
out =
'A' [2]
'B' [0]
'C' [0]
'D' [0]
'E' [0]
'F' [2]
'G' [0]
'H' [1]
'I' [0]
....
Please note that this was a case-sensitive counting for upper-case letters.
For a lower-case letter counting, use this edit to this earlier code -
counts = sum(bsxfun(#eq,[string1-0]',97:122))
For a case insensitive counting, use this -
counts = sum(bsxfun(#eq,[upper(string1)-0]',65:90))
II. Bigram counting case
Let us suppose that you have all the possible bigrams saved in a 1D cell array bigrams1 and the incoming bigrams from the file are saved into another cell array words1. Let us also assume certain values in them for demonstration -
bigrams1 = {
'ar';
'de';
'c3';
'd1';
'ry';
't1';
'p1'}
words1 = {
'de';
'c3';
'd1';
'r9';
'yy';
'de';
'ry';
'de';
'dd';
'd1'}
Now, you can get the counts of the bigrams from words1 that are present in bigrams1 with this code -
[~,~,ind] = unique(vertcat(bigrams1,words1));
bigrams_lb = ind(1:numel(bigrams1)); %// label bigrams1
words1_lb = ind(numel(bigrams1)+1:end); %// label words1
counts = sum(bsxfun(#eq,bigrams_lb,words1_lb'),2)
out = [bigrams1 num2cell(counts)]
The output on code run is -
out =
'ar' [0]
'de' [3]
'c3' [1]
'd1' [2]
'ry' [1]
't1' [0]
'p1' [0]
The result shows that - First element ar from the list of all possible bigrams has no find in words1 ; second element de has three occurrences in words1 and so on.
Hey similar to Dennis solution you can just use histc()
string1 = 'ASHRAFF'
histc(string1,'ABCDEFGHIJKLMNOPQRSTUVWXYZ')
this checks the number of entries in the bins defined by the string 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' which is hopefully the alphabet (just wrote it fast so no garantee). The result is:
Columns 1 through 21
2 0 0 0 0 2 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0
Columns 22 through 26
0 0 0 0 0
Just a little modification of my solution:
string1 = 'ASHRAFF'
alphabet1='A':'Z'; %%// as stated by Oleg Komarov
data=histc(string1,alphabet1);
results=cell(2,26);
for k=1:26
results{1,k}= alphabet1(k);
results{2,k}= data(k);
end
If you look at results now you can easily check rather it works or not :D
This answer creates all bigrams, loads in the file does a little cleanup, ans then uses a combination of unique and histc to count the rows
Generate all Bigrams
note the order here is important as unique will sort the array so this way it is created presorted so the output matches expectation;
[y,x] = ndgrid(['0':'9','a':'z']);
allBigrams = [x(:),y(:)];
Read The File
this removes capitalisation and just pulls out any 0-9 or a-z character then creates a column vector of these
fileText = lower(fileread('d:\loremipsum.txt'));
cleanText = regexp(fileText,'([a-z0-9])','tokens');
cleanText = cell2mat(vertcat(cleanText{:}));
create bigrams from file by shifting by one and concatenating
fileBigrams = [cleanText(1:end-1),cleanText(2:end)];
Get Counts
the set of all bigrams is added to our set (so the values are created for all possible). Then a value ∈{1,2,...,1296} is assigned to each unique row using unique's 3rd output. Counts are then created with histc with the bins equal to the set of values from unique's output, 1 is subtracted from each bin to remove the complete set bigrams we added
[~,~,c] = unique([fileBigrams;allBigrams],'rows');
counts = histc(c,1:1296)-1;
Display
to view counts against text
[allBigrams, counts+'0']
or for something potentially more useful...
[sortedCounts,sortInd] = sort(counts,'descend');
[allBigrams(sortInd,:), sortedCounts+'0']
ans =
or9
at8
re8
in7
ol7
te7
do6 ...
Did not look into the entire code fragment, but from the example at the top of your question, I think you are looking to make a histogram:
string1 = 'ASHRAFF'
nr = histc(string1,'A':'Z')
Will give you:
2 0 0 0 0 2 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
(Got a working solution with hist, but as #The Minion shows histc is more easy to use here.)
Note that this solution only deals with upper case letters.
You may want to do something like so if you want to put lower case letters in their correct bin:
string1 = 'ASHRAFF'
nr = histc(upper(string1),'A':'Z')
Or if you want them to be shown separately:
string1 = 'ASHRaFf'
nr = histc(upper(string1),['a':'z' 'A':'Z'])
bi_freq1 = zeros(1,all_bigrams_len);
for k=1: vocab_len1
for i=1:all_bigrams_len
if char(vocab1(k)) == char(bigrams(i))
bi_freq1(i) = frequencies1(k);
end
end
end

Matlab HDF5: Read DIMENSION_LIST attribute

I'm trying to read HDF5 files with Matlab. I created the files in Fortran, which is only relevant in that I used h5dsattach_scale_f to attached scale datasets to each dimension of my given primary dataset. Most of my logic works well, but I'm having trouble reading the attributes of my primary dataset in order to get at the attached scales.
I start by iterating through each dataset in the file. Once I know I have my primary dataset, I iterate through its attributes with this call:
[status, index_out, SD] = H5A.iterate(dset_id, 'H5_INDEX_NAME', 'H5_ITER_NATIVE', 0, #hdf5_sds_attr_iter, SD);
That calls this function for every attribute:
function [status, SD] = hdf5_sds_attr_iter(dset_id, attr_name, info, SD)
status = 0;
disp(attr_name);
if ~strcmp(attr_name, 'DIMENSION_LIST')
return;
end
attr_id = H5A.open(dset_id, attr_name, 'H5P_DEFAULT');
space = H5A.get_space (attr_id);
[~, dims, ~] = H5S.get_simple_extent_dims(space);
info2 = H5A.get_info(attr_id);
disp(info2);
rdata = H5A.read(attr_id, 'H5ML_DEFAULT');
disp(rdata);
for i = 1:dims
disp(rdata{i});
end
H5S.close(space);
H5A.close(attr_id);
end
This is the output:
DIMENSION_LIST
3
corder_valid: 1
corder: 0
cset: 0
data_size: 48
[8x1 uint8]
[8x1 uint8]
[8x1 uint8]
184
17
0
0
0
0
0
0
32
28
0
0
0
0
0
0
240
29
0
0
0
0
0
0
If I do h5dump on the dataset, this is what that attribute looks like:
ATTRIBUTE "DIMENSION_LIST" {
DATATYPE H5T_VLEN { H5T_REFERENCE { H5T_STD_REF_OBJECT }}
DATASPACE SIMPLE { ( 3 ) / ( 3 ) }
DATA {
(0): (DATASET 1400 /beamdata scale rank 1 ),
(1): (DATASET 6512 /beamdata scale rank 2 ),
(2): (DATASET 6976 /beamdata scale rank 3 )
}
}
Since those numbers (1400, 6512, 6976) do not appear elsewhere in the dump, I don't know how to use them or the output of H5A.read (rdata) to actually get at the scale data. The Matlab HDF5 documentation is rather silent on what to do with attribute data. Does anyone know how to process attribute reference data correctly?

Matlab function calling basic

I'm new to Matlab and now learning the basic grammar.
I've written the file GetBin.m:
function res = GetBin(num_bin, bin, val)
if val >= bin(num_bin - 1)
res = num_bin;
else
for i = (num_bin - 1) : 1
if val < bin(i)
res = i;
end
end
end
and I call it with:
num_bin = 5;
bin = [48.4,96.8,145.2,193.6]; % bin stands for the intermediate borders, so there are 5 bins
fea_val = GetBin(num_bin,bin,fea(1,1)) % fea is a pre-defined 280x4096 matrix
It returns error:
Error in GetBin (line 2)
if val >= bin(num_bin - 1)
Output argument "res" (and maybe others) not assigned during call to
"/Users/mac/Documents/MATLAB/GetBin.m>GetBin".
Could anybody tell me what's wrong here? Thanks.
You need to ensure that every possible path through your code assigns a value to res.
In your case, it looks like that's not the case, because you have a loop:
for i = (num_bins-1) : 1
...
end
That loop will never iterate (so it will never assign a value to res). You need to explicitly specify that it's a decrementing loop:
for i = (num_bins-1) : -1 : 1
...
end
For more info, see the documentation on the colon operator.
for i = (num_bin - 1) : -1 : 1

crystal reports conditional formatting for summary fields

I'm trying to create a decimal formatting formula on my summary fields. The values in the database could have 0, 1, or 2 decimal places. I've started with this:
If (CurrentFieldValue mod 1 = 0) Then
0
Else If (CurrentFieldValue mod .1 = 0) Then
1
Else
2
On a simple single data field, this works and displays the value with 0, 1, or 2 decimal places based on the data coming from my database. The same formula doesn't work for a summary field on my reports with group data. Any ideas?
Edit: Since I don't know how to format code in a comment, I'll address the suggestion of using a formula here:
Didn't work. Formula:
Sum ({myTable.dataValue}, {myTable.groupField})
then I used:
If ({#formula} mod 1 = 0) Then
0
Else If ({#formula} mod .1 = 0) Then
1
Else
2
And I still got whole numbers for everything. My rounding is set to .01 with no formula. Do I need a formula for rounding too? I still don't understand why this works on individual values but not for group summaries.
OK- it turns out this is due to our lack of understanding of the mod function :)
Everything mod 1 actually returns 0. This is the formula you need to use:
if {ER100_ACCT_ORDER.ER100_ORD_TOT} * 100 mod 100 = 0 then
0
else if {ER100_ACCT_ORDER.ER100_ORD_TOT} * 100 mod 10 = 0 then
1
else
2
:)
How about just creating a formula field instead of using the built-in summary field:
sum({mytable.myfield})
Then you can use your conditional formatting:
If ({#formula} mod 1 = 0) Then
0
Else If ({#formula} mod .1 = 0) Then
1
Else
2