I'm getting stuck in what's happening here. This is my understanding, so, C{1} has a column of strings. something like this:
A231
A354
A356
A234
.
.
pattern continues until the end
ids then gets a copy of that column, and idmlh becomes the second element of the cell array which is a matrix in this case. Then an empty array is created in idsCo and idx. Then it goes through all the rows in the column of ids and checks if what is in that row is found in another data structure which has similar dimensions to ids, parIDs. So this is where the first confusion comes in,
if it isnt a member then it stores the index value in idx ? And if it is a member then what happens exactly?
Im most uncertain about this part:
else
[~,~,ii] = intersect(ids{cnt}, parIDs) ;
idsCo = [idsCo ; Lbll(ii) ] ;
end
end
ids(idx) = [] ;
idmlh(idx,:) = [] ;
Below is the full code:
ids = C{1} ;
idmlh = C{2} ;
idsCo = [] ;
idx = [] ; label for
for cnt=1:length(ids)
if ~ismember(strtrim(ids{cnt}), parIDs)
idx = [idx cnt] ;
else
[~,~,ii] = intersect(ids{cnt}, parIDs) ;
idsCo = [idsCo ; Lbl(ii) ] ;
end
end
ids(idx) = [] ;
idmlh(idx,:) = [] ;
1: ids = C{1} ;
2: idmlh = C{2} ;
3: idsCo = [] ;
4: idx = [] ; label for
5: for cnt=1:length(ids)
6: if ~ismember(strtrim(ids{cnt}), parIDs)
7: idx = [idx cnt] ;
8: else
9: [~,~,ii] = intersect(ids{cnt}, parIDs) ;
10: idsCo = [idsCo ; Lbl(ii) ] ;
11: end
12: end
13: ids(idx) = [] ;
14: idmlh(idx,:) = [] ;
It is essential to know what strtrim and intersect do...
In the loop over all elements of ids it checks whether the cnsth element is part of parIDs array.
If it is not (~ismember is evaluated as true) then actual cnt value is appended to idx array.
If it is member (~ismember is evaluated as false) then it asks what elements of ids{cnt} (1x1) and parIDs (Nx1) are in both arrays. Values of the element is ignored ([~,...), index of this element in first array is also ignored (...,~,...). Index of the element in second array is assigned to ii (...,ii]=).
Then the iith element of Lbl is appended to idsCo array.
In the end the elements of ids and their counterparts in idmlh that aren't in parIDs are replaced by empty array.
Related
So I have arr = randi([0,20],20,1). I want to show: If there are numbers less than 5, fprintf('Yes\n') only once. Im using a for loop (for i = 1 : length(arr)) and indexing it.
As your description, maybe you need if statement within for loop like below
for i = 1:length(arr)
if arr(i) < 5
fprintf('Yes\n');
break
end
end
If you want to print Yes once, you can try
if any(arr < 5)
fprintf('Yes\n')
endif
If you don't want to use break, the code below might be an option
for i = 1:min(find(arr <5))
if (arr(i) < 5)
fprintf('Yes\n');
end
end
You can use a break statement upon finding the first value under 5 and printing the Yes statement.
Using a break Statement:
arr = randi([0,20],20,1);
for i = 1: length(arr)
if arr(i) < 5
fprintf("Yes\n");
break;
end
end
Extension:
By Using any() Function:
Alternatively, if you'd like to concise it down without the need for a for-loop the any() function can be used to determine if any values within the array meet a condition in this case arr < 5.
arr = randi([0,20],20,1);
if(any(arr < 5))
fprintf("Yes\n");
end
By Using a While Loop:
Check = 0;
arr = randi([0,20],20,1);
i = 1;
while (Check == 0 && i < length(arr))
if arr(i) < 5
fprintf("Yes\n");
Check = 1;
end
i = i + 1;
end
I have a Matlab function. I need to generalize this function. This code’s aim is to check this IndicMPs are in the TableTemp, if it is there, then we extract relevant age limits, such as: Age_Limite_DC, Age_Limite_IT, Age_Limite_Ch and Transfert_Prime_IT_DC. My idea is to generalize, passing parameters to find out the "Type_pret" is.(May be I'm wrong) Since I'm beginner to Matlab can someone help me to encode a more generic function that can be used in a more general context?
function Structure = optimisation_function()
Data = load('Data.mat');
Structure = Data.Structure;
TableTemp = Data.TableTemp;
Age_Limite_DC = zeros(size(Structure,1),1);
Age_Limite_IT = zeros(size(Structure,1),1);
Age_Limite_CH = zeros(size(Structure,1),1);
Transfert_Prime_IT_DC = zeros(size(Structure,1),1);
for IndexMPAL = 1 : length(Structure.AnneeSouscription)
% Determine Type_Pret (Loan Type)
if ~isempty(strfind(Structure.Type_Pret{IndexMPAL},'A'))
Type_Pret = 'A';
elseif ~isempty(strfind(Structure.Type_Pret{IndexMPAL},'B'))
Type_Pret = 'B';
elseif ~isempty(strfind(Structure.Type_Pret{IndexMPAL},'C'))
Type_Pret = 'C';
elseif ~isempty(strfind(Structure.Type_Pret{IndexMPAL},'D'))
Type_Pret = 'D';
elseif ~isempty(strfind(Structure.Type_Pret{IndexMPAL},'E'))
Type_Pret = 'E';
end
MP_CP = Structure.NomCodeProduit(IndexMPAL);
MP_AnSous = Structure.AnneeSouscription(IndexMPAL);
MP_TypePret = Type_Pret;
IndicCP = strcmp(MP_CP, TableTemp.CodeProduit);
IndicAS = MP_AnSous== TableTemp.AnneeSouscription;
IndicTP = strcmp(MP_TypePret, TableTemp.TypePret);
IndicMP = IndicCP & IndicAS & IndicTP;
if ~any(IndicMP)
Msg = strcat('CodeProduct:',MP_CP{1}, ', Année Souscription:', num2str(MP_AnSous), ', Type Prêt:', MP_TypePret);
error('Error', Msg)
else
Age_Limite_DC(IndexMPAL,1) = TableTemp.Age_Limite_DC(IndicMP,1);
Age_Limite_IT(IndexMPAL,1) = TableTemp.Age_Limite_IT(IndicMP,1);
Age_Limite_CH(IndexMPAL,1) = TableTemp.Age_Limite_CH(IndicMP,1);
Transfert_Prime_IT_DC(IndexMPAL,1)=
TableTemp.Transfert_Prime_IT_DC(IndicMP,1);
end
end
Structure.Age_Limite_DC = Age_Limite_DC;
Structure.Age_Limite_IT = Age_Limite_IT;
Structure.Age_Limite_CH = Age_Limite_CH;
Structure.Transfert_Prime_IT_DC = Transfert_Prime_IT_DC;
end
The if/elseif can be simplified with a cell array:
liststr = {'A','BB','C','D','E'}; % builds a cell array, each cell contains a string
Positive_matches = strfind(liststr,Structure.Type_Pret{IndexMPAL}) % returns a list for each cell of the indices where the element was found (empty if none)
Index = find(~cellfun('isempty', Positive_matches )) % converts the previous list into a logical 0/1 array indicating whether an item was found (1) or not (0)
% if isempty(Index); continue; end % If no index is found, to avoid an error in the next instruction, skips the rest of the code.
Type_Pret = liststr(Index(1));
If the Type_Pret are the same in your list and in Structure? , you can usestrcmp`:
liststr = {'A','B','C','D','E'};
if any(strcmp(Structure.Type_Pret,liststr))
Type_Pret = Structure.Type_Pret
else
% handle error
end
You can also work directly on Structure.Age_Limite_DC without using Age_Limite_DC.
I have a question about removing duplicates in a table (rexx language), I am on netphantom applications that are using the rexx language.
I need a sample on how to remove the duplicates in a table.
I do have a thoughts on how to do it though, like using two loops for these two tables which are A and B, but I am not familiar with this.
My situation is:
rc = PanlistInsertData('A',0,SAMPLE)
TABLE A (this table having duplicate data)
123
1
1234
12
123
1234
I need to filter out those duplicates data into TABLE B like this:
123
1234
1
12
You can use lookup stem variables to test if you have already found a value.
This should work (note I have not tested so there could be syntax errors)
no=0;
yes=1
lookup. = no /* initialize the stem to no, not strictly needed */
j=0
do i = 1 to in.0
v = in.i
if lookup.v <> yes then do
j = j + 1
out.j = v
lookup.v = yes
end
end
out.0 = j
You can eliminate the duplicates by :
If InStem first element, Move the element to OutStem Else check all the OutStem elements for the current InStem element
If element is found, Iterate to the next InStem element Else add InStem element to OutStem
Code Snippet :
/*Input Stem - InStem.
Output Stem - OutStem.
Array Counters - I, J, K */
J = 1
DO I = 1 TO InStem.0
IF I = 1 THEN
OutStem.I = InStem.I
ELSE
DO K = 1 TO J
IF (InStem.I ?= OutStem.K) & (K = J) THEN
DO
J = J + 1
OutStem.J = InStem.I
END
ELSE
DO
IF (InStem.I == OutStem.K) THEN
ITERATE I
END
END
END
OutStem.0 = J
Hope this helps.
Using an ActiveX server from MATLAB, I am trying to highlight many cells in an Excel sheet at once. These are not in specific columns or rows so I use Range('A1,B2,...') to access them. However the string accepted by the Range object has to be less than 255 characters or an error:
Error: Object returned error code: 0x800A03EC
is thrown. The following code reproduces this error with an empty Excel file.
hActX = actxserver('Excel.Application');
hWB = hActX.Workbooks.Open('C:\Book1.xlsx');
hSheet = hWB.Worksheets.Item('Sheet1');
col = repmat('A', 100, 1);
row = num2str((1:100)'); %'
cellInd = strcat(col, strtrim(cellstr(row)));
str1 = strjoin(cellInd(1:66), ','); %// 254 characters
str2 = strjoin(cellInd(1:67), ','); %// 258 characters
hSheet.Range(str1).Interior.Color = 255; %// Works
hSheet.Range(str2).Interior.Color = 255; %// Error 0x800A03EC
hWB.Save;
hWB.Close(false);
hActX.Quit;
How can I get around this? I found no other relevant method of calling Range, or of otherwise getting the cells I want to modify.
If you start with a String, you can test its length to determine if Range() can handle it. Here is an example of building a diagonal range:
Sub DiagonalRange()
Dim BigString As String, BigRange As Range
Dim i As Long, HowMany As Long, Ln As String
HowMany = 100
For i = 1 To HowMany
BigString = BigString & "," & Cells(i, i).Address(0, 0)
Next i
BigString = Mid(BigString, 2)
Ln = Len(BigString)
MsgBox Ln
If Ln < 250 Then
Set BigRange = Range(BigString)
Else
Set BigRange = Nothing
arr = Split(BigString, ",")
For Each a In arr
If BigRange Is Nothing Then
Set BigRange = Range(a)
Else
Set BigRange = Union(BigRange, Range(a))
End If
Next a
End If
BigRange.Select
End Sub
For i = 10, the code will the the direct method, but if the code were i=100, the array method would be used.
The solution, as Rory pointed out, is to use the Union method. To minimize the number of calls from MATLAB to the ActiveX server, this is what I did:
str = strjoin(cellInd, ',');
isep = find(str == ',');
isplit = diff(mod(isep, 250)) < 0;
isplit = [isep(isplit) (length(str) + 1)];
hRange = hSheet.Range(str(1:(isplit(1) - 1)));
for ii = 2:numel(isplit)
hRange = hActX.Union(hRange, ...
hSheet.Range(str((isplit(ii-1) + 1):(isplit(ii) - 1))));
end
I used 250 in the mod to account for the cell names being up to 6 characters long, which is sufficient for me.
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