i have [sentences*words] matrix in which rows are labeled as sentences and columns with words the code i used for that is:
Out = NaN(numel(sentences), numel(out_words));
for i = 1:numel(out_words)
Out(:,i) = cellfun(#(x) numel(strfind(x, out_words{i})), sentences);
end
display(Out)
above code returna a logical matrix a sample example below illustrates the idea:
1 0 1
1 1 0
0 1 1
1 0 1
in above rows are sentences and columns are words, if a word is present in a sentence 1 is written else 0 is written.
Now what i want to do is to compare the rows and save all the locations that have 1 in commmon for example in above row 1 should me compared with all the other remaining rows and row2 with all the remaining till the nth row this operation should be carried out which should save the result in 1D array as follows:
for example:
output=
sentence {1,2} contain red
sentence {1,4} contain red,say
sentence {2,3} contain but
sentence {1,3} contain say
and so on up till n elements
sentence{1,2} 1 is refered to sentence1 and 2 is sentence2 and so on till nth sentence i want to compare the rows and pick the locations on which two words have 1(true) value.
if some one can give a better idea of implementing equality relation for matrices please suggest me, thank you
You can use bsxfun to compare the sentences. Let M be the logical matrix of size #sentences-by-#words, then
cmp = bsxfun( #eq, permute(M,[1 3 2]), permute(M,[3 1 2]) )
Now you have a logical array cmp of size #sentences-by-#sentences-by-#words where the vector v_ij = cmp( ii, jj, : ) has v_ij(k) = true iff sentence ii and sentence jj has the word k in them.
Related
I would like to remove some columns and rows from a big matrix. Those are the columns and the rows which have all zeros values. Is there any function in MATLAB that can do it for you quite fast? My matrices are sparse. I am doing this way:
% To remove all zero columns from A
ind = find(sum(A,1)==0) ;
A(:,ind) = [] ;
% To remove all zeros rows from A
ind = find(sum(A,2)==0) ;
A(ind,:) = [] ;
It would be nice to have a line of code for this as I may do this kind of task repeatedly. Thanks
A single line of code would be:
A=A(any(A,2),any(A,1))
There is no need to use find like you did, you can directly index using logical vectors. The any function finds the rows or columns with any non-zero elements.
1 Dimension:
I'll first show a simpler example based on another duplicate question, asking to to remove only the rows containing zeros elements.
Given the matrix A=[1,2;0,0];
To remove the rows of 0, you can:
sum the absolute value of each rows (to avoid having a zero sum from a mix of negative and positive numbers), which gives you a column vector of the row sums.
keep the index of each line where the sum is non-zero.
in code:
A=[1,2;0,0];
% sum each row of the matrix, then find rows with non-zero sum
idx_nonzerolines = sum(abs(A),2)>0 ;
% Create matrix B containing only the non-zero lines of A
B = A(idx_nonzerolines,:) ;
will output:
>> idx_nonzerolines = sum(abs(A),2)>0
idx_nonzerolines =
1
0
>> B = A(idx_nonzerolines,:)
B =
1 2
2 Dimensions:
The same method can be used for 2 dimensions:
A=[ 1,2,0,4;
0,0,0,0;
1,3,0,5];
idx2keep_columns = sum(abs(A),1)>0 ;
idx2keep_rows = sum(abs(A),2)>0 ;
B = A(idx2keep_rows,idx2keep_columns) ;
outputs:
>> B = A(idx2keep_rows,idx2keep_columns)
B =
1 2 4
1 3 5
Thanks to #Adriaan in comments for spotting the edge case ;)
i have [sentences*words] matrix as shown below
out = 0 1 1 0 1
1 1 0 0 1
1 0 1 1 0
0 0 0 1 0
i want to process this matrix in a way that should tell W1 & W2 in "sentence number 2" and "sentence number 4" occurs with same value i.e 1 1 and 0 0.the output should be as follows:
output{1,2}= 2 4
output{1,2} tells word number 1 and 2 occurs in sentence number 2 and 4 with same values.
after comparing W1 & W2 next candidate should be W1 & W3 which occurs with same value in sentence 3 & sentence 4
output{1,3}= 3 4
and so on till every nth word is compared with every other words and saved.
This would be one vectorized approach -
%// Get number of columns in input array for later usage
N = size(out,2);
%// Get indices for pairwise combinations between columns of input array
[idx2,idx1] = find(bsxfun(#gt,[1:N]',[1:N])); %//'
%// Get indices for matches between out1 and out2. The row indices would
%// represent the occurance values for the final output and columns for the
%// indices of the final output.
[R,C] = find(out(:,idx1) == out(:,idx2))
%// Form cells off each unique C (these will be final output values)
output_vals = accumarray(C(:),R(:),[],#(x) {x})
%// Setup output cell array
output = cell(N,N)
%// Indices for places in output cell array where occurance values are to be put
all_idx = sub2ind(size(output),idx1,idx2)
%// Finally store the output values at appropriate indices
output(all_idx(1:max(C))) = output_vals
You can get a logical matrix of size #words-by-#words-by-#sentences easily using bsxfun:
coc = bsxfun( #eq, permute( out, [3 2 1]), permute( out, [2 3 1] ) );
this logical array is occ( wi, wj, si ) is true iff word wi and word wj occur in sentence si with the same value.
To get the output cell array from coc you need
nw = size( out, 2 ); %// number of words
output = cell(nw,nw);
for wi = 1:(nw-1)
for wj = (wi+1):nw
output{wi,wj} = find( coc(wi,wj,:) );
output{wj,wi} = output{wi,wj}; %// you can force it to be symmetric if you want
end
end
I am new to matlab and I was wondering what it meant to use logical indexing/masking to extract data from a matrix.
I am trying to write a function that accepts a matrix and a user-inputted value to compute and display the total number of values in column 2 of the matrix that match with the user input.
The function itself should have no return value and will be called on later in another loop.
But besides all that hubbub, someone suggested that I use logical indexing/masking in this situation but never told me exactly what it was or how I could use it in my particular situation.
EDIT: since you updated the question, I am updating this answer a little.
Logical indexing is explained really well in this and this. In general, I doubt, if I can do a better job, given available time. However, I would try to connect your problem and logical indexing.
Lets declare an array A which has 2 columns. First column is index (as 1,2,3,...) and second column is its corresponding value, a random number.
A(:,1)=1:10;
A(:,2)=randi(5,[10 1]); //declares a 10x1 array and puts it into second column of A
userInputtedValue=3; //self-explanatory
You want to check what values in second column of A are equal to 3. Imagine as if you are making a query and MATLAB is giving you binary response, YES (1) or NO (0).
q=A(:,2)==3 //the query, what values in second column of A equal 3?
Now, for the indices where answer is YES, you want to extract the numbers in the first column of A. Then do some processing.
values=A(q,2); //only those elements will be extracted: 1. which lie in the
//second column of A AND where q takes value 1.
Now, if you want to count total number of values, just do:
numValues=length(values);
I hope now logical indexing is clear to you. However, do read the Mathworks posts which I have mentioned earlier.
I over simplified the code, and wrote more code than required in order to explain things. It can be achieved in a single-liner:
sum(mat(:,2)==userInputtedValue)
I'll give you an example that may illustrate what logical indexing is about:
array = [1 2 3 0 4 2];
array > 2
ans: [0 0 1 0 1 0]
using logical indexing you could filter elements that fullfil a certain condition
array(array>2) will give: [3 4]
you could also perform alterations to only those elements:
array(array>2) = 100;
array(array<=2) = 0;
will result in "array" equal to
[0 0 100 0 100 0]
Logical indexing means to have a logical / Boolean matrix that is the same size as the matrix that you are considering. You would use this as input into the matrix you're considering, and any locations that are true would be part of the output. Any locations that are false are not part of the output. To perform logical indexing, you would need to use logical / Boolean operators or conditions to facilitate the selection of elements in your matrix.
Let's concentrate on vectors as it's the easiest to deal with. Let's say we had the following vector:
>> A = 1:9
A =
1 2 3 4 5 6 7 8 9
Let's say I wanted to retrieve all values that are 5 or more. The logical condition for this would be A >= 5. We want to retrieve all values in A that are greater than or equal to 5. Therefore, if we did A >= 5, we get a logical vector which tells us which values in A satisfy the above condition:
>> A >= 5
ans =
0 0 0 0 1 1 1 1 1
This certainly tells us where in A the condition is satisfied. The last step would be to use this as input into A:
>> B = A(A >= 5)
B =
5 6 7 8 9
Cool! As you can see, there isn't a need for a for loop to help us select out elements that satisfy a condition. Let's go a step further. What if I want to find all even values of A? This would mean that if we divide by 2, the remainder would be zero, or mod(A,2) == 0. Let's extract out those elements:
>> C = A(mod(A,2) == 0)
C =
2 4 6 8
Nice! So let's go back to your question. Given your matrix A, let's extract out column 2.
>> col = A(:,2)
Now, we want to check to see if any of column #2 is equal to a certain value. Well we can generate a logical indexing array for that. Let's try with the value of 3:
>> ind = col == 3;
Now you'll have a logical vector that tells you which locations are equal to 3. If you want to determine how many are equal to 3, you just have to sum up the values:
>> s = sum(ind);
That's it! s contains how many values were equal to 3. Now, if you wanted to write a function that only displayed how many values were equal to some user defined input and displayed this event, you can do something like this:
function checkVal(A, val)
disp(sum(A(:,2) == val));
end
Quite simply, we extract the second column of A and see how many values are equal to val. This produces a logical array, and we simply sum up how many 1s there are. This would give you the total number of elements that are equal to val.
Troy Haskin pointed you to a very nice link that talks about logical indexing in more detail: http://www.mathworks.com/help/matlab/math/matrix-indexing.html?refresh=true#bq7eg38. Read that for more details on how to master logical indexing.
Good luck!
%% M is your Matrix
M = randi(10,4)
%% Val is the value that you are seeking to find
Val = 6
%% Col is the value of the matrix column that you wish to find it in
Col = 2
%% r is a vector that has zeros in all positions except when the Matrix value equals the user input it equals 1
r = M(:,Col)==Val
%% We can now sum all the non-zero values in r to get the number of matches
n = sum(r)
M =
4 2 2 5
3 6 7 1
4 4 1 6
5 8 7 8
Val =
6
Col =
2
r =
0
1
0
0
n =
1
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)
I have 2 martices of the same size. The first contains values and the second only elements of 0 and 1 (like boolean). I now want all elements of my first Matrix stored in an array where the second Matrix has a 1 at the same index.
Maybe an example makes that clear:
Matrix 1:
a b c
d e f
g h i
Matrix 2:
0 1 1
1 0 0
0 0 1
output:
[b c d i]
I think this will work in two steps, but i cant get it to work.
This will need two steps indeed.
%# transpose Matrix 1 because Matlab iterates by row first
matrix_1 = matrix_1';
%# read values (transpose M2 as well)
%# also transpose the result to get a row-vector
output = matrix_1(matrix_2')';
Note that this indexing operation only works if matrix_2 is logical. If it isn't, cast it by writing logical(matrix_2) instead.
If your arrays are a and b, with b the mask array, try
a(find(b))
This won't produce the output in the order in your question. If order is important resort to #Jonas' approach.