Column A Column B Column C
Row1 1,3,4 4 0
Row2 2,5,6 6 0
Row3 1,2,3 3 1
So, what I want to do is find the easiest way to check when column A, which is supposed to have the values of 1,2,3, and 4, by using column B, and return a 1 if Column A has all of the numbers into a new column C.
So for row 1, and row 2, Column C would be 0, because neither of them are storing all of the values that they are supposed to have. (Row 1 was supposed to have1,2,3,4, and row2 was supposed to have 1,2,3,4,5,6). Row 3 column C would have a 1, because it has all the values it is supposed to have. I can't just count the number of values either, because sometimes there are repeated values in column A.
I am trying to code this in a way that isn't too long, because I have to do this up to Column B = 100.
Thank you all!
Coming from C/C++/Java type languages, the simple approach is for each row look at each value - something like this (not in any particular language)
for (int i = 0; i < NUM_ROWS; i++)
{
int bV = columns['B'][i]
list<int> aV = columns['A'][i]
if (aV.length != bV) {
// If A doesn't have B items in it then
// they can't possibly match.
columns['C'][i] = 0;
}
else {
// have to inspect each element
int ok = 1
for(int j = 1; j <= bV; j++) {
if (aV[j] != j) {
ok = 0
break
}
}
columns['C'][i] = ok
}
}
If your language does list compares and the max allowed in column B isn't too high then you could predefine a list for each value of B and then just compare all the values of A
if (columns['A'][i] == expectedLists[colB]) {
columns['C'][i]= 1
}
else {
columns['C'][i]= 0
}
You could also do something similar with string representations of the list
Related
I want to calculate the mean value of 6 rows each of a matrix 600*9 dimension. The new matrix should be of size 100*9. Can someone help me ?
I saw this code for calculating mean of 2 rows each and it is working fine foe 2 rows :-
x = rand(1028, 18);
result1 = zeros(1028/2, 18);
for ii = 1:1028/2;
result1(ii,:) = mean(x((2*ii-1):(2*ii),:));
end;
The output matrix will have the mean of first 6 rows ( of the input matrix) as the first row and next 6 rows as the 2nd row and so on
Let the data be defined as
x = rand(600, 9); % example data
N = 6; % group size
The desired result can be obtained very easily without loops:
Reshape the data matrix as a 3-D array, where the size along the first dimension is the desired group size;
Compute the mean along the first dimension;
Remove the first dimension, which is now a singleton.
result = reshape(mean(reshape(x, N, [], size(x,2)), 1), [], size(x,2));
Note how this single line contains the three steps described above:
reshape(x, N, [], size(x,2)) % step 1
mean( , 1) % step 2
result = reshape( , [], size(x,2)); % step 3
Using mean and a for loop:
data = rand(600,9);
num_groups = 6;
group_size = size(data,1)/num_groups;
mean_by_group = NaN(num_groups, size(data,2));
for k = 1:num_groups
mean_by_group(k,:) = mean(data((k-1)*group_size+(1:group_size),:),1)
end
Considering having this table in matlab:
t = table([1; 0; 3; 1], [0; 1; 0; 4], 'VariableNames', {'A', 'B'});
A B
_ _
1 0
0 1
3 0
1 4
I want to append a new column Cwith specific value that is based on a condition. Currently I use this loop:
for i=1:height(t)
if t(i, 'A').Variables == 1
t.C(i, 1) = 4;
elseif t(i, 'A').Variables == 3
t.C(i, 1) = 5;
end
end
However, this is a time comsuming operation is case of a table size greater then 100k rows.
What would be the best solution for this?
You can use the following:
[t{t.A==1, 'C'}] = 4;
[t{t.A==3, 'C'}] = 5;
This uses the facts that
Table contents can be indexed via {}, as in cell arrays;
Table columns can be indexed by their name.
Or, as noted by #SardarUsama, you can use the simpler
t.C(t.A==1)=4;
t.C(t.A==3)=5;
This uses dot notation to index the column. The result is a numerical column vector, to which a scalar can be directly assigned.
I have a matrix A of dimension m-by-n composed of zeros and ones, and a matrix J of dimension m-by-1 reporting some integers from [1,...,n].
I want to construct a matrix B of dimension m-by-n such that for i = 1,...,m
B(i,j) = A(i,j) for j=1,...,n-1
B(i,n) = abs(A(i,n)-1)
If sum(B(i,:)) is odd then B(i,J(i)) = abs(B(i,J(i))-1)
This code does what I want:
m = 4;
n = 5;
A = [1 1 1 1 1; ...
0 0 1 0 0; ...
1 0 1 0 1; ...
0 1 0 0 1];
J = [1;2;1;4];
B = zeros(m,n);
for i = 1:m
B(i,n) = abs(A(i,n)-1);
for j = 1:n-1
B(i,j) = A(i,j);
end
if mod(sum(B(i,:)),2)~=0
B(i,J(i)) = abs(B(i,J(i))-1);
end
end
Can you suggest more efficient algorithms, that do not use the nested loop?
No for loops are required for your question. It just needs an effective use of the colon operator and logical-indexing as follows:
% First initialize B to all zeros
B = zeros(size(A));
% Assign all but last columns of A to B
B(:, 1:end-1) = A(:, 1:end-1);
% Assign the last column of B based on the last column of A
B(:, end) = abs(A(:, end) - 1);
% Set all cells to required value
% Original code which does not work: B(oddRow, J(oddRow)) = abs(B(oddRow, J(oddRow)) - 1);
% Correct code:
% Find all rows in B with an odd sum
oddRow = find(mod(sum(B, 2), 2) ~= 0);
for ii = 1:numel(oddRow)
B(oddRow(ii), J(oddRow(ii))) = abs(B(oddRow(ii), J(oddRow(ii))) - 1);
end
I guess for the last part it is best to use a for loop.
Edit: See the neat trick by EBH to do the last part without a for loop
Just to add to #ammportal good answer, also the last part can be done without a loop with the use of linear indices. For that, sub2ind is useful. So adopting the last part of the previous answer, this can be done:
% Find all rows in B with an odd sum
oddRow = find(mod(sum(B, 2), 2) ~= 0);
% convert the locations to linear indices
ind = sub2ind(size(B),oddRow,J(oddRow));
B(ind) = abs(B(ind)- 1);
I have 2 cell arrays which are "celldata" and "data" . Both of them store strings inside. Now I would like to check each element in "celldata" whether in "data" or not? For example, celldata = {'AB'; 'BE'; 'BC'} and data={'ABCD' 'BCDE' 'ACBE' 'ADEBC '}. I would like the expected output will be s=3 and v= 1 for AB, s=2 and v=2 for BE, s=2 and v=2 for BC, because I just need to count the sequence of the string in 'celldata'
The code I wrote is shown below. Any help would be certainly appreciated.
My code:
s=0; support counter
v=0; violate counter
SV=[]; % array to store the support
VV=[]; % array to store the violate
pairs = ['AB'; 'BE'; 'BC']
%celldata = cellstr(pairs)
celldata = {'AB'; 'BE'; 'BC'}
data={'ABCD' 'BCDE' 'ACBE' 'ADEBC '} % 3 AB, 2 BE, 2 BC
for jj=1:length(data)
for kk=1:length(celldata)
res = regexp( data(jj),celldata(kk) )
m = cell2mat(res);
e=isempty(m) % check res array is empty or not
if e == 0
s = s + 1;
SV(jj)=s;
v=v;
else
s=s;
v= v+1;
VV(jj)=v;
end
end
end
If I am understanding your variables correctly, s is the number of cells which the substring AB, AE and, BC does not appear and v is the number of times it does. If this is accurate then
v = cellfun(#(x) length(cell2mat(strfind(data, x))), celldata);
s = numel(data) - v;
gives
v = [1;1;3];
s = [3;3;1];
Hi guys I want to solve sodoku puzzles in matlab. My problem is that I should find same value in every row and every column and every 3*3 sub array.
Our 2d array is 9*9 and populated with value 1-9 randomly.
I wrote this for finding same value in rows, but I don't know how I should do it for columns and 3*3 sub arrays.
conflict_row = 0;
for i=1:9
temp = 0;
for j=1:9
if (temp==A(i,j))
conflict_row = conflict_row+1;
end
temp = A(i,j);
end
end
Sorry I'm a newbie.
Find values that are present in all columns:
v = find(all(any(bsxfun(#eq, A, permute(1:size(A,1), [3 1 2])),1),2));
Find values that are present in all rows:
v = find(all(any(bsxfun(#eq, A, permute(1:size(A,2), [3 1 2])),2),1));
Find values that are present in all 3x3 blocks: reshape the matrix as in this answer by A. Donda to transform each block into a 3D-slice; then reshape each block into a column; and apply 1:
m = 3; %// columns per block
n = 3; %// rows per block
B = permute(reshape(permute(reshape(A, size(A, 1), n, []), [2 1 3]), n, m, []), [2 1 3]);
B = reshape(B,m*n,[]);
v = find(all(any(bsxfun(#eq, B, permute(1:size(B,1), [3 1 2])),1),2));
Probably not the fastest solution but why don't you make a function of it and use it once for rows and once for columns
[conflict_row ] = get_conflict(A)
for i=1:9
temp = 0;
for j=1:9
if (temp==A(i,j))
conflict_row = conflict_row+1;
end
temp = A(i,j);
end
end
And then you call it twice
conflict_row = get_conflict(A); % Rows
Transpose A to get the columns
Convert the columns to rows and use the same code as before
conflict_col = get_conflict(A.');
If you want to work within the same column then you should do something like this (also sorry this is in C# I don't know what language you are working in):
int currentCol = 0;
foreach (var item in myMultiArray)
{
int currentColValue = item[currentCol];
}
This works because myArray is a array of arrays thus to select a specific column can easily be picked out by just allowing the foreach to perform your row iteration, and you just have to select the column you need with the currentCol value.