i have a matrix B of N*3 dim. I want to find the indices of B whose column 3 has value 1.
I used the command [~,id]=ismember(1,B(:,3)). id returns only value 1 even though there are many rows in the matrix which has the column 3 with value 1. Can any one point out what is wrong in the command?
Rather do:
id = find(B(:,3)==1)
but as an aside, to use ismember you should swap your input [~,id]=ismember(B(:,3),1).
Related
I know mapping 2D array into 1D array has been asked many times, but I did not find a solution that would fit a where the column count varies.
So I want get a 1-dimensional index from this 2-dimensional array
Col> _0____1____2__
Row 0 |_0__|_1__|_2__|
V 1 |_3__|_4__|
2 |_5__|_6__|_7__|
3 |_8__|_9__|
4 |_10_|_11_|_12_|
5 |_13_|_14_|
The normal formula index = row * columns + column does not work, since after the 2nd row the index is out of place.
What is the correct formula here?
EDIT:
The specific issue is that I have a list of items in with the layout like in the grid, but a one dimensional array for the data. So while looping through the elements in the UI, I need to get the correct data, but can only get the row and column for that element. I need to find a way to turn a row/column value into an index for the data-array
Bad picture trying to explain it
A truly optimal answer (or even a provably correct one) will depend on the language you are using and how it lays out memory for such arrays.
However, taking your question simply at face value, you have to know what the actual length of each row is in order to calculate a 1D index.
So either the row length follows some pattern that can be inferred from the data, or you have (or can write) a rlen = rowLength( 2dTable, RowNumber) function.
Then, depending on how big the tables are and how fast you need to run, you can calculate a 1D index from the 2d table by adding all the previous row lengths until the current row length is less than the 2d column index.
or build a 1d table of the row lengths (or commulative rowlengths) so you can scan it and so only call your rowlength function for each row only once.
With a better description of your problem, you might get a better answer...
For your example which alternates between 3 and 2 columns you can construct a formula:
index = (row / 2) * (3 + 2) + (row % 2 ? 3 : 0) + column
(C-like syntax, assuming integer division)
In general though, the one and only way to implement what you're doing here, jagged arrays, is to make an array of arrays, a.k.a. an Iliffe vector. That means, use the row number as index into an array of pointers which point to the individual row arrays containing the actual data.
You can have an additional 1D array having the length of the columns say "length". Then your formula is index=sum {length(i)}+column. i runs from 0 to row.
I have a matrix with six columns. I found the max value of a certain column but how would I go about extracting the entire row pertaining to that value?
To extract row 1 of matrix A use A([1],:) to extract row 1 and 2 use A([1,2],:)
Use the max() function as explained here. For example
if A is your matrix
[M, I] = max(A)
Row = A([I(1)],:)
where I(1) is used to find the row containing the max element of the first coloumn
We have a project on a certain math subject and I am done with the computations and it works just fine. So the task is, let's say you have a system of linear equations of certain number of unknowns, you input the number of unknowns, and fill in the values, and using matrix computations, find all the value of unknowns.
To make this short, I already finished the "find the value of unknowns" along with the computation, I checked it, and it seems fine. I can put 6 as the number of unknowns and it automatically computes the inverse of a 6x6 matrix and it will return the 6 unknown values using Index INDIVIDUALLY.
(Note: We aren't allowed to use VBA or Macros since we haven't discussed that yet.
The problem is, I don't know how to automatically fill a RANGE of cells with a VALUE or A FORMULA based on a SINGLE cell value.
For example, In cell A1, I will input 5 (which indicates the number of unknowns), then upon inputting this and hitting enter, let's say a range of cells A2 to A6 (which is 5 cells) will be automatically filled with incremented letters, like for A2 -> A ; A3 -> B ; ... A6 -> E, of which these letters indicate the 5 unknowns.
PROBLEM 2.
Another follow up question, let's say I input again 5, which again stands for the number of missing values/unknowns, in A1, besides the column of the variables A,B,C,D,E (5 unknowns), I want to automatically fill column B respectively with values from an array.
This is just the same with my first problem but this time, instead of incremented letters, it would be Incremented Index function.
For example: I input 5
*Column A will automatically be filled with the variables/letters
*Column B will automatically be filled with the values from an array that's computed using a formula but is not shown independently on cells.
I already have the formula
INDEX(Formula I created, Row number of the answer from the Formula I created , Column number of the answer from the formula I created)
The answers from the formula I made myself is also an array, an "n" rows and 1 column array. If I put the Index formula on a SINGLE cell, it returns specified row number value from the array that resulted in the computation from my formula
What I want is for example, for 5 unknowns
**A | B**
1|.......5..........................
2|.......A..............Some Value 1
3|.......B..............Some Value 2
4|.......C..............Some Value 3
5|.......D..............Some Value 4
6|.......E..............Some Value 5
Wherein the "Some Value" is the Arrayed Answer from my formula and the "1,2,3,4,5" specifies the row number from that arrayed answer.
This is upon inputting the matrix values, inputting the number of unknowns "n" in A1, and automatically filling a range of cells A2 to A"n" with letters A up to what letter "n" corresponds, and automatically filling a range of Cells B2 to B"n" with my formula but with incremented row number for every row in the Index(Formula, Row number , Column number) function.
Note: I hope there's a way to do this using excel functions only since we haven't discussed VBA or Macros yet so we can't use those, and even If we can, I have no knowledge for that. haha. :D
THANK YOU THANK YOU THANK YOU SO MUCH IN ADVANCED! Cheers. :D
Here's a formula for column A:A (write this in cell A2) and drag down:
=IF(ROW()-1<=$A$1,CHAR(ROW()+63),"")
I have three column vectors, then I want to creat matrix from this column victors
A1(:);
A2(:);
A3(:)
each column vectors has 25 element then the new matrix C will be a matrix with 3x25
I want to make A1(:) the first column of matrix c
A2(:) second column
A3(:) third column
Use cat to concatenate along dimension 1 or 2 depending on how you input those three vectors.
Thus, you can use -
C = cat(2,A1(:),A2(:),A3(:)).'
Or
C = cat(1,A1(:).',A2(:).',A3(:).')
Of course, you can skip (:)'s, if you know that all those are column vectors.
The above two approaches assumes that you intend to get an output of size 3 x N, where is N is the number of elements in the column vectors. If you were looking to get an output of size N x 3 , i.e. where each column is formed from the elements of column vectors A1, A2 and so on, just drop the transpose from the first of the two approaches mentioned above. Thus, use this -
C = cat(2,A1(:),A2(:),A3(:))
I have a 60,000-by-2 array. The first column is data 1 and second column is data 2; both of equal length. I'm not sure how to properly write the syntax to compare data 1 to data 2, and if data 1 is larger than data 2 then write that to the third column. Or vice versa if data 2 is larger than data 1. I have begun constructing a for loop, but I'm having syntax issues comparing the columns.
No loops are needed. If you simply want to create a vector containing the largest element in each row of your 60,000-by-2 matrix you can use the max function:
A = rand(6e4,2); % Random demo data
B = max(A,[],2);
Or if you then want to put the result directly in a third column of A:
A(:,3) = max(A,[],2);
Read the documentation for max. You'll see that the 2 in the third argument applied the max function across each row of the input, A.