Excel LINEST function returns an array of outputs (statistic) such as se1, se2, r2, ssresid, etc. I am interested in getting only the ssresid output in one cell. My idea is following:
=CHOOSE(10, LINEST(y,x,,TRUE))
however this does not work. Internet search does not help, but I believe the problem is with my manipulation of the LINEST output array. Could someone please explain what I am doing wrong in the equation, and possibly explain how it works with arrays in this kind of setting (array obtained by function)? Thank you.
Linest returns a matrix. The matrix is n wide x 5 high where n is the number of dependent variables. The format of the output is:
https://support.content.office.net/en-us/media/e0d97b28-95d9-4cb2-888c-78db54378381.gif
Ssresid is in the fifth row, second column. To isolate it, do some matrix math: Eg if n=3 enter the following matrix in cells X21:Z25 or (anywhere else you please.)
X Y Z
21 0 0 0
22 0 0 0
23 0 0 0
24 0 0 0
25 0 1 0
Then enter
{= sum(LINEST(y,x,,TRUE) * x21:z25) }
When you enter the formula be sure to press Ctrl shift enter. This tells excel to evaluate the formula as a matrix formula.
Related
This question is related to this one
I have (3,0), (2,1), and (2,0) in rows 1 to 3. There are ways to generate the array of {0,0,0,1,1,0,0} using cell addresses. (See the reference above.)
Now, my question is can the same array be generated NOT using the cell references but using the three pairs of the numbers only?
NOTE: In the real case, there may be up to six pairs of numbers, A1:B1 to A6:B6, and up to 2880 array elements.
Excel 2010
A B
1 3 0
2 2 1
3 2 0
You can proceed as in the edit to my answer to the original question, by generating an array as follows:
0 0 0
1 0 0
1 1 0
and using Mmult instead of Offset to get the running totals of the first column of numbers.
=SUM(INDEX(INDEX({3,0;2,1;2,0},0,2),N(IF({1},MATCH(ROW(A1:INDEX(A:A,SUM(INDEX({3,0;2,1;2,0},0,1)))),
MMULT(IF(ROW(A1:INDEX($1:$1048576,COUNT(INDEX({3,0;2,1;2,0},0,1)),COUNT(INDEX({3,0;2,1;2,0},0,1))))>
COLUMN(A1:INDEX($1:$1048576,COUNT(INDEX({3,0;2,1;2,0},0,1)),COUNT(INDEX({3,0;2,1;2,0},0,1)))),1,0),INDEX({3,0;2,1;2,0},0,1))+1)))))
If you hard-code the number of pairs of numbers instead of using Countif to count them, the formula is shorter:
=SUM(INDEX(INDEX({3,0;2,1;2,0},0,2),N(IF({1},MATCH(ROW(A1:INDEX(A:A,SUM(INDEX({3,0;2,1;2,0},0,1)))),
MMULT(IF(ROW(A1:INDEX($1:$1048576,3,3))>
COLUMN(A1:INDEX($1:$1048576,3,3)),1,0),INDEX({3,0;2,1;2,0},0,1))+1)))))
Let's say I have a vector A = [-1,2];
Each element in A is described by the actual number and sign. So each element has a 2 dimensional feature-set.
I would like to generate a matrix, in this case 2x2 where the columns correspond to the element, and rows correspond to the presence of a feature. The presence of a feature is described by 1's and 0's. So, if an element is positive, it is 1, if the element is the number 1, then the result is 1 as well. In the case above I would get:
Element 1 Element 2
Is this a 1? 1 0
Is this a positive number? 0 1
What is the smartest way to go about accomplishing this? Obviously if statements would work, but I feel that there should be a faster, much smarter way of going about this. I am coding this in matlab by the way, and I would appreciate any help.
#Benoit_11's solution is a fine one. Here's a similar but maybe simpler solution. You could try both and see which is faster if you care about speed.
features = [abs(A) == 1; A > 0];
this assumes A is a row vector in order to get the output in the format you specified.
Simple way using ismember for the first condition and logical operation for the 2nd condition. ismember outputs a logical array which you can plug into the output you need (here called DescribeA; and likewise when you check for values greater than 0 using the > operator.
%// Test array
A = [-1,2,1,-10,5,-3,1]
%// Initialize output
DescribeA = zeros(2,numel(A));
%// 1st condition. Check if values are 1 or -1
DescribeA(1,:) = ismember(A,1)|ismember(A,-1);
%// Check if they are > 0
DescribeA(2,:) = A>0;
Output in Command Window:
A =
-1 2 1 -10 5 -3 1
DescribeA =
1 0 1 0 0 0 1
0 1 1 0 1 0 1
I feel there is a smarter way for the 1st condition but I can't seem to find it.
I have a table of rows which consist of zeros and numbers like this:
A B C D E F G H I J K L M N
0 0 0 4 3 1 0 1 0 2 0 0 0 0
0 1 0 1 4 0 0 0 0 0 1 0 0 0
9 5 7 9 10 7 2 3 6 4 4 0 1 0
I want to calculate an average of the numbers including zeros, but starting from the first nonzero value and put it into column after tables end. E.g. for the first row first value is 4, so average - 11/11; for the second - 7/13; the last one is 67/14.
How could I using excel formulas do this? Probably OFFSET with nested IF?
This still needs to be entered as an array formula (ctrl-shift-enter) but it isn't volatile:
=AVERAGE(INDEX(($A2:$O2),MATCH(TRUE,$A2:$O2<>0,0)):$O2)
or, depending on location:
=AVERAGE(INDEX(($A2:$O2);MATCH(TRUE;$A2:$O2<>0;0)):$O2)
The sum is the same no matter how many 0's you include, so all you need to worry about is what to divide it by, which you could determine using nested IFs, or take a cue from this: https://superuser.com/questions/671435/excel-formula-to-get-first-non-zero-value-in-row-and-return-column-header
Thank you, Scott Hunter, for good reference.
I solved the problem using a huge formula, and I think it's a bit awkward.
Here it is:
=AVERAGE(INDIRECT(CELL("address";INDEX(A2:O2;MATCH(TRUE;INDEX(A2:O2<>0;;);0)));TRUE):O2)
after trying to understand the bsxfun function I have tried to implement it in a script to avoid looping. I am trying to check if each individual element in an array is contained in one matrix, returning a matrix the same size as the initial array containing 1 and 0's respectively. The anonymous function I have created is:
myfunction = #(x,y) (sum(any(x == y)));
x is the matrix which will contain the 'accepted values' per say. y is the input array. So far I have tried using the bsxfun function in this way:
dummyvar = bsxfun(myfunction,dxcp,X)
I understand that myfunction is equal to the handle of the anonymous function and that bsxfun can be used to accomplish this I just do not understand the reason for the following error:
Non-singleton dimensions of the two input arrays must match each other.
I am using the following test data:
dxcp = [1 2 3 6 10 20];
X = [2 5 9 18];
and hope for the output to be:
dummyvar = [1,0,0,0]
Cheers, NZBRU.
EDIT: Reached 15 rep so I have updated the answer
Thanks again guys, I thought I would update this as I now understand how the solution provided from Divakar works. This might deter confusion from others who have read my initial question and are confused to how bsxfun() works, I think writing it out helps me understand it better too.
Note: The following may be incorrect, I have just tried to understand how the function operates by looking at this one case.
The input into the bsxfun function was dxcp and X transposed. The function handle used was #eq so each element was compared.
%%// Given data
dxcp = [1 2 3 6 10 20];
X = [2 5 9 18];
The following code:
bsxfun(#eq,dxcp,X')
compared every value of dxcp, the first input variable, to every row of X'. The following matrix is the output of this:
dummyvar =
0 1 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
The first element was found by comparing 1 and 2 dxcp = [1 2 3 6 10 20]; X' = [2;5;9;18];
The next along the first row was found by comparing 2 and 2 dxcp = [1 2 3 6 10 20]; X' = [2;5;9;18];
This was repeated until all of the values of dxcp where compared to the first row of X'. Following this logic, the first element in the second row was calculating using the comparison between: dxcp = [1 2 3 6 10 20]; X' = [2;5;9;18];
The final solution provided was any(bsxfun(#eq,dxcp,X'),2) which is equivalent to: any(dummyvar,2). http://nf.nci.org.au/facilities/software/Matlab/techdoc/ref/any.html seems to explain the any function in detail well. Basically, say:
A = [1,2;0,0;0,1]
If the following code is run:
result = any(A,2)
Then the function any will check if each row contains one or several non-zero elements and return 1 if so. The result of this example would be:
result = [1;0;1];
Because the second input parameter is equal to 2. If the above line was changed to result = any(A,1) then it would check for each column.
Using this logic,
result = any(A,2)
was used to obtain the final result.
1
0
0
0
which if needed could be transposed to equal
[1,0,0,0]
Performance- After running the following code:
tic
dummyvar = ~any(bsxfun(#eq,dxcp,X'),2)'
toc
It was found that the duration was:
Elapsed time is 0.000085 seconds.
The alternative below:
tic
arrayfun(#(el) any(el == dxcp),X)
toc
using the arrayfun() function (which applies a function to each element of an array) resulted in a runtime of:
Elapsed time is 0.000260 seconds.
^The above run times are averages over 5 runs of each meaning that in this case bsxfun() is faster (on average).
You don't want every combination of elements thrown into your any(x == y) test, you want each element from dxcp tested to see if it exists in X. So here is the short version, which also needs no transposes. Vectorization should also be a bit faster than bsxfun.
arrayfun(#(el) any(el == X), dxcp)
The result is
ans =
0 1 0 0 0 0
I have a matrix in Matlab(2012) with 3 columns and X number of rows, X is defined by the user, so varies each time. For this example though I will use a fixed 5x3 matrix.
So I would like to perform an iterative function on each row within the matrix, while the value in the third column is below a certain value. Then store the new values within the same matrix, so overwrite the original values.
The code below is a simplified version of the problem.
M=[-2 -5 -3 -2 -4]; %Vector containing random values
Vf_X=M+1; %Defining the first column of the matrix
Vf_Y=M+2; %Defining the secound column of the matrix
Vf_Z=M; %Defining the third column of the matrix
Vf=[Vf_X',Vf_Y',Vf_Z']; %Creating the matrix
while Vf(:,3)<0
Vf=Vf+1;
end
disp(Vf)
The result I get is
1 2 0
-2 -1 -3
0 1 -1
1 2 0
-1 0 -2
Ideally I would like to get this result instead
1 2 0
1 2 0
1 2 0
1 2 0
1 2 0
The while will not start if any value is above zero to begin with and stops as soon as one value goes above zero.
I hope this makes sense and I have supplied enough information
Thank you for your time and help.
Your current problem is that you stop iterating the very moment any of the values in the third row break the condition. Correct me if I'm wrong, but what I think you want is to continue doing iterations on the remaining rows, until the conditions are broken by all third columns.
You could do that like this:
inds = true(size(Vf,1),1);
while any(inds)
Vf(inds,:) = Vf(inds,:)+1;
inds = Vf(:,3) < 0;
end
Of course, for the simple addition you provide, there is a better and faster way:
inds = Vf(:,3)<0;
Vf(inds,:) = bsxfun(#minus, Vf(inds,:), Vf(inds,3));
But for general functions, the while above will do the trick.