Consider the following sheet example:
A1 A2
1 5 10
2 6 12
3 -3 9
4 1 10
5 5 15
6 -4 11
7 9 20
How do I look back from row 6 and sum all A2 rows until a previous negative A1 row.
In this example: 15 + 10 = 25
Assuming -3 is in A3, in C4 and copied down to suit:
=IF(A3<0,0,C3+B3)
This creates a running total, starting immediately after the first negative in the left hand column, that resets after each negative in the left hand column.
Related
I have a 4x9 matrix, and I need to calculate the sum of all numbers in every other column of c starting with the first. Can anyone point me in the right direction? I know we have to use the function sum() but that's about it.
I used Octave rather than MATLAB, but this works for me:
A = randi(10,4,9)
B = A(:, 1:2:9)
C = sum(B)
Generate a 4x9 matrix with random numbers between 1 and 10, then create a sub-matrix with each row, and given columns 1:2:9 means starting from the first column and ending on the 9th, choose every second column, then sum up each column. Example output:
>> A = randi(10,4,9)
A =
1 3 6 8 2 8 4 8 10
3 6 10 4 6 4 6 2 8
4 3 9 2 7 10 6 9 6
8 5 3 9 3 8 4 6 10
>> B = A(:, 1:2:9)
B =
1 6 2 4 10
3 10 6 6 8
4 9 7 6 6
8 3 3 4 10
>> C = sum(B)
C =
16 28 18 20 34
You could also take the sum of matrix C using the sum() first and then select every other element from the result starting from the 1st element.
tmpC = sum(C);
result = tmpC(1:2:end)
If I have this matrix:
A:
X Y Z
1 1 2
0 3 4
0 5 6
2 7 8
7 9 10
8 11 12
3 13 14
12 14 16
15 17 18
How could I create new matrix B, C, D and E which contains:
B:
0 3 4
0 5 6
C:
X Y Z
1 1 2
2 7 8
3 13 14
D:
7 9 10
8 11 12
E:
12 14 16
15 17 18
The idea is to construct a loop asking if 0<A<1 else 1<A<5 else 6<A<10 else 11<A<15. and create new matrix from that condition. Any idea about how to store the results of the loop?
I suggest you an approach that uses the discretize function in order to group the matrix rows into different categories based on their range. Here is the full implementation:
A = [
1 1 2;
0 3 4;
0 5 6;
2 7 8;
7 9 10;
8 11 12;
3 13 14;
12 14 16;
15 17 18
];
A_range = [0 1 5 10 15];
bin_idx = discretize(A(:,1),A_range);
A_split = arrayfun(#(bin) A(bin_idx == bin,:),1:(numel(A_range) - 1),'UniformOutput',false);
celldisp(A_split);
Since you want to consider 5 different ranges based on the first column values, the arguments passed to discretize must be the first matrix column and a vector containing the group limits (first number inclusive left, second number exclusive right, second number inclusive left, third number exclusive right, and so on...). Since your ranges are a little bit messed up, feel free to adjust them to respect the correct output. The latter is returned in the form of a cell array of double matrices in which every element contains the rows belonging to a distinct group:
A_split{1} =
0 3 4
0 5 6
A_split{2} =
1 1 2
2 7 8
3 13 14
A_split{3} =
7 9 10
8 11 12
A_split{4} =
12 14 16
15 17 18
Instead of using a loop, use logical indexing to achieve what you want. Use the first column of A and check for the ranges that you want to look for, then use this to subset into the final matrix A to get what you want.
For example, to create the matrix C, find all locations in the first column of A that are between 1 and 5, then subset the matrix along the rows using these locations:
m = A(:,1) >= 1 & A(:,1) <= 5;
C = A(m,:);
You can repeat this in a similar way for the rest of the matrices you want to create.
Say I have a matrix A whose first column contains item IDs with repetition and second column contains their weights.
A= [1 40
3 33
2 12
4 22
2 10
3 6
1 15
6 29
4 10
1 2
5 18
5 11
2 8
6 25
1 14
2 11
4 28
3 38
5 35
3 9];
I now want to find the difference of each instance of A and its associated minimum weight. For that, I make a matrix B with its first column containing the unique IDs from column 1 of A, and its column 2 containing the associated minimum weight found from column 2 of A.
B=[1 2
2 8
3 6
4 10
5 11
6 25];
Then, I want to store in column 3 of A the difference of each entry and its associated minimum weight.
A= [1 40 38
3 33 27
2 12 4
4 22 12
2 10 2
3 6 0
1 15 13
6 29 4
4 10 0
1 2 0
5 18 7
5 11 0
2 8 0
6 25 0
1 14 12
2 11 3
4 28 18
3 38 32
5 35 24
3 9 3];
This is the code I wrote to do this:
for i=1:size(A,1)
A(i,3) = A(i,1) - B(B(:,1)==A(i,2),2);
end
But this code takes a long time to execute as it needs to loop through B every time it loops through A. That is, it has a complexity of size(A) x size(B). Is there a better way to do this without using loops, that would execute faster?
You can use accumarray to first compute the minimum value in the second column of A for each unique value in the first column of A. We can then index into the result using the first column of A and compare to the second column of A to create the third column.
% Compute the mins
min_per_group = accumarray(A(:,1), A(:,2), [], #min);
% Compute the difference between the second column and the minima
A(:,3) = A(:,2) - min_per_group(A(:,1));
How do I sort a column based on the values in another column in MATLAB?
Column A shows position data (it is neither ascending or descending in order) Column B contains another column of position data. Finally column C contains numerical values. Is it possible to link the first position value in B with its numerical value in the first cell of C? Then after this I want to sort B such that it is in the same order as column A with the C values following their B counterparts?The length of my columns would be 1558 values.
Before case;
A B C
1 4 10
4 1 20
3 5 30
5 2 40
2 3 50
After Case;
A B C
1 1 20
4 4 10
3 3 50
5 5 30
2 2 40
Basically A and B became the same and Column C followed B.
Since you don't want things necessarily in ascending or descending order, I don't think any built-in sorting functions like sortrows() will help here. Instead you are matching elements in one column with elements in another column.
Using [~,idx]=ismember(A,B) will tell you where each element of B is in A. You can use that to sort the desired columns.
M=[1 4 10
4 1 20
3 5 30
5 2 40
2 3 50];
A=M(:,1); B=M(:,2); C=M(:,3);
[~,idx]=ismember(A,B);
sorted_matrix = [A B(idx) C(idx)]
Powerful combo of bsxfun and matrix-multiplication solves it and good for code-golfing too! Here's the implementation, assuming M as the input matrix -
[M(:,1) bsxfun(#eq,M(:,1),M(:,2).')*M(:,2:3)]
Sample run -
>> M
M =
1 4 10
4 1 20
3 5 30
5 2 40
2 3 50
>> [M(:,1) bsxfun(#eq,M(:,1),M(:,2).')*M(:,2:3)]
ans =
1 1 20
4 4 10
3 3 50
5 5 30
2 2 40
Given M = [A B C]:
M =
1 4 10
4 1 20
3 5 30
5 2 40
2 3 50
You need to sort the rows of the matrix excluding the first column:
s = sortrows(M(:,2:3));
s =
1 20
2 40
3 50
4 10
5 30
Then use the first column as the indices to reorder the resulting submatrix:
s(M(:,1),:);
ans =
1 20
4 10
3 50
5 30
2 40
This would be used to build the output matrix:
N = [M(:,1) s(M(:,1),:)];
N =
1 1 20
4 4 10
3 3 50
5 5 30
2 2 40
The previous technique will obviously only work if A and B are permutations of the values (1..m). If this is not the case, then we need to find the ranking of each value in the array. Let's start with new values for our arrays:
A B C
1 5 60
6 1 80
9 6 60
-4 9 40
5 -4 30
We construct s as before:
s = sortrows([B C]);
s =
-4 30
1 80
5 60
6 60
9 40
We can generate the rankings one of two ways. If the elements of A (and B) are unique, we can use the third output of unique as in this answer:
[~, ~, r] = unique(A);
r =
2
4
5
1
3
If the values of A are not unique, we can use the second return value of sort, the indices in the original array of the elements in sorted order, to generate the rank of each element:
[~, r] = sort(A);
r =
4
1
5
2
3
[~, r] = sort(r);
r =
2
4
5
1
3
As you can see, the resulting r is the same, it just takes 2 calls to sort rather than 1 to unique. We then use r as the list of indices for s above:
M = [A s(r, :)];
M =
1 1 80
6 6 60
9 9 40
-4 -4 30
5 5 60
If you must retain the order of A then use something like this
matrix = [1 4 10; 4 1 20; 3 5 30; 5 2 40; 2 3 50];
idx = arrayfun(#(x) find(matrix(:,2) == x), matrix(:,1));
sorted = [matrix(:,1), matrix(idx,2:3)];
I have a table rCom which has various columns. I would like to sum across each row..
for example:
Date TypeA TypeB TypeC TypeD
date1 40.5 23.1 45.1 65.2
date2 23.3 32.2 56.1 30.1
How can I write a q query to add a fourth column 'Total' that sums across each row?
why not just:
update Total: TypeA+TypeB+TypeC+TypeD from rCom
?
Sum will work just fine:
q)flip`a`b`c!3 3#til 9
a b c
-----
0 3 6
1 4 7
2 5 8
q)update d:sum(a;b;c) from flip`a`b`c!3 3#til 9
a b c d
--------
0 3 6 9
1 4 7 12
2 5 8 15
Sum has map reduce which will be better for a huge table.
One quick point regarding summing across rows. You should be careful about nulls in 1 column resulting in a null result for the sum. Borrowing #WooiKent Lee's example.
We put a null into the first position of the a column. Notice how our sum now becomes null
q)wn:.[flip`a`b`c!3 3#til 9;(0;`a);first 0#] //with null
q)update d:sum (a;b;c) from wn
a b c d
--------
3 6
1 4 7 12
2 5 8 15
This is a direct effect of the way nulls in q are treated. If you sum across a simple list, the nulls are ignored
q)sum 1 2 3 0N
6
However, a sum across a general list will not display this behavior
q)sum (),/:1 2 3 0N
,0N
So, for your table situation, you might want to fill in with a zero beforehand
q)update d:sum 0^(a;b;c) from wn
a b c d
--------
3 6 9
1 4 7 12
2 5 8 15
Or alternatively, make it s.t. you are actually summing across simple lists rather than general lists.
q)update d:sum each flip (a;b;c) from wn
a b c d
--------
3 6 9
1 4 7 12
2 5 8 15
For a more complete reference on null treatment please see the reference website
This is what worked:
select Answer:{[x;y;z;a] x+y+z+a }'[TypeA;TypeB;TypeC;TypeD] from
([] dt:2014.01.01 2014.01.02 2014.01.03; TypeA:4 5 6; TypeB:1 2 3; TypeC:8 9 10; TypeD:3 4 5)