Consider the matrix:
1 2 3
4 5 6
7 8 9
I'd like to take the middle column, assign it to a variable, and replace the middle row with it, giving me
1 2 3
2 5 8
7 8 9
I'm extracting the middle column using
a:m[;enlist1]
which returns
2
5
8
How do I replace the middle row with a? Is a flip necessary?
Thanks.
If you want to update the matrix in place you can use
q)show m:(3;3)#1+til 10
1 2 3
4 5 6
7 8 9
q)a:m[;1]
q)m[1]:a
q)show m
1 2 3
2 5 8
7 8 9
q)
cutting out "a" all you need is:
m[1]:m[;1]
You can use dot amend -
q)show m:(3;3)#1+til 10
1 2 3
4 5 6
7 8 9
q)show a:m[;1]
2 5 8
q).[m;(1;::);:;a]
1 2 3
2 5 8
7 8 9
Can see documentation here:
http://code.kx.com/wiki/Reference/DotSymbol
http://code.kx.com/wiki/JB:QforMortals2/functions#Functional_Forms_of_Amend
Making it slightly more generic where you can define the operation, row, and column
q)m:3 cut 1+til 9
1 2 3
4 5 6
7 8 9
Assigning the middle column to middle row :
q){[ m;o;i1;i2] .[m;enlist i1;o; flip[m] i2 ] }[m;:;1;1]
1 2 3
2 5 8
7 8 9
Adding the middle column to middle row by passing o as +
q){[ m;o;i1;i2] .[m;enlist i1;o; flip[m] i2 ] }[m;+;1;1]
1 2 3
6 10 14
7 8 9
Related
I have a table
t: flip `ref`a`b`c`d`e!(til 10;10?10;10?10;10?10;10?10;10?10)
ref a b c d e
0 5 3 3 9 1
1 1 9 0 0 0
2 5 9 4 1 7
3 0 0 5 1 3
4 2 6 8 9 3
5 3 2 0 6 6
6 7 6 4 9 8
7 4 8 9 7 2
8 7 0 8 8 3
9 7 9 0 4 8
how can I set all values in columns a,b,c ,.. to 0Ni if their value equals the value in column ref without having to do a single line update for every columns?
So something taht would look a bit like (which returns ERROR:type)
{update x:?[x=t;0Ni;x] from t} each `a`b`c`....
The previous answer involves working with strings, this can often get very messy. To avoid this it is possible to build up the query by passing column names as symbols instead. The dictionary for the functional select can be built up using the following function:
q){y!enlist[({?[y=x;0Ni;y]};x)],/:y:(),y}[`ref;`a`b`c]
a| ({?[y=x;0Ni;x]};`ref) `a
b| ({?[y=x;0Ni;x]};`ref) `b
c| ({?[y=x;0Ni;x]};`ref) `c
The initial column is x and it allows to any number of columns to be passed as y for comparison.
This can then be added into the functional select:
q)![t;();0b;{y!enlist[({?[y=x;0Ni;y]};x)],/:y:(),y}[`ref;`a`b`c]]
ref a b c d e
-------------
0 4 5 8 4 8
1 2 6 9 1
2 8 4 7 2 9
3 0 1 2 7 5
4 5 3 0 4
5 8 3 1 6
6 5 7 4 9 6
7 2 8 2 2 1
8 2 7 1 8
9 6 1 8 8 5
You may reuse the next code snippet
t: flip `ref`a`b`c`d`e!(til 10;10?10;10?10;10?10;10?10;10?10);
columns: `a`b`c`d`e;
![t;();0b;columns!{parse "?[",x,"=ref;0Ni;",x,"]" }each string columns]
Where updated columns are put in columns list.
And functional update, which maps every column X to value ?[x=ref;0Ni;x], is used
Given two parameters:
n %number of repetitions per value
k %max value to repeat
I would like to create a vector of size n*k, which is a concatenation of k vectors of size n, such that the i'th vector contains the value i at each coordinate.
Example:
n = 5;
k = 9;
Desired result:
[1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5,6,6,6,6,6,7,7,7,7,7,8,8,8,8,8,9,9,9,9,9];
Is there an elegant way to achieve this?
Thanks!
quite a few ways to do it:
method 1:
A=1:k
repelem(A',n,1)'
method 2:
A=1:k
kron(A', ones(n,1))'
method 3:
A=1:k
B=repmat(A, n, 1)
B(:)'
method 4:
A=1:k
B=ones(n,1)*A
B(:)'
Here is an alternative method
A = reshape(mtimes((1:k).',ones(1,n)).',1,n*k)
A =
Columns 1 through 22
1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5
Columns 23 through 44
5 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9
Column 45
9
It multiplies each element by ones n times
>> mtimes((1:k).',ones(1,5)).'
ans =
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
and then reshapes the whole matrix to one vector
I have generated a table of data with time in one column with attempts 1-10 in the next series of columns. I want to be able to extract the max value in each attempt for further analysis.
I have tried for table MGA
max = max(MGA(:, []))
I get the following error -- "You cannot subscript a table using only one subscript. Table subscripting requires both row and variable subscripts."
First off: Never do max = max();, you'll overload max, and you won't be able to use it again.
And to answer the question, you can do as follows (notice that I've kept the values in the first columns):
MGA
MGA =
1 5 3 8 9
2 4 7 3 3
3 8 7 6 9
4 8 2 7 3
5 2 2 9 10
6 5 5 10 4
7 5 10 6 2
8 7 4 2 3
9 8 6 2 7
10 8 3 3 5
max_values = [MGA(:,1), max(MGA(:,2:end),[],2)]
max_values =
1 9
2 7
3 9
4 8
5 10
6 10
7 10
8 7
9 8
10 8
I currently have a 3 by 3 matrix "m":
1 2 3
4 5 6
7 8 9
I would like to add a row to matrix 'm' to have a resultant matrix of:
1 2 3
4 5 6
7 8 9
10 11 12
A matrix in q is just a list of lists where inner lists represent rows.
m: ((1 2 3);(4 5 6);(7 8 9))
In order to add one more row all you have to do is add one more inner list to it:
m: m,enlist 10 11 12
enlist is important here, without it you'll end up with this:
q)((1 2 3);(4 5 6);(7 8 9)),10 11 12
1 2 3
4 5 6
7 8 9
10
11
12
I agree; using 0N!x to view the structure is very useful.
To achieve what you want then you can simply do;
q)show m:3 cut 1+til 9 /create matrix
1 2 3
4 5 6
7 8 9
q)show m,:10 11 12 /join new 'row'
1 2 3
4 5 6
7 8 9
10 11 12
q)
Anybody know a fast way to produce a matrix consisting of a linspace for each row? For example, the sort of pattern I'm looking for in this matrix is:
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
...
1 2 3 4 5 6 7 8 9 10
Anyone know any fast tricks to produce this WITHOUT using a for loop?
I just figured this out, so just in case anyone else was troubled by this, we can achieve this exact pattern by:
a=linspace(1,10,10);
b=ones(3,1)*a;
This will give:
>> a = 1 2 3 4 5 6 7 8 9 10
>> b = 1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
You need to use repmat.
Example:
>> B = repmat(1:10,[3 1])
B =
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
You can vary the value of 3 there. You can change it accordingly.
Another shortcut I can recommend is similar to repmat, but you specify a base array first of a = 1:10;. Once you do this, you specify a series of 1s in the first dimension when indexing which should produce a matrix of the same vectors with many rows as you want, where each row consists of the base array a. As such:
%// number of times to replicate
n = 4;
a = 1:10;
a = a(ones(1,n),:);
Result:
a =
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
Insert this command: transpose(ndgrid(1:10,1:n));, where n is the number of rows desired in the result.
You can consider these solutions:
With basic matrix indexing (taken from here)
b=a([1:size(a,1)]' * ones(1,NumToReplicate), :) %row-wise replication
b=a(:, ones(NumToReplicate, 1)) %column-wise replication
With bsxfun:
bsxfun(#times,a,(ones(1,NumToReplicate))') %row-wise replication
bsxfun(#times,a',(ones(1,NumToReplicate))) %column-wise replication
You are welcome to benchmark above two solutions with repmat.