I have matrix A
A =
5 10 15 20 25
10 9 8 7 6
-5 -15 -25 -35 -45
1 2 3 4 5
28 91 154 217 280
And i need to make a matrix B of the first, fourth and fifth row and the first and fifth column from matrix A.
How can i do it?
>> B = A([1,4,5],[1,5])
B =
5 25
1 5
28 280
You should look up how to use index expressions in the Matlab and Octave language to extract and work with submatrices.
See the Octave help on Index expressions: https://octave.org/doc/latest/Index-Expressions.html
Related
Example input:
3 4
10 20 30
40 10 30
20 10 0
5 15 5
Example output:
1 3
2 1
4 2
85
30 + 40 + 15 = 85
I need get the result of a matrix with thousand rows and columns
I have some data:
P = [3 10 25 32 43 1 3
6 12 35 39 49 4 9
2 9 23 36 47 2 9
...
7 20 35 42 44 3 7
15 18 19 41 42 4 6
10 18 32 35 46 3 10];
Data is always between 1 and 50.
I am selecting left 5 columns and 2 right columns:
L=P(:,1:5);
R=P(:,6:7);
I am counting occurrences:
a=tabul(L);
b=tabul(R);
In this moment, in a I am getting:
50. 3.
49. 4.
48. 3.
which tells me, that value 50 occurs 3 times, 49 occurs 4 times and so on.
What I need now is sort matrix a by second column but the first column should be arranged with the second column values. So it would look like this:
49. 4.
50. 3.
48. 3.
How can I sort matrix a this way (later I will sort b the same way)?
I was trying something like:
[a,idx]=gsort(a(:,2),"g","d");
a=a(idx,:);
but this not does what I need.
It does not work because you are overwriting a in the gsort call although you just need the index here. The following does what you want:
[dummy,idx]=gsort(a(:,2),"g","d");
a=a(idx,:);
I have a crosstab and create custom grand total for the row level in each column dimension, by using a data element expression.
Crosstab Example:
Cat 1 Cat 2 GT
ITEM C F % VALUE C F % VALUE
A 101 0 0.9 10 112 105 93.8 10 20
B 294 8 2.7 6 69 66 95.7 10 16
C 211 7 3.3 4 212 161 75.9 6 10
------------------------------------------------------------------
GT 606 15 2.47 6 393 332 84.5 8 **14**
Explanation for GT row:
Those C and F column is summarized from the above. But the
% column is division result of F/C.
Create a data element to fill the VALUE column, which comes from range of value definition, varies for each Cat (category). For instance... in Cat 1, if the value is between 0 - 1 the value will be 10, or between 1 - 2 = 8, etc. And condition for Cat 2, between 85 - 100 = 10, and 80 - 85 = 8, etc.
The GT row (with the value of 14), is gathered by adding VALUE of Cat 1 + Cat 2.
I am able to work on point 1 and 2 above, but I can't seem to make it working for GT row. I don't know the code/expression to sum up the VALUE data element for this 2 categories. Because those VALUE field comes from one data element in design mode.
I have found the solution for my problem. I can show the result by using a report variable. I am assigning 2 report variables in % field expression, based on the category in data cube dimension (by using if statement). And then in data element expression, I am calling both of the expressions and add them.
The third item in the FinnAPL Library is called “Cumulative maxima (⌈) of subvectors of Y indicated by X ” where X is a binary vector and Y os a vector of numbers. Here's an example of its usage:
X←1 0 0 0 1 0 0 0
Y←9 78 3 2 50 7 69 22
Y[A⍳⌈\A←⍋A[⍋(+\X)[A←⍋Y]]] ⍝ output 9 78 78 78 50 50 69 69
You can see that beginning from either the beginning or from any 1 value in the X array, the cumulave maximum is found for all corresponding digits in Y until another 1 is found in X. In the example given, X is divding the array into two equal parts of 4 numbers each. In the first part, 9 is the maxima until 78 is encountered, and in the second part 50 is the maxima until 69 is encountered.
That's easy enough to understand, and I could blindly use it as is, but I'd like to understand how it works, because APL idioms are essentially algorithms made up of operators and functions. To understand APL well, it's important to understand how the masters were able to weave it all together into such compact and elegant lines of code.
I find this particular idiom especially hard to understand because of the indexing nested two layers deep. So my question is, what makes this idiom tick?
This idiom can be broken down into smaller idioms, and most importantly, it contains idiom #11 from the FinnAPL Library entitled:
Grade up (⍋) for sorting subvectors of Y indicated by X
Using the same values for X and Y given in the question, here's an example of its usage:
X←1 0 0 0 1 0 0 0
Y←9 78 3 2 50 7 69 22
A[⍋(+\X)[A←⍋Y]] ⍝ output 4 3 1 2 6 8 5 7
As before, X is dividing the vector into two halves, and the output indicates, for each position, what digit of Y is needed to sort each of the halves. So, the 4 in the output is saying that it needs the 4th digit of Y (2) in the 1st position; the 3 indicates the 3rd digit (3) in the 2nd position; the 1 indicates the 1st digit (9) in the third position; etc. Thus, if we apply this indexing to Y, we get:
Y[A[⍋(+\X)[A←⍋Y]]] ⍝ output 2 3 9 78 7 22 50 69
In order to understand the indexing within this grade-up idiom, consider what is happening with the following:
(+\X)[A←⍋Y] ⍝ Sorted Cumulative Addition
Breaking it down step by step:
A←⍋Y ⍝ 4 3 6 1 8 5 7 2
+\X ⍝ 1 1 1 1 2 2 2 2
(+\X)[A←⍋Y] ⍝ 1 1 2 1 2 2 2 1 SCA
A[⍋(+\X)[A←⍋Y]] ⍝ 4 3 1 2 6 8 5 7
You can see that sorted cumulative addition (SCA) of X 1 1 2 1 2 2 2 1 applied to A acts as a combination of compress left and compress right. All values of A that line up with a 1 are moved to the left, and those lining up with a 2 move to the right. Of course, if X had more 1s, it would be compressing and locating the compressed packets in the order indicated by the values of the SCA result. For example, if the SCA of X were like 3 3 2 1 2 2 1 1 1, you would end up with the 4 digits corresponding to the 1s, followed by the 3 digits corresponding to the 2s, and finally, the 2 digits corresponding to the 3s.
You may have noticed that I skipped the step that would show the effect of grade up ⍋:
(+\X)[A←⍋Y] ⍝ 1 1 2 1 2 2 2 1 SCA
⍋(+\X)[A←⍋Y] ⍝ 1 2 4 8 3 5 6 7 Grade up
A[⍋(+\X)[A←⍋Y]] ⍝ 4 3 1 2 6 8 5 7
The effect of compression and rearrangement isn't accomplised by SCA alone. It effectively acts as rank, as I discussed in another post. Also in that post, I talked about how rank and index are essentially two sides of the same coin, and you can use grade up to switch between the two. Therefore, that is what is happening here: SCA is being converted to an index to apply to A, and the effect is grade-up sorted subvectors as indicated by X.
From Sorted Subvectors to Cumulative Maxima
As already described, the result of sorting the subvectors is an index, which when applied to Y, compresses the data into packets and arranges those packets according to X. The point is that it is an index, and once again, grade up is applied, which converts indexes into ranks:
⍋A[⍋(+\X)[A←⍋Y]] ⍝ 3 4 2 1 7 5 8 6
The question here is, why? Well, the next step is applying a cumulative maxima, and that really only makes sense if it is applied to values for rank which represent relative magnitude within each packet. Looking at the values, you can see that 4 is is the maxima for the first group of 4, and 8 is for the second group. Those values correspond to the input values of 78 and 69, which is what we want. It doesn't make sense (at least in this case) to apply a maxima to index values, which represent position, so the conversion to rank is necessary. Applying the cumulative maxima gives:
⌈\A←⍋A[⍋(+\X)[A←⍋Y]] ⍝ 3 4 4 4 7 7 8 8
That leaves one last step to finish the index. After doing a cumulative maxima operation, the vector values still represent rank, so they need to be converted back to index values. To do that, the index-of operator is used. It takes the value in the right argument and returns their position as found in the left argument:
A⍳⌈\A←⍋A[⍋(+\X)[A←⍋Y]] ⍝ 1 2 2 2 5 5 7 7
To make it easier to see:
3 4 2 1 7 5 8 6 left argument
3 4 4 4 7 7 8 8 right argument
1 2 2 2 5 5 7 7 result
The 4 is in the 2nd position in the left argument, so the result shows a 2 for every 4 in the right argument. The index is complete, so applying it to Y, we get the expected result:
Y[A⍳⌈\A←⍋A[⍋(+\X)[A←⍋Y]]] ⍝ 9 78 78 78 50 50 69 69
My implementation:
X←1 0 0 0 1 0 0 0
Y←9 78 3 2 50 7 69 22
¯1+X/⍳⍴X ⍝ position
0 4
(,¨¯1+X/⍳⍴X)↓¨⊂Y
9 78 3 2 50 7 69 22 50 7 69 22
(1↓(X,1)/⍳⍴X,1)-X/⍳⍴X ⍝ length
4 4
(,¨(1↓(X,1)/⍳⍴X,1)-X/⍳⍴X)↑¨(,¨¯1+X/⍳⍴X)↓¨⊂Y
9 78 3 2 50 7 69 22
⌈\¨(,¨(1↓(X,1)/⍳⍴X,1)-X/⍳⍴X)↑¨(,¨¯1+X/⍳⍴X)↓¨⊂Y
9 78 78 78 50 50 69 69
∊⌈\¨(,¨(1↓(X,1)/⍳⍴X,1)-X/⍳⍴X)↑¨(,¨¯1+X/⍳⍴X)↓¨⊂Y
9 78 78 78 50 50 69 69
Have a nice day.
Imagine I've defined the following name in J:
m =: >: i. 2 4 5
This looks like the following:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
26 27 28 29 30
31 32 33 34 35
36 37 38 39 40
I want to create a monadic verb of rank 1 that applies to each list in this list of lists. It will double (+:) or add 1 (>:) to each alternate item in the list. If we were to apply this verb to the first row, we'd get 2 3 6 5 10.
It's fairly easy to get a list of booleans which alternate with each item, e.g., 0 1 $~{:$ m gives us 0 1 0 1 0. I thought, aha! I'll use something like +:`>: #. followed by some expression, but I could never quite get it to work.
Any suggestions?
UPDATE
The following appears to work, but perhaps it can be refactored into something more elegant by a J pro.
poop =: monad define
(($ y) $ 0 1 $~{:$ y) ((]+:)`(]>:) #. [)"0 y
)
I would use the oblique verb, with rank 1 (/."1)- so it applies to successive elements of each list in turn.
You can pass a gerund into /. and it applies them in order, extending cyclically.
+:`>: /."1 m
2
3
6
5
10
12
8
16
10
20
22
13
26
15
30
32
18
36
20
40
42
23
46
25
50
52
28
56
30
60
62
33
66
35
70
72
38
76
40
80
I spent a long time and I looked at it, and I believe that I know why ,# works to recover the shape of the argument.
The shape of the arguments to the parenthesized phrase is the shape of the argument passed to it on the right, even though the rank is altered by the " conjugate (well, that is what trace called it, I thought it was an adverb). If , were monadic, it would be a ravel, and the result would be a vector or at least of a lower rank than the input, based on adverbs to ravel. That is what happens if you take the conjunction out - you get a vector.
So what I believe is happening is that the conjunction is making , act like a dyadic , which is called an append. The append alters what it is appending to what it is appending to. It is appending to nothing but that thing still has a shape, and so it ends up altering the intermediate vector back to the shape of the input.
Now I'm probably wrong. But $,"0#(+:>:/.)"1 >: i. 2 4 5 -> 2 4 5 1 1` which I thought sort of proved my case.
(,#(+:`>:/.)"1 a) works, but note that ((* 2 1 $~ $)#(+ 0 1 $~ $)"1 a) would also have worked (and is about 20 times faster, on large arrays, in my brief tests).