Please I need help to find the relations between these numbers , I hope if some can help me
3 22 25 30 49
1 12 27 42 47
22 37 43 44 45
8 21 25 38 50
Thank you
Related
I want to change the base of a multiplication table to another base.
If I use
disp(dec2base((1:10).*(1:10)',7))
the numbers come flowing out individually. However I want them to stay in the exact position in the given matrix.
The numerical base is a display issue, numbers are always stored and manipulated in base 2 internally. So all you need to do is write a loop that displays the numbers in they way you want to. For example:
for ii=1:10
for jj=1:10
fprintf('%6s',dec2base(ii*jj,7));
end
fprintf('\n');
end
Output:
1 2 3 4 5 6 10 11 12 13
2 4 6 11 13 15 20 22 24 26
3 6 12 15 21 24 30 33 36 42
4 11 15 22 26 33 40 44 51 55
5 13 21 26 34 42 50 55 63 101
6 15 24 33 42 51 60 66 105 114
10 20 30 40 50 60 100 110 120 130
11 22 33 44 55 66 110 121 132 143
12 24 36 51 63 105 120 132 144 156
13 26 42 55 101 114 130 143 156 202
Storing base-7 representation of numbers as string array:
M = (1:10).*(1:10)';
out = strings(size(M));
for jj = 1:size(M,2)
for ii = 1:size(M,1)
out(ii,jj) = dec2base(M(ii,jj) ,7);
end
end
This question already has answers here:
Sorting entire matrix according to one column in matlab
(2 answers)
Closed 4 years ago.
I have multiple columns in my dataset and column 2 contains value from 1 till 7. I want to sort my dataset with respect to second column . Thanks in advance
The command you need is sortrows
By default this sorts with respect to the first column, but an additional argument can be used to change this to the 2nd (or 5th, 17th etc)
If A is your original array:
B = sortrows(A,2);
will give you the sorted array B w.r.t 2nd column
What did you mean by sort with respect to second column? You should be more specific or at least give us an example.
If you need a simple sort on each column use the following
A =
95 45 92 41 13 1 84
23 1 73 89 20 74 52
60 82 17 5 19 44 20
48 44 40 35 60 93 67
89 61 93 81 27 46 83
76 79 91 0 19 41 1
Sort each column of A in ascending order:
c = sort(A, 1)
c =
23 1 17 0 13 1 1
48 44 40 5 19 41 20
60 45 73 35 19 44 52
76 61 91 41 20 46 67
89 79 92 81 27 74 83
95 82 93 89 60 93 84
numb is a list of numbers:
q))input
42 58 74 51 63 23 41 40 43 16 64 29 35 37 30 3 34 33 25 14 4 39 66 49 69 13..
31 41 39 27 9 21 7 25 34 52 60 13 43 71 10 42 19 30 46 50 17 33 44 28 3 62..
15 57 4 55 3 28 14 21 35 29 52 1 50 10 39 70 43 53 46 68 40 27 13 69 20 49..
3 34 11 53 6 5 48 51 39 75 44 32 43 23 30 15 19 62 64 69 38 29 22 70 28 40..
18 30 60 56 12 3 47 46 63 19 59 34 69 65 26 61 50 67 8 71 70 44 39 16 29 45..
I want to iterate through each row and calculate the sum of the first 2 and then 3 and then 4 numbers etc. If that sum is greater than 1000 I want to stop the iteration on that particualr row and jump on the next row and do the same thing. This is my code:
{[input]
tot::tot+{[x;y]
if[1000<sum x;:count x;x,y]
}/[input;input]
}each numb
My problem here is that after the count of x is added to tot the over keeps going on the same row. How can I exit over and jump on the next row?
UPDATE: (QUESTION STILL OPEN) I do appreciate all the answers so far but I am not looking for an efficient way to sum the first n numbers. My question is how do I break the over and jump on the next line. I would like to achieve the same thing as with those small scripts:
C++
for (int i = 0; i <= 100; i++) {
if (i = 50) { printf("for loop exited at: %i ", i); break; }
}
Python
for i in range(100):
if i == 50:
print(i);
break;
R
for(i in 1:100){
if(i == 50){
print(i)
break
}
}
I think this is what you are trying to accomplish.
sum {(x & sums y) ? x}[1000] each input
It takes a cumulative sum of each row and takes an element wise minimum between that sum and the input limit thereby capping the output at the limit like so:
q)(100 & sums 40 43 16 64 29)
40 83 99 100 100
It then uses the ? operator to find the first occurance of that limit (i.e the element where this limit was equaled or passed) adding one as it is 0 indexed. In the example the first 100 occurs after 3 elements. You might want add one to include the first element after the limit in the count.
q)40 83 99 100 100 ? 100
3
And then it sums this count over all rows of the input.
You could use coverage in this case to exit when you fail to satisfy a condition
https://code.kx.com/q/ref/adverbs/#converge-repeat
The first parameter would be a function that does your check based on the current value of x which will be the next value to be passed in the main function.
For your example ive made a projection using the main input line then increase the indexes of what i am summing each time:
q)numb
98 11 42 97 89 80 73 35 4 30
86 33 38 86 26 15 83 71 21 22
23 43 41 80 56 11 22 28 47 57
q){[input] {x+1}/[{100>sum (y+1)#x}[input;];0] }each numb
1 1 2
this returns the first index of each where running sum is over 100
However this isn't really an ideal use case of KDB
could instead be done with something like
(sums#/:numb) binr\: 100
maybe your real example makes more sense
You can use while loops in KDB although all KDB developers are generally too afraid of being openly mocked and laughed at for doing so
q){i:0;while[i<>50;i+:1];:"loop exited at ",string i}`
"loop exited at 50"
Kdb does have a "stop loop" mechanism but only in the case of a monadic function with single seed value
/keep squaring until number is no longer less than 1000, starting at 2
q){x*x}/[{x<1000};2]
65536
/keep dealing random numbers under 20 until you get an 18 (seed value 0 is irrelevant)
q){first 1?20}\[18<>;0]
0 19 17 12 15 10 18
However this doesn't really fit your use case and as other people have pointed out, this is not how you would/should solve this problem in kdb.
This question already has answers here:
Collapsing matrix into columns
(8 answers)
Closed 6 years ago.
Let's say I have A = [1:8; 11:18; 21:28; 31:38; 41:48] Now I would like to move everything from column 4 onward to the row position. How do I achieve this?
A =
1 2 3 4 5 6 7 8
11 12 13 14 15 16 17 18
21 22 23 24 25 26 27 28
31 32 33 34 35 36 37 38
41 42 43 44 45 46 47 48
to
A2 =
1 2 3 4
11 12 13 14
21 22 23 24
31 32 33 34
41 42 43 44
5 6 7 8
15 16 17 18
35 36 37 38
45 46 47 48
reshape doesn't seem to do the trick
Here's a vectorized approach with reshape and permute -
reshape(permute(reshape(a,size(a,1),4,[]),[1,3,2]),[],4)
Making it generic, we could introduce the number of columns as a parameter. Hence, let ncols be that one. So, the solution becomes -
ncols = 4
reshape(permute(reshape(a,size(a,1),ncols,[]),[1,3,2]),[],ncols)
Sample run -
>> a
a =
20 79 18 82 27 23 59 66 46 21 48 95
96 83 46 49 34 88 23 42 17 27 15 54
11 88 34 92 23 62 86 56 32 32 91 54
>> reshape(permute(reshape(a,size(a,1),4,[]),[1,3,2]),[],4)
ans =
20 79 18 82
96 83 46 49
11 88 34 92
27 23 59 66
34 88 23 42
23 62 86 56
46 21 48 95
17 27 15 54
32 32 91 54
More info on the intuition behind such a General idea for nd to nd transformation, which even though originally was meant for NumPy/Python, extends to any programming paradigm in general.
Use Matrix indexing!
B=[A(:,1:4);A(:,5:8)]
In a loop...
for ii=0:floor(size(A,2)/4)-1
B([1+5*ii:5*(ii+1)],:)=A(:,[1+4*ii:4*(ii+1)] );
end
One more... perhaps unoptimized way would be to decompose the matrix into cells row-wise, transpose the cell array then concatenate everything back together:
B = cell2mat(mat2cell(A, size(A, 1), 4 * ones((size(A, 2) / 4), 1)).');
The above first uses mat2cell to decompose the matrix into non-overlapping cells. Each cell has the same number of rows as A but the total number of columns is 4 and there are exactly size(A, 2) / 4 of them. As such, we need to indicate a vector of ones where each element is 4 and there are size(A, 2) / 4 of these to tell us the number of columns for each cell. This creates a row-wise cell array and so we transpose this cell array and merge all of the cells together into one final matrix with cell2mat.
What's the fastest way to create a 8x8 matrix filled with 1-64 by row. The help docs say i should even be able to fill a matrix with an array, but i can't seem to make it work. I've been told it can be done more easily than i do it, but I've not seen it done. Here's an idea of what i'm looking for...
v26 =
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
41 42 43 44 45 46 47 48
49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64
but to get it to do this, I had to do a row-by-row fill with ...
v26 = [1:8; 9:16; 17:24; 25:32; 33:40; 41:48; 49:56; 57:64]
make a sequence, then you reshape it:
m = reshape(1:64, [8 8])';
You have to transpose it in the end b/c matlab is column major.