I want to see whether a column X in a table T is in sorted order. class(x) is double. I compare X to sort(x) using isequal and find that the answer is false.
I visually inspect, and find no entries that differ. Further, by making a table [T.x sort(T.x)], I see that the first 2011 entries are equal to each other. The 2012th row is some how not equal, using the isequal function. Here is a dput of this row
ans =
reshape([2080857.000000 2080857.000000 ],[1 2])
Converting to int64 also does not give equality.
This is not the same as the linked duplicate question. I have tried techniques from there, including format long g and my values are still the same. In addition, the two columns of the table are the 'same', one is just the sorted ordering of the other, and to isequal, the first 2011 entries are equal and the 2012th entry is equal to my naked eye, or converted to integer.
Related
I have a matrix and I am interested in changing values that satisfy a certain condition inside that matrix differently, depending on where they are. Say I have a matrix smallPic. How do I obtain a matrix smallPicB with the same dimensions that changed all values that are above 50 in the first two columns to a 255, while those that are in the third and fourth column are changed to a 180?
I have this code which works, but it is pretty ugly and requires splitting the matrix and concatenating it again:
smallPic1=smallPic(:,1:2);smallPic1(smalllPic1>50)=255;
smallPic2=smallPic(:,3:4);smallPic2(smalllPic2>50)=180;
smallPicB = [smalllPic1 smalllPic2];
How would you combine the logical index with the scalar index in one command?
What doesn't work is this:
smallPic(:,smallPic(:,3:4)>50) = 180
Here, the second mention of smallPic inside the brackets does not allow indexing into the correct position of smallPic because it doesn't have the same dimensions as smallPic. So this command actually replaces values in the first two columns of smallPic that are in the same row as those values that are above 50 in the third and fourth column, instead of replacing the values in the third and fourth column themselves.
Any other suggestions?
It probably is not what you're looking for, but it can help if you have lots of assignments like that:
J = repmat(1:size(smallPic, 2), size(smallPic, 1), 1)
smallPic((J<3)&(smallPic>50))=255
smallPic((J>2)&(J<5)&(smallPic>50))=180
You can also call ismember function if column indices are not consecutive:
smallPic(ismember(J, [[1:2 5:6]])&(smallPic>50))=255
I want to test my D trigger scheme and I don't know how to do that. I don't even know how "truth table" works and maybe someone can help/explain how to make a test for my D trigger scheme?
A logical table works as follows:
Each row represents a case, each column represents a statement. A cell represents the logical value of a statement. It is wise to differentiate mentally atomic columns from molecular columns. In your case, you have three atomic columns, they represent the cases. These are the first three columns in your logical table. The other columns are molecular columns, that is, their value is composed by other columns.
D is said to be the value of XOR-ing the third and the fifth column. If you look in each row the value of the third and the fifth column in the table, you will see that D4 is true (1) if they differ and it is false (0) if they are similar.
I am not sure what do you mean by testing the D trigger scheme, but if by that you mean that we should test whether the formula matches to the values, then we can say that yes, it matches, since the general concept of XOR is matched in all cases in your logical table.
My goal is to create a random, 20 by 5 array of integers, sort them by increasing order from top to bottom and from left to right, and then calculate the mean in each of the resulting 20 rows. This gives me a 1 by 20 array of the means. I then have to find the column whose mean is closest to 0. Here is my code so far:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
MeanArray= mean(transpose(NewArray(:,:)))
X=min(abs(x-0))
How can I store the column number whose mean is closest to 0 into a variable? I'm only about a month into coding so this probably seems like a very simple problem. Thanks
You're almost there. All you need is a find:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
% MeanArray= mean(transpose(NewArray(:,:))) %// gives means per row, not column
ColNum = find(abs(mean(NewArray,1))==min(abs(mean(NewArray,1)))); %// gives you the column number of the minimum
MeanColumn = RandomArray(:,ColNum);
find will give you the index of the entry where abs(mean(NewArray)), i.e. the absolute values of the mean per column equals the minimum of that same array, thus the index where the mean of the column is closest to 0.
Note that you don't need your MeanArray, as it transposes (which can be done by NewArray.', and then gives the mean per column, i.e. your old rows. I chucked everything in the find statement.
As suggested in the comment by Matthias W. it's faster to use the second output of min directly instead of a find:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
% MeanArray= mean(transpose(NewArray(:,:))) %// gives means per row, not column
[~,ColNum] = min(abs(mean(NewArray,1)));
MeanColumn = RandomArray(:,ColNum);
So I'm working with sparse matrix and I have to find out different info about a very big one (10^6 size) and I need to find out the mean of the outlinks. Just to be sure I think mean is what you get from 3+4+5/3=4, 4 is the mean.
I thought of something like this:
[row,col] = find(A(:,2),1,'first')
and then I would do 1/numberInThatIndex or something similar, since it's a S-matrix (pretty sure it's called that).
And I would iterate column by column but for some reason it's not giving me the first number in each column, if I do find(A(:,1),1,'first') it does give me the first in the first column, but not in the second if I change it to A(:,2).
I'd also need something to store that index to access the value, I thought of a 2xN vector but I guess it's not the best idea. I mean, find is going to give me index, but I need the value in that index, and then store that or show it. Not sure if I'm explaining myself properly but I'm trying, sorry about that.
Just to be clear both when I input A(:,1) and A(:,2) it gives me index from the first column, and I do not want that, I want first element found from each column, so I can calculate the mean out of the number in that index.
edit: allright it seems like that indeed does work, but when I was checking the results I was putting 3817 instead of 3871 that was the given answer and so I found a 0 when I wanted something that's not a zero. Not sure if I should delete all of this.
To solve your problem, you can do the following:
numberNonZerosPerColumn = sum(S~=0,1);
meanValue = nanmean(1./numberNonZerosPerColumn);
Count the number of nonzero elements in every column n(i)
Compute the values v(i) that are stored there, which are defined by v(i) := 1/n(i)
Take the mean of those values where n(i) is not zero (i.e. summing all those values, where v(i) is not NaN and divide by the number of columns that contain at least one zero)
If you want to treat columns without any nonzero entry as v(i):= 0, but still use them in your mean, you can use:
numberNonZerosPerColumn = sum(S~=0,1);
meanValue = nansum(1./numberNonZerosPerColumn)/size(S,2);
Suppose I have 121 elements and want to get all combinations of 4 elements taken at a time, i.e. 121c4.
Since combnk(1:121, 4) takes a lot of time, I want to go for 2% of that combination by providing:
z = 1:50:length(121c4(:, 1))
For example: 1st row, 5th row, 100th row and so on, up to 121c4, picking only those rows from a 121c4 matrix without generating the complete combination (it's consuming too much for large numbers like 625c4).
If you haven't defined an ordering on the combinations, why not just use
randi(121,p,4)
where p is the number of combinations you want in your set ? With this approach you may, or may not, want to replace duplicates.
If you have defined an ordering on the combinations, tell us what it is.