I have two columns in .std file. I want average of the second column values corresponding to all values ranging from some value (eg. 1.0- 1.9) in first column how can I program in Matlab?
Say, a is the name of your two column matrix. If you want to find all of the values in the first column in the range of 1.0 - 1.9 and then use those entries to find the mean in the second column you can do this:
f = find(a(:,1)>=1 & a(:,1)<=1.9)
m = mean(a(f,2))
find will find the values that lie within this range and return the index, and a(f,2) accesses those indices in the in the second column and takes the mean. You can also do it with one line like so:
m = mean(a((a(:,1)>=1 & a(:,1)<=1.9),2))
Related
In Matlab, I am given a table with two columns. Now I want to find the corresponding value of the left column to the maximum value of the right column:
P1_sat = P1(ismember(P2,max(P2)))
This works, however, the maximum is identical at 3 values of the left column, P1. These 3 values are right next to each other. So I want to consider the mid value. Is there a "consider the mid value" - command?
Adding the following code line will complete my goal:
P1_sat = sum(P1_sat,1) / length(P1_sat)
The three different values of P1, whose corresponding value of P2 is the maximum of P2, are added and then divided by 3. This gives the average value, which is also the mid point.
I am having matrix with approx 3000 rows(changing) and 3 columns.
I have count of both rows and columns.
I am trying to plot the graph:
x=1:3000;
plot(matrix(x,1))
is there any way that I can include all rows in the plot instruction itself so that I can remove 'x=1:3000' ?
Also, I want to divide, 1st column of matrix which have 3000 rows into another matrix of 3 columns each with 1000 rows. Any specific instruction for this ?
I have made for loop for this and then i am placing individually the elements in the new array. But its taking long time.
As to the plotting issue, using the colon operator will plot all rows for your desired column:
plot(matrix(:,1));
EDIT: You mentioned you were a beginner. In case you haven't seen the colon operator used like this before, a colon operator all by itself when indexing into a matrix essentially means "all __", either "all rows" if in the first position or "all columns" if in the second position.
As for the second question, of splitting one column into a new matrix with multiple columns, you can use the reshape() function, which takes the input matrix to be reshaped and a number of output rows and columns. For example, to split the first column of matrix into 3 columns and put them into newMatrix, use the following:
newMatrix = reshape(matrix(:,1),[],3);
Note that the above code uses [] in the second argument (the number of rows argument) to mean "automatically determine number of rows".This is automatically determined based on the number of columns, which is defined in the third argument here as 3. The reshape function requires that the number of output rows * output columns be equal to input rows * input columns. So in the above case this will only work if the starting matrix has a number of rows which is divisible by 3.
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
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);
I have created a vector containing zeros and 1's using the following command in a for loop.
G(:,i)=rand(K,1)<rand;
Since this is part of a larger problem at a particular stage I need to count the number of 1's that are present in each column.
I have tried to find the count using a for loop which is very messy and takes too long.
I found that histc can be used for this but I get an error
histc(G(:,1),1)
First input must be non-sparse numeric array.
Is there a better way to do this or am I missing something here ?
If you have a matrix G containing zeroes and ones, and you want to know how many ones are in each column, all you need is SUM:
nZeroes = sum(G);
This will give you a vector containing a total for each column in G.