Related
I have this piece of code
N=10^4;
for i = 1:N
[E,X,T] = fffun(); % Stochastic simulation. Returns every time three different vectors (whose length is 10^3).
X_(i,:)=X;
T_(i,:)=T;
GRID=[GRID T];
end
GRID=unique(GRID);
% Second part
for i=1:N
for j=1:(kmax)
f=find(GRID==T_(i,j) | GRID==T_(i,j+1));
s=f(1);
e=f(2)-1;
counter(X_(i,j), s:e)=counter(X_(i,j), s:e)+1;
end
end
The code performs N different simulations of a stochastic process (which consists of 10^3 events, occurring at discrete moments (T vector) that depends on the specific simulation.
Now (second part) I want to know, as a function of time istant, how many simulations are in a particular state (X assumes value between 1 and 10). The idea I had: create a grid vector with all the moments at which something happens in any simulation. Then, looping over the simulations, loop over the timesteps in which something happens and incrementing all the counter indeces that corresponds to this particular slice of time.
However this second part is very heavy (I mean days of processing on a standard quad-core CPU). And it shouldn't.
Are there any ideas (maybe about comparing vectors in a more efficient way) to cut the CPU time?
This is a standalone 'second_part'
N=5000;
counter=zeros(11,length(GRID));
for i=1:N
disp(['Counting sim #' num2str(i)]);
for j=1:(kmax)
f=find(GRID==T_(i,j) | GRID==T_(i,j+1),2);
s=f(1);
e=f(2)-1;
counter(X_(i,j), s:e)=counter(X_(i,j), s:e)+1;
end
end
counter=counter/N;
stop=find(GRID==Tmin);
stop=stop-1;
plot(counter(:,(stop-500):stop)')
with associated dummy data ( filedropper.com/data_38 ). In the real context the matrix has 2x rows and 10x columns.
Here is what I understand:
T_ is a matrix of time steps from N simulations.
X_ is a matrix of simulation state at T_ in those simulations.
so if you do:
[ut,~,ic]= unique(T_(:));
you get ic which is a vector of indices for all unique elements in T_. Then you can write:
counter = accumarray([ic X_(:)],1);
and get counter with no. of rows as your unique timesteps, and no. of columns as the unique states in X_ (which are all, and must be, integers). Now you can say that for each timestep ut(k) the number of time that the simulation was in state m is counter(k,m).
In your data, the only combination of m and k that has a value greater than 1 is (1,1).
Edit:
From the comments below, I understand that you record all state changes, and the time steps when they occur. Then every time a simulation change a state you want to collect all the states from all simulations and count how many states are from each type.
The main problem here is that your time is continuous, so basically each element in T_ is unique, and you have over a million time steps to loop over. Fully vectorizing such a process will need about 80GB of memory which will probably stuck your computer.
So I looked for a combination of vectorizing and looping through the time steps. We start by finding all unique intervals, and preallocating counter:
ut = unique(T_(:));
stt = 11; % no. of states
counter = zeros(stt,numel(ut));r = 1:size(T_,1);
r = 1:size(T_,1); % we will need that also later
Then we loop over all element in ut, and each time look for the relevant timestep in T_ in all simulations in a vectorized way. And finally we use histcounts to count all the states:
for k = 1:numel(ut)
temp = T_<=ut(k); % mark all time steps before ut(k)
s = cumsum(temp,2); % count the columns
col_ind = s(:,end); % fins the column index for each simulation
% convert the coulmns to linear indices:
linind = sub2ind(size(T_),r,col_ind.');
% count the states:
counter(:,k) = histcounts(X_(linind),1:stt+1);
end
This takes about 4 seconds at my computer for 1000 simulations, so it adds to a little more than one hour for the whole process. Not very quick...
You can try also one or two of the tweaks below to squeeze run time a little bit more:
As you can read here, accumarray seems to work faster in small arrays then histcouns. So may want to switch to it.
Also, computing linear indices directly is a quicker method than sub2ind, so you may want to try that.
implementing these suggestions in the loop above, we get:
R = size(T_,1);
r = (1:R).';
for k = 1:K
temp = T_<=ut(k); % mark all time steps before ut(k)
s = cumsum(temp,2); % count the columns
col_ind = s(:,end); % fins the column index for each simulation
% convert the coulmns to linear indices:
linind = R*(col_ind-1)+r;
% count the states:
counter(:,k) = accumarray(X_(linind),1,[stt 1]);
end
In my computer switching to accumarray and or removing sub2ind gain a slight improvement but it was not consistent (using timeit for testing on 100 or 1K elements in ut), so you better test it yourself. However, this still remains very long.
One thing that you may want to consider is trying to discretize your timesteps, so you will have much less unique elements to loop over. In your data about 8% of the time intervals a smaller than 1. If you can assume that this is short enough to be treated as one time step, then you could round your T_ and get only ~12.5K unique elements, which take about a minute to loop over. You can do the same for 0.1 intervals (which are less than 1% of the time intervals), and get 122K elements to loop over, what will take about 8 hours...
Of course, all the timing above are rough estimates using the same algorithm. If you do choose to round the times there may be even better ways to solve this.
I would like to identify the largest possible contiguous subsample of a large data set. My data set consists of roughly 15,000 financial time series of up to 360 periods in length. I have imported the data into MATLAB as a 360 by 15,000 numerical matrix.
This matrix contains a lot of NaNs due to some of the financial data not being available for the entire period. In the illustration, NaN entries are shown in dark blue, and non-NaN entries appear in light blue. It is these light blue non-NaN entries which I would like to ideally combine into an optimal subsample.
I would like to find the largest possible contiguous block of data that is contained in my matrix, while ensuring that my matrix contains a sufficient number of periods.
In a first step I would like to sort my matrix from left to right in descending order by the number of non-NaN entries in each column, that is, I would like to sort by the vector obtained by entering sum(~isnan(data),1).
In a second step I would like to find the sub-array of my data matrix that is at least 72 entries along the first dimension and is otherwise as large as possible, measured by the total number of entries.
What is the best way to implement this?
A big warning (may or may not apply depending on context)
As Oleg mentioned, when an observation is missing from a financial time series, it's often missing for reason: eg. the entity went bankrupt, the entity was delisted, or the instrument did not trade (i.e. illiquid). Constructing a sample without NaNs is likely equivalent to constructing a sample where none of these events occur!
For example, if this were hedge fund return data, selecting a sample without NaNs would exclude funds that blew up and ceased trading. Excluding imploded funds would bias estimates of expected returns upwards and estimates of variance or covariance downwards.
Picking a sample period with the fewest time series with NaNs would also exclude periods like the 2008 financial crisis, which may or may not make sense. Excluding 2008 could lead to an underestimate of how haywire things could get (though including it could lead to overestimate the probability of certain rare events).
Some things to do:
Pick a sample period as long as possible but be aware of the limitations.
Do your best to handle survivorship bias: eg. if NaNs represent delisting events, try to get some kind of delisting return.
You almost certainly will have an unbalanced panel with missing observations, and your algorithm will have to be deal with that.
Another general finance / panel data point, selecting a sample at some time point t and then following it into the future is perfectly ok. But selecting a sample based upon what happens during or after the sample period can be incredibly misleading.
Code that does what you asked:
This should do what you asked and be quite fast. Be aware of the problems though if whether an observation is missing is not random and orthogonal to what you care about.
Inputs are a T by n sized matrix X:
T = 360; % number of time periods (i.e. rows) in X
n = 15000; % number of time series (i.e. columns) in X
T_subsample = 72; % desired length of sample (i.e. rows of newX)
% number of possible starting points for series of length T_subsample
nancount_periods = T - T_subsample + 1;
nancount = zeros(n, nancount_periods, 'int32'); % will hold a count of NaNs
X_isnan = int32(isnan(X));
nancount(:,1) = sum(X_isnan(1:T_subsample, :))'; % 'initialize
% We need to obtain a count of nans in T_subsample sized window for each
% possible time period
j = 1;
for i=T_subsample + 1:T
% One pass: add new period in the window and subtract period no longer in the window
nancount(:,j+1) = nancount(:,j) + X_isnan(i,:)' - X_isnan(j,:)';
j = j + 1;
end
indicator = nancount==0; % indicator of whether starting_period, series
% has no NaNs
% number of nonan series of length T_subsample by starting period
max_subsample_size_by_starting_period = sum(indicator);
max_subsample_size = max(max_subsample_size_by_starting_period);
% find the best starting period
starting_period = find(max_subsample_size_by_starting_period==max_subsample_size, 1);
ending_period = starting_period + T_subsample - 1;
columns_mask = indicator(:,starting_period);
columns = find(columns_mask); %holds the column ids we are using
newX = X(starting_period:ending_period, columns_mask);
Here's an idea,
Assuming you can rearrange the series, calculate the distance (you decide the metric, but if looking at is nan vs not is nan, Hamming is ok).
Now hierarchically cluster the series and rearrange them using either a dendrogram
or http://www.mathworks.com/help/bioinfo/examples/working-with-the-clustergram-function.html
You should probably prune any series that doesn't have a minimum number of non nan values before you start.
First I have only little insight in financial mathematics. I understood it that you want to find the longest continuous chain of non-NaN values for each time series. The time series should be sorted depending on the length of this chain and each time series, not containing a chain above a threshold, discarded. This can be done using
data = rand(360,15e3);
data(abs(data) <= 0.02) = NaN;
%% sort and chop data based on amount of consecutive non-NaN values
binary_data = ~isnan(data);
% find edges, denote their type and calculate the biggest chunk in each
% column
edges = [2*binary_data(1,:)-1; diff(binary_data, 1)];
chunk_size = diff(find(edges));
chunk_size(end+1) = numel(edges)-sum(chunk_size);
[row, ~, id] = find(edges);
num_row_elements = diff(find(row == 1));
num_row_elements(end+1) = numel(chunk_size) - sum(num_row_elements);
%a chunk of NaN has a -1 in id, a chunk of non-NaN a 1
chunks_per_row = mat2cell(chunk_size .* id,num_row_elements,1);
% sort by largest consecutive block of non-NaNs
max_size = cellfun(#max, chunks_per_row);
[max_size_sorted, idx] = sort(max_size, 'descend');
data_sorted = data(:,idx);
% remove all elements that only have block sizes smaller then some number
some_number = 20;
data_sort_chop = data_sorted(:,max_size_sorted >= some_number);
Note that this can be done a lot simpler, if the order of periods within a time series doesn't matter, aka data([1 2 3],id) and data([3 1 2], id) are identical.
What I do not know is, if you want to discard all periods within a time series that don't correspond to the biggest value, get all those chains as individual time series, ...
Feel free to drop a comment if it has to be more specific.
I am accessing 10 images from a folder "c1" and I have query image. I have implemented code for loading images in cell array and then I'm calculating histogram intersection between query image and each image from folder "c1" one-by-one. Now i want to draw precision-recall curve but i am not sure how to write code for getting "precision-recall curve" using the data obtained from histogram intersection.
My code:
Inp1=rgb2gray(imread('D:\visionImages\c1\1.ppm'));
figure, imshow(Inp1), title('Input image 1');
srcFiles = dir('D:\visionImages\c1\*.ppm'); % the folder in which images exists
for i = 1 : length(srcFiles)
filename = strcat('D:\visionImages\c1\',srcFiles(i).name);
I = imread(filename);
I=rgb2gray(I);
Seq{i}=I;
end
for i = 1 : length(srcFiles) % loop for calculating histogram intersections
A=Seq{i};
B=Inp1;
a = size(A,2); b = size(B,2);
K = zeros(a, b);
for j = 1:a
Va = repmat(A(:,j),1,b);
K(j,:) = 0.5*sum(Va + B - abs(Va - B));
end
end
Precision-Recall graphs measure the accuracy of your image retrieval system. They're also used in the performance of any search engine really, like text or documents. They're also used in machine learning evaluation and performance, though ROC Curves are what are more commonly used.
Precision-Recall graphs are more suitable for document and data retrieval. For the case of images here, given your query image, you are measuring how similar this image is with the rest of the images in your database. You then have similarity measures for each of the database images in relation to your query image and then you sort these similarities in descending order. For a good retrieval system, you would want the images that are the most relevant (i.e. what you are searching for) to all appear at the beginning, while the irrelevant images would appear after.
Precision
The definition of Precision is the ratio of the number of relevant images you have retrieved to the total number of irrelevant and relevant images retrieved. In other words, supposing that A was the number of relevant images retrieved and B was the total number of irrelevant images retrieved. When calculating precision, you take a look at the first several images, and this amount is A + B, as the total number of relevant and irrelevant images is how many images you are considering at this point. As such, another definition Precision is defined as the ratio of how many relevant images you have retrieved so far out of the bunch that you have grabbed:
Precision = A / (A + B)
Recall
The definition of Recall is slightly different. This evaluates how many of the relevant images you have retrieved so far out of a known total, which is the the total number of relevant images that exist. As such, let's say you again take a look at the first several images. You then determine how many relevant images there are, then you calculate how many relevant images that have been retrieved so far out of all of the relevant images in the database. This is defined as the ratio of how many relevant images you have retrieved overall. Supposing that A was again the total number of relevant images you have retrieved out of a bunch you have grabbed from the database, and C represents the total number of relevant images in your database. Recall is thus defined as:
Recall = A / C
How you calculate this in MATLAB is actually quite easy. You first need to know how many relevant images are in your database. After, you need to know the similarity measures assigned to each database image with respect to the query image. Once you compute these, you need to know which similarity measures map to which relevant images in your database. I don't see that in your code, so I will leave that to you. Once you do this, you then sort on the similarity values then you go through where in the sorted similarity values these relevant images occur. You then use these to calculate your precision and recall.
I'll provide a toy example so I can show you what the graph looks like as it isn't quite clear on how you're calculating your similarities here. Let's say I have 5 images in a database of 20, and I have a bunch of similarity values between them and a query image:
rng(123); %// Set seed for reproducibility
num_images = 20;
sims = rand(1,num_images);
sims =
Columns 1 through 13
0.6965 0.2861 0.2269 0.5513 0.7195 0.4231 0.9808 0.6848 0.4809 0.3921 0.3432 0.7290 0.4386
Columns 14 through 20
0.0597 0.3980 0.7380 0.1825 0.1755 0.5316 0.5318
Also, I know that images [1 5 7 9 12] are my relevant images.
relevant_IDs = [1 5 7 9 12];
num_relevant_images = numel(relevant_IDs);
Now let's sort the similarity values in descending order, as higher values mean higher similarity. You'd reverse this if you were calculating a dissimilarity measure:
[sorted_sims, locs] = sort(sims, 'descend');
locs will now contain the image ranks that each image ranked as. Specifically, these tell you which position in similarity the image belongs to. sorted_sims will have the similarities sorted in descending order:
sorted_sims =
Columns 1 through 13
0.9808 0.7380 0.7290 0.7195 0.6965 0.6848 0.5513 0.5318 0.5316 0.4809 0.4386 0.4231 0.3980
Columns 14 through 20
0.3921 0.3432 0.2861 0.2269 0.1825 0.1755 0.0597
locs =
7 16 12 5 1 8 4 20 19 9 13 6 15 10 11 2 3 17 18 14
Therefore, the 7th image is the highest ranked image, followed by the 16th image being the second highest image and so on. What you need to do now is for each of the images that you know are relevant, you need to figure out where these are located after sorting. We will go through each of the image IDs that we know are relevant, and figure out where these are located in the above locations array:
locations_final = arrayfun(#(x) find(locs == x, 1), relevant_IDs)
locations_final =
5 4 1 10 3
Let's sort these to get a better understand of what this is saying:
locations_sorted = sort(locations_final)
locations_sorted =
1 3 4 5 10
These locations above now tell you the order in which the relevant images will appear. As such, the first relevant image will appear first, the second relevant image will appear in the third position, the third relevant image appears in the fourth position and so on. These precisely correspond to part of the definition of Precision. For example, in the last position of locations_sorted, it would take ten images to retrieve all of the relevant images (5) in our database. Similarly, it would take five images to retrieve four relevant images in the database. As such, you would compute precision like this:
precision = (1:num_relevant_images) ./ locations_sorted;
Similarly for recall, it's simply the ratio of how many images were retrieved so far from the total, and so it would just be:
recall = (1:num_relevant_images) / num_relevant_images;
Your Precision-Recall graph would now look like the following, with Recall on the x-axis and Precision on the y-axis:
plot(recall, precision, 'b.-');
xlabel('Recall');
ylabel('Precision');
title('Precision-Recall Graph - Toy Example');
axis([0 1 0 1.05]); %// Adjust axes for better viewing
grid;
This is the graph I get:
You'll notice that between a recall ratio of 0.4 to 0.8 the precision is increasing a bit. This is because you have managed to retrieve a successive chain of images without touching any of the irrelevant ones, and so your precision will naturally increase. It goes way down after the last image, as you've had to retrieve so many irrelevant images before finally hitting a relevant image.
You'll also notice that precision and recall are inversely related. As such, if precision increases, then recall decreases. Similarly, if precision decreases, then recall will increase.
The first part makes sense because if you don't retrieve that many images in the beginning, you have a greater chance of not including irrelevant images in your results but at the same time, the amount of relevant images is rather small. This is why recall would decrease when precision would increase
The second part also makes sense because as you keep trying to retrieve more images in your database, you'll inevitably be able to retrieve all of the relevant ones, but you'll most likely start to include more irrelevant images, which would thus drive your precision down.
In an ideal world, if you had N relevant images in your database, you would want to see all of these images in the top N most similar spots. As such, this would make your precision-recall graph a flat horizontal line hovering at y = 1, which means that you've managed to retrieve all of your images in all of the top spots without accessing any irrelevant images. Unfortunately, that's never going to happen (or at least not for now...) as trying to figure out the best features for CBIR is still an on-going investigation, and no image search engine that I have seen has managed to get this perfect. This is still one of the most broadest and unsolved computer vision problems that exist today!
Edit
You retrieved this code to compute histogram intersection from this post. They have a neat way of computing histogram intersection as:
n is the total number of bins in your histogram. You'll have to play around with this to get good results, but we can leave that as a parameter in your code. The code above assumes that you have two matrices A and B where each column is a histogram. You'll generate a matrix that is of a x b, where a is the number of columns in A and b is the number of columns in b. The row and column of this matrix (i,j) tells you the similarity between the ith column in A with the b jth column in B. In your case, A would be a single column which denotes the histogram of your query image. B would be a 10 column matrix that denotes the histograms for each of the database images. Therefore, we will get a 1 x 10 array of similarity measures through histogram intersection.
As such, we need to modify your code so that you're using imhist for each of the images. We can also specify an additional parameter that gives you how many bins each histogram will have. Therefore, your code will look like this. Each new line that I have placed will have a %// NEW comment beside each line.
Inp1=rgb2gray(imread('D:\visionImages\c1\1.ppm'));
figure, imshow(Inp1), title('Input image 1');
num_bins = 32; %// NEW - I'm specifying 32 bins here. Play around with this yourself
A = imhist(Inp1, num_bins); %// NEW - Calculate histogram
srcFiles = dir('D:\visionImages\c1\*.ppm'); % the folder in which images exists
B = zeros(num_bins, length(srcFiles)); %// NEW - Store histograms of database images
for i = 1 : length(srcFiles)
filename = strcat('D:\visionImages\c1\',srcFiles(i).name);
I = imread(filename);
I=rgb2gray(I);
B(:,i) = imhist(I, num_bins); %// NEW - Put each histogram in a separate
%// column
end
%// NEW - Taken directly from the website
%// but modified for only one histogram in `A`
b = size(B,2);
Va = repmat(A, 1, b);
K = 0.5*sum(Va + B - abs(Va - B));
Take note that I have copied the code from the website, but I have modified it because there is only one image in A and so there is some code that isn't necessary.
K should now be a 1 x 10 array of histogram intersection similarities. You would then use K and assign sims to this variable (i.e. sims = K;) in the code I have written above, then run through your images. You also need to know which images are relevant images, and you'd have to change the code I've written to reflect that.
Hope this helps!
I have production (q) values from 4 different methods stored in the 4 matrices. Each of the 4 matrices contains q values from a different method as:
Matrix_1 = 1 row x 20 column
Matrix_2 = 100 rows x 20 columns
Matrix_3 = 100 rows x 20 columns
Matrix_4 = 100 rows x 20 columns
The number of columns indicate the number of years. 1 row would contain the production values corresponding to the 20 years. Other 99 rows for matrix 2, 3 and 4 are just the different realizations (or simulation runs). So basically the other 99 rows for matrix 2,3 and 4 are repeat cases (but not with exact values because of random numbers).
Consider Matrix_1 as the reference truth (or base case ). Now I want to compare the other 3 matrices with Matrix_1 to see which one among those three matrices (each with 100 repeats) compares best, or closely imitates, with Matrix_1.
How can this be done in Matlab?
I know, manually, that we use confidence interval (CI) by plotting the mean of Matrix_1, and drawing each distribution of mean of Matrix_2, mean of Matrix_3 and mean of Matrix_4. The largest CI among matrix 2, 3 and 4 which contains the reference truth (or mean of Matrix_1) will be the answer.
mean of Matrix_1 = (1 row x 1 column)
mean of Matrix_2 = (100 rows x 1 column)
mean of Matrix_3 = (100 rows x 1 column)
mean of Matrix_4 = (100 rows x 1 column)
I hope the question is clear and relevant to SO. Otherwise please feel free to edit/suggest anything in question. Thanks!
EDIT: My three methods I talked about are a1, a2 and a3 respectively. Here's my result:
ci_a1 =
1.0e+008 *
4.084733001497999
4.097677503988565
ci_a2 =
1.0e+008 *
5.424396063219890
5.586301025525149
ci_a3 =
1.0e+008 *
2.429145282593182
2.838897116739112
p_a1 =
8.094614835195452e-130
p_a2 =
2.824626709966993e-072
p_a3 =
3.054667629953656e-012
h_a1 = 1; h_a2 = 1; h_a3 = 1
None of my CI, from the three methods, includes the mean ( = 3.454992884900722e+008) inside it. So do we still consider p-value to choose the best result?
If I understand correctly the calculation in MATLAB is pretty strait-forward.
Steps 1-2 (mean calculation):
k1_mean = mean(k1);
k2_mean = mean(k2);
k3_mean = mean(k3);
k4_mean = mean(k4);
Step 3, use HIST to plot distribution histograms:
hist([k2_mean; k3_mean; k4_mean]')
Step 4. You can do t-test comparing your vectors 2, 3 and 4 against normal distribution with mean k1_mean and unknown variance. See TTEST for details.
[h,p,ci] = ttest(k2_mean,k1_mean);
EDIT : I misinterpreted your question. See the answer of Yuk and following comments. My answer is what you need if you want to compare distributions of two vectors instead of a vector against a single value. Apparently, the latter is the case here.
Regarding your t-tests, you should keep in mind that they test against a "true" mean. Given the number of values for each matrix and the confidence intervals it's not too difficult to guess the standard deviation on your results. This is a measure of the "spread" of your results. Now the error on your mean is calculated as the standard deviation of your results divided by the number of observations. And the confidence interval is calculated by multiplying that standard error with appx. 2.
This confidence interval contains the true mean in 95% of the cases. So if the true mean is exactly at the border of that interval, the p-value is 0.05 the further away the mean, the lower the p-value. This can be interpreted as the chance that the values you have in matrix 2, 3 or 4 come from a population with a mean as in matrix 1. If you see your p-values, these chances can be said to be non-existent.
So you see that when the number of values get high, the confidence interval becomes smaller and the t-test becomes very sensitive. What this tells you, is nothing more that the three matrices differ significantly from the mean. If you have to choose one, I'd take a look at the distributions anyway. Otherwise the one with the closest mean seems a good guess. If you want to get deeper into this, you could also ask on stats.stackexchange.com
Your question and your method aren't really clear :
Is the distribution equal in all columns? This is important, as two distributions can have the same mean, but differ significantly :
is there a reason why you don't use the Central Limit Theorem? This seems to me like a very complex way of obtaining a result that can easily be found using the fact that the distribution of a mean approaches a normal distribution where sd(mean) = sd(observations)/number of observations. Saves you quite some work -if the distributions are alike! -
Now if the question is really the comparison of distributions, you should consider looking at a qqplot for a general idea, and at a 2-sample kolmogorov-smirnov test for formal testing. But please read in on this test, as you have to understand what it does in order to interprete the results correctly.
On a sidenote : if you do this test on multiple cases, make sure you understand the problem of multiple comparisons and use the appropriate correction, eg. Bonferroni or Dunn-Sidak.
just wondering if anyone has any ideas about an issue I'm having.
I have a fair amount of data that needs to be displayed on one graph. Two theoretical lines that are bold and solid are displayed on top, then 10 experimental data sets that converge to these lines are graphed, each using a different identifier (eg the + or o or a square etc). These graphs are on a log scale that goes up to 1e6. The first few decades of the graph (< 1e3) look fine, but as all the datasets converge (> 1e3) it's really difficult to see what data is what.
There's over 1000 data points points per decade which I can prune linearly to an extent, but if I do this too much the lower end of the graph will suffer in resolution.
What I'd like to do is prune logarithmically, strongest at the high end, working back to 0. My question is: how can I get a logarithmically scaled index vector rather than a linear one?
My initial assumption was that as my data is lenear I could just use a linear index to prune, which lead to something like this (but for all decades):
//%grab indicies per decade
ind12 = find(y >= 1e1 & y <= 1e2);
indlow = find(y < 1e2);
indhigh = find(y > 1e4);
ind23 = find(y >+ 1e2 & y <= 1e3);
ind34 = find(y >+ 1e3 & y <= 1e4);
//%We want ind12 indexes in this decade, find spacing
tot23 = round(length(ind23)/length(ind12));
tot34 = round(length(ind34)/length(ind12));
//%grab ones to keep
ind23keep = ind23(1):tot23:ind23(end);
ind34keep = ind34(1):tot34:ind34(end);
indnew = [indlow' ind23keep ind34keep indhigh'];
loglog(x(indnew), y(indnew));
But this causes the prune to behave in a jumpy fashion obviously. Each decade has the number of points that I'd like, but as it's a linear distribution, the points tend to be clumped at the high end of the decade on the log scale.
Any ideas on how I can do this?
I think the easiest way to do this would be to use the LOGSPACE function to generate a set of indices into your data. For example, to create a set of 100 points logarithmically spaced from 1 to N (the number of points in your data), you can try the following:
indnew = round(logspace(0,log10(N),100)); %# Create the log-spaced index
indnew = unique(indnew); %# Remove duplicate indices
loglog(x(indnew),y(indnew)); %# Plot the indexed data
Creating a logarithmically-spaced index like this will result in fewer values being chosen from the end of the vector relative to the start, thus pruning values more severely towards the end of the vector and improving the appearance of the log plot. It would therefore be most effective with vectors that are sorted in ascending order.
The way I understand the problem is that your x-values are linearly spaced, so that if you plot them logarithmically, there are way more data points in 'higher' decades, so that markers lie extremely close to one another. For example, if x goes from 1 to 1000, there are 10 points in the first decade 90 in the second, and 900 in the third. You want to have, say, 3 points per decade instead.
I see two ways to solve the problem. The easier one is to use differently colored lines instead of different markers. Thus, you don't sacrifice any data points, and you can still distinguish everything.
The second solution is to create an unevenly spaced index. Here's how you can do that.
%# create some data
x = 1:1000;
y = 2.^x;
%# plot the graph and see the dots 'coalesce' very quickly
figure,loglog(x,y,'.')
%# for the example, I use a step size of 0.7, which is `log(1)`
xx = 0.7:0.7:log(x(end)); %# this is where I want the data to be plotted
%# find the indices where we want to plot by finding the closest `log(x)'-values
%# run unique to avoid multiples of the same index
indnew = unique(interp1(log(x),1:length(x),xx,'nearest'));
%# plot with fewer points
figure,loglog(x(indnew),y(indnew),'.')