Extracting data from histogram plots

Hello. I'm trying to process some data from some chemical analyses I did a while ago. I have 3 types of data: particle diameter, nitrogen content (%), and sulfur content (%). I've already managed to organize the particle diameter data into a histogram plot with something like 50 bins. Now, I'd like to figure out the average nitrogen and sulfur content of the particles in each bin. I'm not sure how to do this, though, and I haven't found any obvious tutorials to explain how to do this. Any advice?

Answer 1

[Adam Danz]

3 methods to group data and compute mean for each group

Each method deals with empty bins differently.

discretize + splitapply

Use discretize to group each value into the bins used in histogram and then splitapply to compute the mean for each group. Note that each bin must contain at least one data point.

Example: compute the mean of data in bins defined by edges.

rng default  % for reproducibility of this demo
data = rand(1,100)*100;
edges = 0:10:100; 
binID = discretize(data,edges)

binID = 1×100

9 10 2 10 7 1 3 6 10 10 2 10 10 5 9 2 5 10 8 10 7 1 9 10 7 8 8 4 7 2

a = splitapply(@mean,data,binID)

a = 1×10

5.3838 15.2780 26.0259 35.6310 46.5284 55.4195 66.1338 75.5438 83.3041 94.1885

discretize + groupsummary

Use discretize to group each value into the bins and then groupsummary to compute the mean of each group. When working with vectors, the first two arguments must be column vectors.

Note that the output vector skips empty bins. See additional outputs to groupsummary to identify which bins are represented in the first output.

s = groupsummary(data(:),binID(:),'mean')

s = 10×1

5.3838 15.2780 26.0259 35.6310 46.5284 55.4195 66.1338 75.5438 83.3041 94.1885

discretize + accumarray

Use discretize to group each value into the bins and then accumarray to compute the mean of all bins.

Note that empty bins are represented by a 0.

m = accumarray(binID(:),data,[],@mean)

m = 10×1

5.3838 15.2780 26.0259 35.6310 46.5284 55.4195 66.1338 75.5438 83.3041 94.1885

Comparison of these methods when some bins are empty

data = randn(100,1)+10;  % expected range: ~6 : ~13
edges = 0:3:15;  % 5 bins but the first two will be empty
binID = discretize(data, edges);

m = accumarray(binID,data,[],@mean)

m = 5×1

0 0 8.4699 10.1766 12.7170

s = groupsummary(data,binID(:),'mean')

s = 3×1

8.4699 10.1766 12.7170

a = splitapply(@mean,data,binID)

Error using splitapply
For N groups, every integer between 1 and N must occur at least once in the vector of group numbers.

[Haley Royer]

Is there any way to discretize based on a lograrithmic scale? My data is organized so that the smallest bins have the smallest size (e.g., bin 1 ranges from 0.1 to 0.175 microns) and the largest bins have the largest size (e.g., bin 60 ranges from 7.499 to 8.058 microns). Each bin size is different and increases in size from the smallest to the largest bin.

[Adam Danz]

Yes, the second input argument for discretize is a vector of bin edges which you can define any way you'd like. If you're happy with your histogram bin edges, you can use the same edges to discretize your data.

[Torsten]

If you had read the documentation of "discretize", you would know that the parameter "edges" is exactly what you are asking for.

[Haley Royer]

I already read the discretize documentation. I just missed the edges parameter. No need to be a dick about it.

[Haley Royer]

Thank you, Adam. That's helpful advice. I think I've got what I need now. I appreciate the guidance.

[Haley Royer]

Hi again. I've run into another issue. When trying to use splitapply I get the following error

"Group numbers must be a vector of positive integers, and cannot be a sparse vector."

My understanding is that because I have values in my column that are zero, splitapply cannot be used. Some of the particles I'm looking at don't have nitrogen or sulfur, but I still have to average a group such as 0 0 0 5.0 2.5. Any way to get around this?

[Adam Danz]

Let's keep it civil here.

As you mentioned, if one of the bins have no values, then splitapply won't work.

I'll add alternatives to my answer.