Histogram Only Plotting 1 Point

Hello,

I have a 1x100 table.

If I print 'error', it does the math, by taking the field 'constant' from each row of the table and subtracts by 94, and then prints all 100 results.

This is working correctly, but the plotting is not.

When I go to create the histogram of the 100 results, it only prints 1. Any idea why?

Thank you!

Answer 1

[Image Analyst]

Don't use error as a variable name since it's a built-in function. Also, what is the badly-named "constant"? Is it an array or table? If it's a table, you need to extract the column and subtract 94 from it. Why are you using a loop when nothing inside the loop depends on the badly-named "i"?

This works just fine:

% Create sample data.
% column1 = rand(1000, 1);
column2 = rand(1000, 1);
t = table(column1, column2);
% Now that we have our data, let's get the histogram of one of the columns
theColumn = t{:, 2} - 94; % Extract column 2 and subtract 94
% Take the histogram of it
h = histogram(theColumn);
grid on;
xlabel('Value', 'FontSize', 20);
ylabel('Count', 'FontSize', 20);

[Sclay748]

Sorry, It is because this code is for work, so I can't share the table or actual file, so I tried to recreate it with simple names for the sake of asking for help.

I'll give yours a shot though.

[Sclay748]

I use a loop because it would only print one calulation out of 100 if I got rid of it.

[Image Analyst]

Yeah but the badly-named error is not error(i) so the same error scalar is just getting overwritten every single time. After the loop you have only one single value, so you will have only one histogram bin. But like I said, if you have a table, you can extract the column all in one shot and can pass that to histogram() and not even use a loop at all.

Answer 2

[Binbin Qi]

I think you can use bar, not histogram

bar(error)