Histogram Only Plotting 1 Point

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);

3 个评论
显示 1更早的评论隐藏 1更早的评论

Sclay748 2020-8-18

编辑：Sclay748 2020-8-18

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

Image Analyst 2020-8-19

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.

请先登录，再进行评论。