# Histogram Only Plotting 1 Point

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Sclay748 2020 年 8 月 18 日
コメント済み: Image Analyst 2020 年 8 月 19 日
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!

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### 採用された回答

Image Analyst 2020 年 8 月 18 日

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 件のコメント表示非表示 2 件の古いコメント
Image Analyst 2020 年 8 月 19 日
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.

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### その他の回答 (1 件)

Binbin Qi 2020 年 8 月 18 日
I think you can use bar, not histogram
bar(error)

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