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Suppose I have some people and they are of different weight, e.g.

weight in Kq: 40-45 45-50 50-55 55-60 60-65

No. of Persons: 4 12 13 6 5

I want to represent the weight on x-axis and No. of persons on y-axis. This will tell us that there are 4 persons whose weight lies within 40-45 and 12 persons whose weight lies in the interval 45-50 kilo and so on. How can we do that?

Adam Danz
on 21 Oct 2020

Edited: Adam Danz
on 21 Oct 2020

For continuous intervals

Since your intervals are continuous, you should be using histogram() instead of bar().

kg = 40:5:65;

n = [4 12 13 6 5];

histogram('BinEdges',kg,'BinCounts',n)

xlabel('Weight (kg)')

ylabel('Number of participants')

For discountinuous or categorical intervals

For any future visitors who are not using continuous x-bins and want to label bin edges, follow this demo.

% Create data

bins = [10 20;

50 60;

75 85];

y = [20;10;30];

% Create bar plot

% The .5 specifies bin width, making room for labels

h = bar(y, .5);

% Get bar centers and bar widths

xCnt = h.XData + h.XOffset; % XOffset is undocumented!

width = h.BarWidth;

% Get x-val of bar-edges

barEdgesX = xCnt + width.*[-.5;.5];

% Set new xtick and xticklabels, rotate by 90deg.

ax = h.Parent; % axis handle, if you don't have it already

ax.XTick = barEdgesX(:);

ax.XTickLabel = string(reshape(bins',[],1));

ax.XTickLabelRotation = 90;

the cyclist
on 21 Oct 2020

For "discontinuous" bins, it might be simpler to use your original syntax, but with a zero bin count for empty bins.

For example,

kg = 40:5:75;

n = [4 0 12 13 0 6 5];

figure

histogram('BinEdges',kg,'BinCounts',n)

xlabel('Weight (kg)')

ylabel('Number of participants')

the cyclist
on 21 Oct 2020

Edited: the cyclist
on 21 Oct 2020

Here is one way:

w = 42.5 : 5.0 : 62.5;

n = [4 12 13 6 5];

bar(w,n,'BarWidth',1)

FYI, if you have the underlying individual weight data, rather than the bin counts, you will definitely want to use histogram as in Adam's solution (although the syntax will be different).

Sadiq Akbar
on 22 Oct 2020

Thank you very much to both of you Adam Danz and the cyclist.

Adam Danz I ran your 1st code. It works very well. But when I try it on the following data, it gives me error

I replaced Kg and n vectors by:

Kg=[2.E-03

1.E-03

4.E-04

3.E-04

3.E-04

2.E-04

2.E-04

2.E-04

2.E-04

2.E-04

2.E-04

1.E-04

1.E-04

9.E-05

8.E-05

8.E-05

8.E-05

8.E-05

8.E-05

6.E-05

5.E-05

4.E-05

3.E-05

2.E-05

2.E-05

2.E-05

2.E-05

1.E-05

1.E-05

1.E-05

1.E-05

9.E-06

9.E-06

7.E-06

7.E-06

6.E-06

5.E-06

5.E-06

4.E-06

4.E-06

2.E-06

2.E-06

1.E-06

1.E-06

1.E-06

4.E-07

3.E-07

3.E-07

2.E-07

2.E-07

2.E-07

2.E-07

2.E-07

2.E-07

2.E-07

1.E-07

1.E-07

1.E-07

7.E-08

6.E-08

4.E-08

1.E-08

1.E-08

9.E-09

4.E-09

2.E-09

2.E-10

3.E-11

7.E-12

8.E-13

7.E-13

6.E-13

5.E-13

3.E-13

9.E-14

1.E-14

5.E-15

5.E-17

2.E-22

5.E-23

2.E-23

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

];

n=1:100;

histogram('BinEdges',kg,'BinCounts',n)

xlabel('Weight (kg)')

ylabel('Number of participants')

But it gives me the following error:

>> histogramByAdamDanz

Error using histogram

Expected input number 2, BinEdges, to be non-decreasing valued.

Error in histogram>parseinput (line 267)

validateattributes(value,{'numeric','logical'},{'vector', ...

Error in histogram (line 124)

[opts,passthrough,dispatch] = parseinput(args,firstaxesinput);

Error in histogramByAdamDanz (line 16)

histogram('BinEdges',fitness2sn0,'BinCounts',Runs)

>>

the cyclist
on 22 Oct 2020

As I mentioned, because you have the individual data, not the bin counts, you need a different syntax.

You can just do

figure

histogram(Kg,'BinEdges',0:1e-5:1e-3)

xlabel('Weight (kg)')

ylabel('Number of participants')

Adam Danz
on 22 Oct 2020

Yeah, in my example, N are the counts and KG are the bin edges. The KG data in your comment above do not appear to be bin edges, as the cyclist mentioned. They appear to be the raw data in which case, what are N?

Perhaps you're just looking for,

histogram(kg)

Sadiq Akbar
on 22 Oct 2020

Thank you very much the cyclist. It works now. But I have certain points to ask you:

(1) If I want to change the data, i.e. instead of one Kg, I have three Kg's like Kg1, Kg2 and Kg3, then how will I do this histogram. i.e.

Kg1=[2.E-03

1.E-03

4.E-04

3.E-04

3.E-04

2.E-04

2.E-04

2.E-04

2.E-04

2.E-04

2.E-04

1.E-04

1.E-04

9.E-05

8.E-05

8.E-05

8.E-05

8.E-05

8.E-05

6.E-05

5.E-05

4.E-05

3.E-05

2.E-05

2.E-05

2.E-05

2.E-05

1.E-05

1.E-05

1.E-05

1.E-05

9.E-06

9.E-06

7.E-06

7.E-06

6.E-06

5.E-06

5.E-06

4.E-06

4.E-06

2.E-06

2.E-06

1.E-06

1.E-06

1.E-06

4.E-07

3.E-07

3.E-07

2.E-07

2.E-07

2.E-07

2.E-07

2.E-07

2.E-07

2.E-07

1.E-07

1.E-07

1.E-07

7.E-08

6.E-08

4.E-08

1.E-08

1.E-08

9.E-09

4.E-09

2.E-09

2.E-10

3.E-11

7.E-12

8.E-13

7.E-13

6.E-13

5.E-13

3.E-13

9.E-14

1.E-14

5.E-15

5.E-17

2.E-22

5.E-23

2.E-23

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

0.E+00

];

Kg2=[2.E-23

9.E-05

3.E-10

9.E-05

2.E-03

4.E-03

1.E-02

8.E-20

2.E-02

2.E-03

2.E-03

2.E-01

2.E-09

1.E-19

3.E-14

1.E+00

3.E-03

4.E-03

5.E-07

6.E-25

3.E-25

8.E-05

1.E-21

5.E-05

2.E-16

3.E-03

2.E-05

2.E-21

1.E-03

7.E-19

2.E-04

6.E-05

0.E+00

6.E-03

2.E-04

1.E-04

2.E-03

2.E-04

2.E-02

9.E-04

7.E-23

1.E+00

1.E-10

2.E-03

2.E-17

4.E-03

4.E-21

7.E-04

0.E+00

1.E-08

2.E-03

5.E-17

2.E-05

4.E-09

2.E-02

9.E-04

1.E-03

2.E-06

6.E-02

3.E-07

1.E-05

3.E-08

2.E-03

3.E-07

2.E-05

1.E-26

2.E-08

1.E-09

1.E-01

0.E+00

3.E-03

2.E-19

5.E-03

1.E-20

4.E-04

9.E-04

2.E-22

0.E+00

7.E-23

2.E-23

2.E-17

2.E-02

9.E-03

4.E-04

7.E-04

1.E-04

6.E-05

3.E-02

2.E-02

1.E-02

2.E-19

5.E-04

6.E-03

3.E-04

7.E-02

2.E-28

3.E-04

3.E-04

2.E-04

];

Kg3=[4.E-02

8.E-02

1.E-01

2.E-02

2.E-05

8.E-03

8.E-02

2.E-01

2.E-01

4.E-02

6.E-05

2.E-06

2.E-02

7.E-02

7.E-03

1.E-02

1.E-03

1.E-01

2.E-02

9.E-01

1.E-02

1.E-07

6.E-02

1.E-01

8.E-02

3.E-02

6.E-02

2.E-01

2.E-01

5.E-02

7.E-04

2.E-03

5.E-02

5.E-02

3.E-02

9.E-03

3.E-03

9.E-01

1.E-03

4.E-02

1.E-03

1.E-01

4.E-03

4.E-01

2.E-01

7.E-02

4.E-04

9.E-02

2.E-01

5.E-02

4.E-02

1.E-02

2.E-01

1.E-02

1.E-05

1.E-02

4.E-02

2.E-01

3.E-02

8.E-02

9.E-01

7.E-03

1.E-05

2.E-03

5.E-02

2.E-03

9.E-02

3.E-01

1.E-02

3.E-02

6.E-02

1.E-03

6.E-02

9.E-02

1.E-02

1.E-01

6.E-02

2.E-02

1.E-01

2.E-01

2.E-01

3.E-01

9.E-02

3.E-03

5.E-05

2.E-02

1.E-01

2.E-04

1.E-01

3.E-02

2.E-01

1.E-01

1.E-02

4.E-02

3.E-02

6.E-03

3.E-02

8.E-02

8.E-02

]

(2) 2ndly if I want to give colours to each bar, how to do that?

(3) If I want to plot the three Kgs data in one graph, then (a) can we make all three Kgs as one Kg and plot it like above? If yes then how will we decide the range as you did inside histogeram command which is 0:1e-5:1e-3

(b)If I plot one histogram for Kg1, then I want to plot Kg2 and Kg3 histogram over the 1st, then how will we do that?

(c) I want to show x-axis in log scale

Thanks once again that you listened to me.

Sadiq Akbar
on 22 Oct 2020

Adam Danz
on 22 Oct 2020

The histogram function does two main things.

- counts the number of data points within each bin
- plots the results

The first method in my answer uses,

histogram('BinEdges',kg,'BinCounts',n)

which bypasses the first step and the user supplies the bin-counts (and bin edges). In that case, the histogram function is only plotting the results.

So yes, when your supplying raw data, the histogram function does the bin-counting for you.

Sadiq Akbar
on 22 Oct 2020

the cyclist
on 22 Oct 2020

Sadiq Akbar
on 22 Oct 2020

Sadiq Akbar
on 22 Oct 2020

Sadiq Akbar
on 22 Oct 2020

Adam Danz
on 22 Oct 2020

That's partially helpful. What we're still missing is,

- The data used to generate the histogram
- The syntax you used (the line that contains the histogram() function)
- And what the x-axis should represent (I believe it's weight in kg but I don't know why the xlim is 0-1).

Sadiq Akbar
on 22 Oct 2020

sorry for that. Ok I I am attaching the three mat files and write my code here.

% Multiple Histogram of my data

%1st keep the attached mat files 2sn0_sorted, 3sn0_sorted and 4sn0_sorted in the

%same directory as this m file, then run it and observe my plots. If you

%see, the markings on horizontal axis of the plots, it is in floating point

%numbers while in theirs, its integers.

load 2sn0_sorted

fitness2sn0=one;

% Runs=tt;

load 3sn0_sorted

fitness3sn0=one;

load 4sn0_sorted

fitness4sn0=one;

h1 = histogram(fitness2sn0);

hold on

h2 = histogram(fitness3sn0);

h3 = histogram(fitness4sn0);

Sadiq Akbar
on 22 Oct 2020

Sadiq Akbar
on 22 Oct 2020

Sadiq Akbar
on 22 Oct 2020

Adam Danz
on 22 Oct 2020

If you plot the data from fitness2sn0.mat (variable 'one'), you'll see that is spans from x=0 to x=0.0022349. That's closer to the range of data in your histogram.jpg image (though it has a different distribution).

"Then why mine is not like that? or how can I get like that?"

I'm not sure if that's the right question to be asking.

The way I see it, the question isn't how to make your data look like that plot. The questions could be,

- (not really a question) Perhaps I am plotting it corretly and my data have a different result.
- Am I using at the right data in the first place? Should I be binning and counting this variable?
- I am using the correct data but the outcome is unexpected. Is that because of an error in the analysis, an error when importing the data, or is it a true reflection of whatever I was measuring?

We're at a crossroads here where no amount of technical help is going to answer these questions (without more background informat). You as the owner of the data and the person analyzing it needs to take a step back think about it from big-picture perspective.

I'll point out that the values in your original question (kilograms etc) don't seem to be relevant to the jpg image or the mat data you shared so I sense some disorientation here.

Sadiq Akbar
on 22 Oct 2020

Sadiq Akbar
on 23 Oct 2020

I found one such code on this site today. The code is as follows:

[~,edges] = histcounts(log10(x));

histogram(x,10.^edges)

set(gca, 'xscale','log')

In this code, when I replace x by my data, then it plots well. Now I want to insert a preview in this graph. How can I insert the same graph as a preview pane in the same figure window as is in the attachment figure.

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