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Hi tina123;
I do not know. But aren't we still working on the histogram?
yes we are still working on histogram .
Can we get energy from histogram ?
thanks
Hy bobbym,
How can we get energy from histogram ?
like if we have high frequecny --> high energy
low frequency --> low energy .
Also how can we get probability from frequency distribution histogram
but how can we get energy and probability from histogram ?
Any idea .
please help thanks
Sorry .. probably its a stupid question but important for me .
I always have confusion that in real line.
-1 < -2 is true or
-1 > -2
Which is true ?
Thanks
hmm i see . i got it .
So how about to get the 1st moent,.
I think the quantity that i computed that 1st moment not area . sorry .
also how can we get variance ?
okkkk Well i think if we use " area under the curve " from calculus .
then area calculation would be from bin 5 to bin 10 ..
how ???
Well we have interval between bin 5 to bin 10 .
so we have total 5 , 6 , 7 , 8 , 9 , 10 intervals in bin range 5 to 10 .
and frequency corsponding to each bin is calculated.
so we get area .
A = frequecny1 * bin [5] + frequecny2 * bin[6] + .....frequecnt5*bin[10]
the sum will be our area .
Right ? ?
If we have data of 352* 288 dimension. and data is represented as histogram with 20 bins . where Each bin have from 0 to 9 . and the frequency of each bin is given below.
255 , 70 , 60 , 50 ,10 , 5 ,4 , 4 , 60, 50 , 30 , 10 , 5 , 2, 2 , 1 , 0 , 5 , 3, 3
The issue is : I need to find area from
bin 7 to 13 : frequency are 60 ,50,,30,10,5
How to do that ?
Area meant number of data. like 3*3 . area will be 9.
That meant area is number of data . Since frequency represent the occurence of pixels .Can we say that area will be sum of frequency ?
Please help me to get the area of bin from 7 to 13 in which frequency are :
60 ,50,30,10,5
Please guide me thanks
Edited:
one way is defined in calculus i-e " Area under the curve . "
see Finding Areas using Definite Integration in this link : http://www.intmath.com/integration/3-area-under-curve.php
Now in our case we have low bound is i = 7 and upper bound i = 13 .
then area would be f(b) - f(a) = 13 - 7 = 6
Area could not be 6 .. its wrong .
How can we get area under the histogram curve .
thanks
i have solved that issue thanks .
hi tina123
Sorry but some of your posts don't seem to be getting the responses you want.
It would help if you gave us some background to explain what you are trying to do.
For this problem I think you want to identify the maximum frequency from a frequency table and make it zero instead. If I'm correct then you are asking for a computing code routine to find a maximum.
What computer language are you using ?
Bob
Well its my research. I do't wanna post any question regarding it . But since i need discussion and help to take a good decision. so i am here to post my queries .
If any one have interest in computer vision then please help me via email. Its my humble request.
Anyway, I am using c++ . I can't explain my queries in detail . I would like via email . So please if any one have interest then please let me know ,.
thanks
I have histogram like shown in the image : http://i1069.photobucket.com/albums/u472/tina12354/l.png
That is bimodel histogram. What wanna do is . i want to remove the maximum bar histogram until it comes to flat .
Does any one have idea how to do that ?
Thanks
Hi;
4.5625336119761198e+002 that means
In integers it is now 456
Thanks bobby
Can any one convert the following data in integers .
thanks
<data> 4.5625336119761198e+002 0. 1.5886571681507525e+002 0. 4.5625336119761198e+002 1.3768139354666781e+002 0. 0. 1.</data>
hi tina123,
If you look at this page:
http://www.mathsisfun.com/definitions/histogram.html
you will see you require a range of data for each frequency to make a histogram.
The best you can do with these data is a bar graph.
Please explain where the data comes from.
Bob
Well you make me more confuse .
As far as i understand , histogram is the occurrence Of Pixels .
so ,
0 0 0 0 1 1 5 4 666 1 1 0 0 0 0 0 0 0 0
Now histogram would be like this
Number of occurrence of 0 ==> 12
Number of occurrence of 1 ==> 4
Number of occurrence of 4 ==> 1
Number of occurrence of 5 ==> 1
Number of occurrence of 666 ==> 1
to show it in bar graph,
0 is 12 times .
1 is 4 timse .
4 is 1 times
5 is 1 times
and 666 is 1 times .
so if we wanna get mean of hitogram . we can do like
sum of all occurence / totalbins.
That's what i know .
Hy its 666 ..
Hy wait .. I have to get histogram 1st .
and then get mean from histogram ..
Ok please help me to find histogram of above image .
thanks
Let say we have image data like this .
0 0 0 0 1 1 5 4 666 1 1 0 0 0 0 0 0 0 0
Now histogram would be like this
Number of occurrence of 0 ==> 12
Number of occurrence of 1 ==> 4
Number of occurrence of 4 ==> 1
Number of occurrence of 5 ==> 1
Number of occurrence of 666 ==> 1
mean : 12 +4 +1 +1 +1 / total number of points.
mean = 19/19 == > 1
Now issue is i wanna show mean in color form in histogram .
but how can we get location of mean in histogram like in x bins and y probability ?
Where will be mean locate?
Hello guys,
I have histogram of colors from 0 to 20 .
My image is like blackground and two the most nearest object exist in the image.
I need to segment skin colored object like hand or face from the image .
Conceptually, the static background object would be at color 0 with high probability because my background is black. and some where the two most nearest object found at hue shown in the images
http://i1069.photobucket.com/albums/u472/tina12354/fig.png
I want to segment hand or face by mean .
but i mostly have high probability at color 0 . then mean would be at color 0 . right ?
what is the value if we add infinity in any real number .
say 1- infinity = ?
Thanks
ok fine
ok see the image from this link : http://i1069.photobucket.com/albums/u472/tina12354/img.png
Hello guys,
I have an image like this http://i1069.photobucket.com/albums/u472/tina12354/img.png
In the image , hand trajectory is tracked by assigning codes .
but when both hands goes on same hand direction then hand tracking get failed because the motion with the same direction in the time intervel are connected.
Could any one give me idea to solve it .
thanks
Hy bob 0255 is data value.
What do you meant by this
you want (i) compute the nearest neighbour for every value (ii) then find the maximum?
I did not clearly understand .
Do you mean find the maximum of (i) and (ii) . Is it ?
what is maximum nearest valuse in this data set ?
0255
15.15876
20.351796
30.170401
40.316067
50.404016
61.05814
70.491965
82.8501
93.38054
101.46215
114.14735
120.508455
130.439745
140.170401
150.175898
161.06363
170.0687102
180.406764
190.0632134
202.32831e-010
210.0219873
220.36279
230.0247357
240.181395
0255
15.13722
20.340832
30.137432
40.36557
50.310597
61.09671
70.409548
85.3791
91.19016
104.29888
111.13519
120.503002
130.247378
140.170416
150.0439783
161.06647
170.0302351
180.412297
190.00274864
204.65661e-010
210.145678
220.206148
230.159421
0255
15.20038
20.310593
30.332582
40.151174
50.324337
61.0857
73.27085
83.36705
91.20939
104.36205
110.489253
120.470013
130.123688
140.142928
151.03898
160.0632181
170.415041
180.00549723
199.31323e-010
200.151174
210.208895
220.162168
0255
12.3228
20.296123
30.285256
40.184737
50.339591
61.52408
72.82539
83.61868
94.37121
101.14646
110.589529
120.279823
130.184737
141.05409
150.0733516
160.402075
170.0706349
180
190.187454
200.233638
210.182021
0255
12.32821
20.369471
30.239069
40.440105
51.16003
60.431955
75.36004
82.03209
94.33585
100.573223
110.448255
120.222769
130.0706341
141.13015
150.407504
160.0733508
172.32831e-010
180.192885
190.236353
200.184735
0255
11.31238
20.257065
30.351773
40.316595
51.48286
65.31177
72.21076
84.29704
90.592602
100.232711
110.127179
121.08779
130.368008
140.0351773
156.98492e-010
160.327419
170.178592
0255
10.90731
20.353743
30.426652
41.11794
53.20799
63.66704
75.09012
80.639978
90.434753
100.159319
111.03963
120.418551
130.0378046
140
150.33214
160.180922
0255
10.612203
20.350601
30.48275
41.51298
55.27789
62.01191
74.52275
80.725474
90.213057
101.09226
110.0620294
120.442296
13-2.79397e-00
140.345207
150.191482
0255
10.658002
20.450354
30.407206
41.51826
56.12427
65.47975
70.671486
80.469231
90.177984
101.11645
110.469231
124.65661e-010
130.388329
140.210345
0255
10.533698
20.320758
31.76821
45.26151
56.20492
60.695425
70.407012
81.04583
90.404317
105.12227e-009
110.315367
120.140163
0255
10.326201
20.520304
31.33446
46.30026
55.18686
60.593092
70.0673968
80.932772
90.264196
10-1.14087e-00
110.21567
0255
10.495721
21.31204
36.36355
45.56339
50.684311
61.03724
70.35832
8-7.68341e-009
90.231696
0255
10.570801
21.75548
37.45003
44.83027
50.323095
61.41354
76.75209e-009
80.468488
0255
10.390794
26.34434
36.08831
40.495904
50.711515
62.04891e-008
70.0943297
0255
11.34664
27.27994
34.83444
40.934569
52.32831e-009
60.131971
thanks
Hi tina123;
Where is the image located?
I uploaded my link in photobucket website.
I an not able to post a link of my image here l.
even i am not able to use image tag too.;
then how can i show you my histigram .
How to find neighbouring maximum peaks in the image histogram
Hi;
This is just a euclidean distance between points. It uses the pythagorean theorem. For a plane it looks like this:
ok but Is that L2 norm ?
I heard new term for that simple distance method .