How to count how many x are greater than (x>0.2, 03 and 0.4) in a specific time?

Hi Everybody,

I need to know how many (x>0.2,0.3&0.4)occurred at each hour. First, count it to make me able tabulate it (i.e. in time between 5-6, we observed 5 event(x>0.2) and so on) and then plot it as a bar. Please assume, x and y are defined as a column data in .xls format.

x             y
0.05          10 
0.1           10.1 (1 min)
0.02          10.2
0.2           10.3
0.012         10.4 
.             . 
.             . 
.             .
0.2           11

Thank you in advance,

[Jan]

@Ara: Without knowing any detail about the available input, it is impossible to guess what you are asking for. Perhaps the inputs are written on papaer, in an XLS or text file, a vector in the DATEVEC or DATENUM format, or anything else. In addition it is not clear, what an "event" is. So please edit the origianl question (not in comment or answer section) and add all required details. Usually it helps to remember, that for an answer a question is mandatory.

[Image Analyst]

What is y? Is it hours? Minutes? Why does 10.1 say 1 minute next to it?

[Ara]

Please assume, y is a column data of time that look like this. For example, 10.1 means time in 10am and 1min and 10. 2 min and so on.The y column (time) is start from 0-24, while I only need to count time between 0-10hrs and 22-0hrs.

[Walter Roberson]

How do you tell the difference between 10:01 and 10:10? 10.1 and 10.10 become the same number.

[Ara]

@ Walter: Exactly. As you said, It should be like 10.01 and 10.10. And the signal varied usually during an hour. It means between each minutes or even more minutes hard to find (x>0.2,0.3 or 0.4).

[Image Analyst]

You'll have to read them in as text then and parse them into two separate numbers.

Answer 1

[Cedric]

The following could be a solution..

 >> x = rand(1,1e3)/2 ;             % Fake x, for the example. 
 >> y = sort(rand(1,1e3)*23.99) ;   % Fake y, for the example.

Build a cell array of distributions

 >> bins  = 0.05 : 0.1 : 0.45 ;              
 >> dists = arrayfun(@(h) hist(x(floor(y)==h), bins), 0:23, ...
                     'UniformOutput', false) ;

Test..

 >> size(dists)
 ans =
     1    24
 >> dists{1}                      % Distribution for the period [0-1h[.
 ans =
     6     6     9    12     6
 >> dists{12}                     % Distribution for the period [11-12h[.
 ans =
     3    12     4     7     5

so for the period [11-12h[, 3 values are in the range [0,0.1[, 12 in the range [0.1, 0.2[, etc..

Note: you can achieve the same thing using a more basic approach, that you would have to put in a loop..

 >> id_hr = y >= 11 & y < 12 ;    % Index of y values (hrs) in the range 11-12h.
 >> sum(x(id_hr) < 0.1)           % Count # of corresponding x values below 0.1.
 ans =
      3

here we find the 3 that we found above as 1st element of dists{12}.

[Ara]

Thanks Cedric Wannaz for the code. This work properly. Just last question, Is it wrong if I define bins as below? In this case, it's supposed to count range between[0.2-0.3]and [0.3-0.4]and greater than 0.4, right?

    bins  = 0.2:0.3:0.45 ;

[Cedric]

It seems that I brought a few simplifications while you were testing my first version; sorry for this. You made just one little mistake actually, with respect to what I had done, when you defined bins: the way you wrote it means 0.2 to 0.4 by steps of 0.3, which is not what you want. Also, note that HIST takes the center of bins and not left boundaries, and that boundaries inclusion might not be exactly what you want. At this point, I guess that it is better to avoid using HIST if you want to have a total control over boundaries, and to build your own hist function.

The following should work (that I updated according to the simplification that I brought in my edited answer):

 s4   = rand(1, 1e3)/2 ;            % Fake x, for the example. 
 time = sort(rand(1,1e3)*23.99) ;   % Fake y, for the example.

 % Build an anonymous function called myHist which builds a vector of 
 % three counts, for ranges ]0.2,0.3], ]0.3,0.4], and >0.4.
 myHist = @(v) [sum(v>0.2 & v<=0.3), sum(v>0.3 & v<=0.4), sum(v>0.4)] ;

 % Build a cell array of distributions
 dists = arrayfun(@(h) myHist(s4(floor(time)==h)), 0:23, ...
                  'UniformOutput', false) ;

If you evaluate

 bar([0.25,0.35,0.45], dists{2}, 1)

you get a barplot for content of cell #2 which represents the distribution during the period [1-2h[. Now what do you want to plot against time? If it is the evolution of the # of occurrences of values in the range ]0.2,0.3] for example, I would say that the simplest is to convert to cell array into a matrix, which makes indexing simpler:

 >> dists_mat = cell2mat(dists.')
 dists_mat =
     7     8     7
     9    12     6
     4    11     7
    13     4     6
     8     7     0
     7     8     4
     7     9     5
    16     7    10
     6    13     8
     8    13    10
     8     5     4
     7     7     7
    15     7     4
     6     9    12
     2    11     8
     5    10    10
     6    12     7
     7     5    12
     5    10     9
    13     9     6
     8     5     8
     5    10    15
    10    14     5
     8    10     7

and here you can plot/bar/etc each column as a function of e.g. the hour

 plot(0:23, dists_mat(:,1)) ;

[Cedric]

Last comment before leaving for the night: while it is fun to play with arrayfun and anonymous functions, I think that a solution based on a FOR loop would be more appropriate in the sense that it would be simpler to understand later ..

 dists_mat2 = zeros(24, 3) ;
 for h = 0 : 23
    hId = time >= h & time < h+1 ;
    dists_mat2(h+1, 1) = sum(s4(hId)>0.2 & s4(hId)<=0.3) ;
    dists_mat2(h+1, 2) = sum(s4(hId)>0.3 & s4(hId)<=0.4) ;
    dists_mat2(h+1, 3) = sum(s4(hId)>0.4) ;    
 end

Test:

 >> dists_mat2
 dists_mat2 =
     7     8     7
     9    12     6
     4    11     7
    13     4     6
     8     7     0
     7     8     4
     7     9     5
    16     7    10
     6    13     8
     8    13    10
     8     5     4
     7     7     7
    15     7     4
     6     9    12
     2    11     8
     5    10    10
     6    12     7
     7     5    12
     5    10     9
    13     9     6
     8     5     8
     5    10    15
    10    14     5
     8    10     7

 >> all(dists_mat2(:) == dists_mat(:))
 ans =
      1

OK.

Note that this solution can be optimized. One way is to build an array of sums before the loop. Another way would be to use ACCUMARRAY.

[Ara]

Thank you so so much. The code you wrote is exactly what I want and it's completely correct. Just When I run with my data it shows in y coordinator, the value of 14 for the periods between 5-6hrs which make me surprise because the value should be around 4 or 5. So, the question is because of building Hist function? Sorry to ask lots of questions because I have to describe the method below the figure.

[Ara]

Thank you VERY much. Let me play around with the plot and if you do not mind I will ask any question that I faced with.

[Ara]

Hi Cedric,

You wrote a FOR loop in my code. How to get rid of the loop and read data for one day of interest only without going to a circles again and again.

• I would like to write my CV in Latex. Can you help me or do you have a template of it so that I can use it?

If I find the answer of this question, I will be the happiest woman in this World.

Please tell me how to correct it?

Best,

Ara

Answer 2

[Cedric]

You're welcome!

Well, at this point you should build a test dataset just to check, e.g.

 >> s4   = [0.21, 0.21, 0.21, 0.32, 0.45, 0.45, 0.25] ;
 >> time = [1.1,  1.2,  1.3,  1.4,  1.5,  1.6,  2] ;

so there should be [3, 1, 2] associated with period [1-2h[, and [1, 0, 0] associated with period [2-3h[.

 >>  myHist = @(v) [sum(v>0.2 & v<=0.3), sum(v>0.3 & v<=0.4), sum(v>0.4)] ;
 >> dists = arrayfun(@(h) myHist(s4(floor(time)==h)), 0:23, ...
                  'UniformOutput', false) ;

 >> dists{1}          % [0-1h[
 ans =
     0     0     0
 >> dists{2}          % [1-2h[
 ans =
     3     1     2
 >> dists{3}          % [2-3h[
 ans =
     1     0     0

It seems to be working. Then, using your dataset, you can check manually what happens with the period 5-6h. I show you here with the period 1-2h:

 >> id_time = time >= 1 & time < 2            % Flag relevant times.
 id_time =
     1     1     1     1     1     1     0

this is a vector of logicals that we use then for indexing s4

 >> s4_selection = s4(id_time)
 s4_selection =
    0.2100    0.2100    0.2100    0.3200    0.4500    0.4500

as you can see, these are the elements of s4 that correspond to a time in the range 1-2h[. Now we test myHist on this set of values

 >> myHist(s4_selection)
 ans =
     3     1     2

It seems to be working.. let's check manually that the first count is correct

 >> s4_selection > 0.2 & s4_selection <= 0.3
 ans =
     1     1     1     0     0     0

again, vector of logicals flagging relevant elements; counting them just means summing the vector of logicals (there is a conversion to numeric)

 >> sum(s4_selection > 0.2 & s4_selection <= 0.3)
 ans =
     3

So it's working. Now ARRAYFUN repeats the same kind of operations with the small difference that instead of checking whether time is greater than h and smaller than h+1 for h=0:23, we test whether floor(time)==h.

Hope it helps; let me know if you are having troubles using this material!

[Ara]

It works correctly.Thanks for all of comments and comprehensive answer and sorry for the late response. I have two more questions;

1- If the time comes out during each 2 min instead of 1min, the result would be different?

2- Another column of data is related to the number of satellites(PRN), am wondering to know in that time your code count s4 in a specific time, for example time between [1-2], how many satellites observed it? In this way, I will know whether satellites number 5, 6 and 10 observed simultaneously in the same time or not?

Please let me know if I have to open in another question format?

[Cedric]

No need for another question/answer; I added a second answer yesterday by mistake actually (was tired ;-)), but I should just have gone on using comments under my first answer.

1. I have no idea; the only thing that I can say is that the code would work as well if it is working now, as the FLOOR operation reduces any positive floating point number to the nearest integer below. So whether you have 10, 10.1, 10.2, 10.3, 10.4, etc, or 10, 10.2, 10.4, 10.6, etc, doesn't matter: they will all be floor'ed to 10 (which is what you want as you are producing an hourly statistics). But my guess is that you would have just half the data points, which might "screw" a bit your hourly distributions if you have as little as 29 points/hour and not all of them are >0.2.

2. I am not sure that I understand what you want to do, so let's discuss an example: assume that you have s5, a vector of satellite IDs..

 >> time = [1.1,  1.2,  1.3,  1.4,  1.5,  1.6,  2] ;
 >> s4   = [0.21, 0.21, 0.21, 0.32, 0.45, 0.45, 0.25] ;
 >> s5   = [3,    5 ,   3,    6,    6,    2,    5] ;

and you want to get the IDs of all satellites that gave a measure during each hour. You can get this with:

 >> sats = arrayfun(@(h) unique(s5(floor(time)==h)), 0:23, ...
                    'UniformOutput', false) ;

Let's check the output:

 >> sats{1}
 ans =
   Empty matrix: 0-by-1
 >> sats{2}
 ans =
     2     3     5     6
 >> sats{3}
 ans =
     5

It seems to be fine. Here we are doing something simpler than before; for each hour h in 0:23 we get all satellite IDs that produced a measurement during [h-h+1[. As some satellites can produce multiple measures during an hour (I guess), we ask for a list of unique IDs.

Again, you can perform one step manually to understand..

 >> id_time = floor(time) == 1      % Flag relevant times.
 id_time =
     1     1     1     1     1     1     0

(this time I used floor() instead of time >= 1 & time <2, just for the example)

 >> s5_selection = s5(id_time)      % All satellite IDs relevant to 
                                    % period [1-2h[.
 s5_selection =
     3     5     3     6     6     2

we just want a lit of unique IDs actually, so..

 >> unique(s5_selection)
 ans =
     2     3     5     6

which is what we got in sats{2}.

[Ara]

Thanks for the time and the codes. You are really helpful and genius person. Just want to say, all of my data are defined as a column like this and I read it (for example for PRN)which are different with the data you assumed.

PRN=data(:,3);

S4                         Time        PRN
0.0593550051806922         10.2        28
0.220171162246558          10.4         3 
0.206488387535474          10.5         2

[Cedric]

Thank you for the positive feedback! I see now; it looks like you get the num output of XLSREAD.

So your code looks like the following I imagine:

 >> sats = arrayfun(@(h) unique(PRN(floor(time)==h)), 0:23, ...
                    'UniformOutput', false) ;

or if you want to avoid using an intermediary variable:

 >> sats = arrayfun(@(h) unique(data(floor(time)==h,3)), 0:23, ...
                    'UniformOutput', false) ;

[Ara]

Sorry to bother you again. I face difficulty when I try to change the time to count all only for daytime. It means to not count the time between 10-22 but as I am dealing with huge data so difficult to check it manually. Could you please help me how to define time only for the period of time<=10 and time>=22? Thank you.

[Cedric]

Hi Ara, no problem. Looks like you want night time indeed! The simplest solution is probably to use what we've done already leading to two cell arrays: dists and sats, and to take all the cells but cells # 11 to 22. You could for example define

 >> dists_NT = dists([1:10, 23, 24]) ;       % Suffix _NT for Night Time. 
 >> sats_NT  = sats([1:10, 23, 24]) ;

Be aware though that these reduced arrays have 12 cells, and cells #11 and #12 correspond to periods [22-23h[ and [23-24h[, so you cannot use the index anymore to obtain directly the hour at the end of the period (which you can do on dists and sats as e.g. dists{24} contains the distribution for period [23-24h[).

Another solution if you really didn't want to get results for the period 10h-22h, would be to update a little calls to ARRAYFUN and replace 0:23 with [0:9,22,23]. For example:

 >> dists = arrayfun(@(h) myHist(s4(floor(time)==h)), [0:9,22,23], ...
                     'UniformOutput', false) ;

A third solution would be to reduce the data almost from the beginning. Say you have your three vectors S4, Time, and PRN as defined below

        S4        Time        PRN
        0.5       2           28
        0.1       6           2
        0.3       10          3
        0.2       14          17   
        0.6       18          8  
        0.4       22          1

You can locate Time elements that are in the range that you want as follows (I use the suffix _NT for Night Time);

 >> id_NT = Time <= 10 & Time >= 22
 id_NT =
     1
     1
     1
     0
     0
     1

you see that it is a vector of logicals that are false for Time = 14h or 18h. Then you can use this vector for logical indexing all vectors of data:

 >> S4_NT   = S4(id_NT) ;
 >> Time_NT = Time(id_NT) ;
 >> PRN_NT  = PRN(id_NT) ;

You could then use these reduces vectors in your computations.

As you can see, there are several ways to proceed; the difficulty is actually to understand what happens the size of reduced output and how to index it properly. If you just want to sum up values or compute a mean, there is no problem, but if you want to access the cell that corresponds to the period [22-23h[, you have to be clear about how the index was shifted when part of the data was discarded.

[Ara]

Thank you, you are so helpful. Just, Could you please again look at this last version of the code to make me sure is the correct one?

clear all;clc;close all;
data=xlsread('1608_1.xls');
data_filterr=find(data(:,25)>60);
data_filtered=data(data_filterr,:);
elev_cutof20=find(data_filtered(:,6)>=15);
data_cutoff15=data_filtered(elev_cutof20,:);
r=data_cutoff15(:,2);
time=((r./3600)-24*1);
s4r=data_cutoff15(:,8);
s4cor=data_cutoff15(:,9);
s4=sqrt(s4r.^2-s4cor.^2);
PRN=data_cutoff15(:,3);

s4 = rand(1,1e3)/2 ;             % Fake x, for the example. 
time = sort(rand(1,1e3)*23.99) ;   % Fake y, for the example.
%Build a cell array of distributions

   bins  = 0.05 : 0.1 : 0.45 ;              
  %  dists = arrayfun(@(h) hist(x(floor(y)==h), bins), 0:23, ...
  %                      'UniformOutput', false) ;

    dists_mat2 = zeros(24, 3) ;
   for h = 0 : 23
      hId = time >= h & time < h+1 ;
      dists_mat2(h+1, 1) = sum(s4(hId)>0.2 & s4(hId)<=0.3) ;
      dists_mat2(h+1, 2) = sum(s4(hId)>0.3 & s4(hId)<=0.4) ;
      dists_mat2(h+1, 3) = sum(s4(hId)>0.4) ;    
   end

   myHist = @(v) [sum(v>0.2 & v<=0.3), sum(v>0.3 & v<=0.4), sum(v>0.4)] ;
    dists = arrayfun(@(h) myHist(s4(floor(time)==h)), 0:23, ...
        'UniformOutput', false) ;

dists_NT = dists([1:10, 23, 24]) ;       % Suffix _NT for Night Time. 
 sats_NT  = sats([1:10, 23, 24]) ;

   dists = arrayfun(@(h) myHist(s4(floor(time)==h)), [0:9,22,23], ...
                       'UniformOutput', false) ;
  %    sats = arrayfun(@(h) unique(PRN(floor(time)==h)),[0:9,22,23], ...
  %                     'UniformOutput', false) ;

        dists_mat = cell2mat(dists.');
   figure;
        bar([0.25,0.35,0.45], dists{2}, 1);
    figure;plot(0:9, dists_mat(:,1)) ;           
     hold on; plot(0:9, dists_mat(:,2)) ;
     hold on;Plot(0:9, dists_mat(:,3)) ;

[Cedric]

I cannot check/guarantee that it is working, because it's a lengthy process that requires you to experiment with sets of known values (so you can check "by hand" that computations are correct), then maybe with larger sets of random values (based on relevant distributions), and finally with your dataset. The code that you pasted contains a bit of all the examples that we discussed, so you might want to eliminate duplicates..

The only thing that I can do at this point is to format it a little better and to add comments (after that I'm off for the night!).

 s4 = rand(1,1e3)/2 ;               % Fake s4, for the example. 
 time = sort(rand(1,1e3)*23.99) ;   % Fake time, for the example.

 % -- This was the first example, using bins that are not exactly
 %    what you need.
 % bins  = 0.05 : 0.1 : 0.45 ;              
 % dists = arrayfun(@(h) hist(x(floor(y)==h), bins), 0:23, ...
 %                      'UniformOutput', false) ;

 % -- This is an example showing how to proceed with a FOR loop instead 
 %    of anonymous functions and ARRAYFUN. This was just in case you would
 %    not understand the latter two mentioned mechanism/function.
 % dists_mat2 = zeros(24, 3) ;
 % for h = 0 : 23
 %     hId = time >= h & time < h+1 ;
 %     dists_mat2(h+1, 1) = sum(s4(hId)>0.2 & s4(hId)<=0.3) ;
 %     dists_mat2(h+1, 2) = sum(s4(hId)>0.3 & s4(hId)<=0.4) ;
 %     dists_mat2(h+1, 3) = sum(s4(hId)>0.4) ;    
 %  end

 % -- This is the example that computes distributions for all hours in
 %    the range 0:23.
 myHist = @(v) [sum(v>0.2 & v<=0.3), sum(v>0.3 & v<=0.4), sum(v>0.4)] ;
 dists = arrayfun(@(h) myHist(s4(floor(time)==h)), 0:23, ...
                  'UniformOutput', false) ;

 % -- Here you should put code for computing sats!
 % ...

 % -- This shows you how to build smaller cell arrays, with a selection
 %    of cells that correspond to your time range. Note that this could
 %    NOT work because you do not compute sats in the code above, so I
 %    commented out the line computing sats_NT and you'll have to add the 
 %    code for defining it above.
 dists_NT = dists([1:10, 23, 24]) ;       % Suffix _NT for Night Time. 
 % sats_NT  = sats([1:10, 23, 24]) ;

 % -- This is an example that shows you how to compute only the 
 %    distributions that correspond to the relevant time range.
 %    It should output the same result as dists_NT above.
 dists = arrayfun(@(h) myHist(s4(floor(time)==h)), [0:9,22,23], ...
                       'UniformOutput', false) ;
 % sats = arrayfun(@(h) unique(PRN(floor(time)==h)),[0:9,22,23], ...
 %                     'UniformOutput', false) ;

 % See, dists and dists_NT are equal: the evaluation of the following 
 % returns 1 (true).
 isequal(dists, dists_NT)

 % -- Here you seem to display dists (more precisely counts) for the 
 %    period [1-2h[.
 figure;
 bar([0.25,0.35,0.45], dists{2}, 1);

 % -- Here you seem to display the evolution over the first 10 hours of
 %    3 counts/bins. 
 %  Note: you made a mistake; the first argument when you call plot has 
 %    10 elements and the second had 12 elements as dists_mat is a 12x3
 %    array (1:10 and 23, 24 means 12 times). I brought a little 
 %    correction for selecting only rows 1:10 of dists_mat.
 %    You also need to execute "hold on" only once after the figure is 
 %    created; there is no need to repeat it after each plot. I added a
 %    few colors, a grid, and a legend, just for the example.
 dists_mat = cell2mat(dists.');
 figure;
 plot(0:9, dists_mat(1:10,1), 'g') ;           
 hold on; 
 plot(0:9, dists_mat(1:10,2), 'b') ;
 plot(0:9, dists_mat(1:10,3), 'r') ;
 grid on ;
 legend('bin ]0.2,0.3]', 'bin ]0.3,0.4]', 'bin ]0.4,\infty[') ;

[Ara]

Thank you so much. Have a nice day!

[Ara]

Hi Cedric, sorry again bothering you(feeling ashamed). Just one more question regarding the number of satellites, I made a table as below so want to describe but I am not sure all the (9 PRNs) during 006-007 LT observed s4>0.2 simultaneously or just give the number of PRNs not necessarily greater than 0.2? In other words, if during each hrs all satellites observed simultaneously the event (s4>0.2)so the result would be very interesting for my work.However, you assumed some example and then check it out to let me understand but in that case all s4 assumed greater than 0.2. Perhaps, another Hist should be define? I do not know so please you tell me. Here is one line of the table.

Date  Time(hrs)  0.2<S4≤0.3  0.3<S4≤ 0.4  S4>0.4   PRNs
1Nov2010  006-007  11    3    0       3
                                                                6
                                                                14
                                                                16
                                                                19
                                                                20 
                                                                23
                                                                24    
                                                                31 
sats = arrayfun(@(h) unique(PRN(floor(time)==h)),[0:9,22,23], ...
                     'UniformOutput', false) ;

[Cedric]

Hi Ara, no problem at all. I don't fully understand the question though. At this point, you get a histogram of counts and a list of satellites for each hour. In you example, 11 is the number/count of measurements of S4 in the period [6h-7h[ that were in the range ]0.2,0.3]. Do you need the list of PRNs that produced theses measures, or the count of different PRNs, or something else?

[Ara]

Hi Cedric, thank you for your help and patience. I am SO sorry for the late response. I just want to know which PRNs observed range between ]0.2,0.3]. Now by using your code, I know 11 PRNs are in the range between]0.2-0.3] and 3 are in the range ]0.3-0.4] but I do not know the exact number of the PRNs for each range. I want to know for example, PRNs 19,20&31 observed s4 between ]0.3,0.4]. Another problem is I made a table that shows different range of s4 for each hrs in LT daytime, so to get the histogram I need to get the mean and show by histogram? Now I am looking for a way to show a plot for month Nov that show the no of observation vs time(006-0018)in different range and another plot show with specific PRNs. I hope not to make you so confuse.

[Cedric]

No problem, Ara. So you want to be able to get the ID of all satellites that provided a measurement e.g. in the bin ]0.2,0.3] and during the period [1h,2h[. Will you need then to have a count of measurements for these satellites? For example, is it possible that satellite #16 generates 7 measurements in the range ]0.2,0.3] during the period [1h,2h[ (?) and if so, do you just need the information that satellite #16 provided at least a measurement, or do you need the count (7)?

My impression at this point is that if it becomes more complicated, it would be interesting to re-structure the approach into something that might be less efficient, but simpler for you to understand. Also, it would help if you could send me part of your data by email (available on my profile), e.g. 2 days of data if not too big, so I can perform tests using valid data.

[Ara]

I emailed you. Thanks.

[Ara]

Hi Dr. Cedric Wannaz,

Thank you for providing useful explanation and the best solution back in 2013.

I still read through your code, programming and all you tought me about MATLAB and it is very useful so thanks again for being helpful all the time. However, I wish to contact you as I amfacing some questions about MATLAB programming and I need to send it to your email address. I wrote to your email address as you previuosly provided for me but I guess you no longer check your email!!!

I need to install Mozilla File the one that you have asked me to use in Australia. How can I install that file again?

I have some more questions and I wish to email it to you to discuss more details about MATLAB solution and I need to send the code to you but I could not find your emal address in your profile. Would you mind to provide your email address in your profile or here so that I can send my questions along with my code to you? You have my email address so you can write back to me in any of my email.

Thanks.

Ara