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select data from specific range

Asked by joo tan on 25 Dec 2011
Latest activity Commented on by Mitson Monteiro on 6 Dec 2013
Dear all,
My file have 3 column and more than hundred line..First column represent latitude, second longitude and lastly sea level. My plan is to average the sea level data based on specific range of latitude and longitude..I try design the program but still not work..Let say my file name is 200012.txt and i try to extract sea level data in range of (2<lat<3) & (95<lon<97)..But still not success. Help me
f=load('200012.txt'); lat=f(:,1); lon=f(:,2); sla=f(:,3);
x=sla(find((2<lat<3) & (95<lon<97)))

  1 Comment

Would you upload your txt file?

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5 Answers

Answer by Chandra Kurniawan on 25 Dec 2011
 Accepted Answer

Hello,
I have a text file formated :
1 90 100
1 96 101
1 95 102
2 91 103
2 96 104
1 89 105
3 87 106
3 96 107
2 90 108
4 97 109
2 76 110
And Here the script :
clear; clc;
f = load('200012.txt');
lat = f(:,1);
lon = f(:,2);
sla = f(:,3);
%x = sla(find((2<lat<3) & (95<lon<97)))
r = sla(find(lat >=2 & lat <= 3 & lon >=95 & lon <= 97))
The Result :
r =
104
107

  7 Comments

joo tan on 25 Dec 2011
i meant, to find the sla value in the specific range lat and lon but the lat and lon changes 0.25..let say first operation is to find and mean the sla value in the range
2=<lat<=2.25 and 95=<lon<=95.25
second operation is
2.25=<lat<=2.5 and 95.25=<lon<=95.5
and it changes until lat=15 and lon=126
i try use this program but the processing non-stop..i dont know why..the file size become more big
f=load('200012.txt');
lat=f(:,1);
lon=f(:,2);
sla=f(:,3);
F='mean.TXT';
for i=2:0.25:14.75;
for j=2.25:0.25:15;
for k=95:0.25:125.75;
for l=95.25:0.25:126;
r = sla(find(lat >=(i) & lat <= (j) & lon >=(k) & lon <= (l)));
b=mean(r);
fid=fopen(F,'a');
fprintf(fid,'\n%15.6f%15.6f%15.6f%15.6f%15.6f%15.6f\n',b);
fclose(fid);
end
end
end
end
if you want try run, this is link for the file
http://www.mediafire.com/file/jy6t6yjs9jxpeno/200012.txt
No, I think you can't do that.
i = 2 : 0.25 : 15; %size 1x53
j = 95 : 0.25 : 126; %size 1x125
Coz, the size of i and j must be same.
suppose if i want the answer as r = 2 96 104 3 96 107 what will it be

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Answer by Chandra Kurniawan on 25 Dec 2011

If size of i and j are same, then
f = load('200012.txt');
lat = f(:,1);
lon = f(:,2);
sla = f(:,3);
i = 2 : 0.25 : 15;
j = 95 : 0.25 : 126;
for x = 1 : numel(i)-1
r = sla(find(lat >=i(x) & lat <= i(x+1) & lon >= j(x) & lon <= j(x+1)))
b = mean(r)
end
This code I just use x = 1 : 53

  1 Comment

joo tan on 25 Dec 2011
tq chandra...i am sorry..i mistake give wrong information..the true is to find the sla value in the specific range lat and lon but the lat and lon changes 0.25..let say first operation is to find and mean the sla value in the range
2=<lat<=2.25 and 95=<lon<=95.25
second operation is
2=<lat<=2.25 and 95.25=<lon<=95.5
third
2=<lat<=2.25 and 95.5=<lon<=95.75
.............until the longitude 126..after that, the operation will continue with
latitude 2.25 and longitude same as above..it is like to grid bin by bin..the size of each bin 0.25
But, tq because your idea is good..i will try to find solution..I am so sorry about this..take your time

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Answer by joo tan on 25 Dec 2011

already solve problem..thanks
f = load('D1.ASC'); lat = f(:,1); lon = f(:,2); sla = f(:,3); i=2:0.25:15 j = 95 : 0.25 : 126; F='meanBIN.TXT' for y=1:52 for x = 1 : 124 r1 = sla(find(lat >i(y) & lat <= i(y+1) & lon >= j(x) & lon <= j(x+1))) b = mean(r1)
fid=fopen(F,'a'); fprintf(fid,'\n%15.6f%15.6f%15.6f%15.6f%15.6f%15.6f\n',i(y+1),j(x+1),b); fclose(fid); end end

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Answer by Image Analyst
on 25 Dec 2011

Try something like this (I didn't test it):
rowsToAverage = lat>2 & lat<3 & lon>95 & lon<97;
meanValue = mean(sla(rowsToAverage));

  1 Comment

joo tan on 25 Dec 2011
yes..it is works..but my problem is, i want to make the latitude and longitude spacing 0.25..

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Answer by joo tan on 25 Dec 2011

yes..it is works..but my problem is, i want to make the latitude and longitude spacing 0.25..

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