select data from specific range

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joo tan
joo tan 2011-12-25
评论： Mitson Monteiro 2013-12-6
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)))

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采纳的回答

Chandra Kurniawan
Chandra Kurniawan 2011-12-25
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 个评论
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Chandra Kurniawan
Chandra Kurniawan 2011-12-25
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.
Mitson Monteiro
Mitson Monteiro 2013-12-6
suppose if i want the answer as r = 2 96 104 3 96 107 what will it be

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更多回答（4 个）

Chandra Kurniawan
Chandra Kurniawan 2011-12-25
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 个评论
joo tan
joo tan 2011-12-25
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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joo tan
joo tan 2011-12-25
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
Image Analyst
Image Analyst 2011-12-25
Try something like this (I didn't test it):
rowsToAverage = lat>2 & lat<3 & lon>95 & lon<97;
meanValue = mean(sla(rowsToAverage));
  1 个评论
joo tan
joo tan 2011-12-25
yes..it is works..but my problem is, i want to make the latitude and longitude spacing 0.25..

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joo tan
joo tan 2011-12-25
yes..it is works..but my problem is, i want to make the latitude and longitude spacing 0.25..

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