select data from specific range

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)))

[Chandra Kurniawan]

Would you upload your txt file?

Answer 1

[Chandra Kurniawan]

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

[joo tan]

also work..but,how to spacing for lat and long 0.25 to find the sla value.

Thanks

[joo tan]

i try to design back the program to make loop and spacing 0.25..but not work..any idea??

f=load('200012.txt');

lat=f(:,1);

lon=f(:,2);

sla=f(:,3);

for i=2:0.25:14.75

j=2.25:0.25:15

k=95:0.25:125.75

l=95.25:0.25:126

r = sla(find(lat >=(i) & lat <= (j) & lon >=(k) & lon <= (l)))

a=mean(r)

end

[Chandra Kurniawan]

I don't understand what you want to do.

Did you mean you need to find lat at 2.00, 2.25, 2.50, ... , 14.75??

[joo tan]

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

[joo tan]

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

[Chandra Kurniawan]

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]

suppose if i want the answer as r = 2 96 104 3 96 107 what will it be

Answer 2

[Chandra Kurniawan]

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

[joo tan]

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

Answer 3

[joo tan]

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

Answer 4

[Image Analyst]

Try something like this (I didn't test it):

rowsToAverage = lat>2 & lat<3 & lon>95 & lon<97;
meanValue = mean(sla(rowsToAverage));

[joo tan]

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

Answer 5

[joo tan]

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