How to repeat calculations and store answers

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Dylan Mecca
Dylan Mecca 2018 年 2 月 20 日
コメント済み: Dylan Mecca 2018 年 2 月 20 日
I have a set of data that I want to fin the distance between two points for every point. d = ((x2-x1)^2 + (y2-y1)^2)^(1/2). Where x2 and y2 are in the next row and that d gets stored into another column vector. Does anybody have any suggestions for how to do this? I will include sample data that has x and y points.
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Bob Thompson
Bob Thompson 2018 年 2 月 20 日
I'm sure somebody knows a better way, but what about a for loop?
for i=2:size(data,1)
d(i) = ((x(i)-x(i-1))^2+(y(i)-y(i-1))^2)^0.5;
end

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Stephen23
Stephen23 2018 年 2 月 20 日
編集済み: Stephen23 2018 年 2 月 20 日
You should write simple vectorized code, e.g.:
X = [-83.55705;-83.55705;-83.557078;-83.557078;-83.557078;-83.557105;-83.557105;-83.557133;-83.557133;-83.55716;-83.55716;-83.55716;-83.557188;-83.557188;-83.557215;-83.557215;-83.557243;-83.557272;-83.557272;-83.557272;-83.557298;-83.557298;-83.557327;-83.557327;-83.557353;-83.557353;-83.557353;-83.55738;-83.55738;-83.557408;-83.557408;-83.557435;-83.557435;-83.557463;-83.557463;-83.557463;-83.55749;-83.55749;-83.557517;-83.557517;-83.557543;-83.557543;-83.557543;-83.557572;-83.557572;-83.557598;-83.557598;-83.557625;-83.557625;-83.557652;-83.557652;-83.557652;-83.557678;-83.557678;-83.557703;-83.557703;-83.55773;-83.55773;-83.55773;-83.557757;-83.557757;-83.557783;-83.557783;-83.55781;-83.55781;-83.557837;-83.557837;-83.557837;-83.557862;-83.557862;-83.557888;-83.557888;-83.557915;-83.557915;-83.557915;-83.557942;-83.557942;-83.557968;-83.557968;-83.557995;-83.557995;-83.557995;-83.558022;-83.558022;-83.558048;-83.558048;-83.558073;-83.558073;-83.5581;-83.5581;-83.5581;-83.558127;-83.558127;-83.558153;-83.558153;-83.55818;-83.55818;-83.55818;-83.558207;-83.558207;-83.558232;-83.558232;-83.558258;-83.558258;-83.558285;-83.558285;-83.558285;-83.55831;-83.55831;-83.558337;-83.558337;-83.558363;-83.558363;-83.558363;-83.558388;-83.558388;-83.558415;-83.558415;-83.558442;-83.558442;-83.558467;-83.558467;-83.558493;-83.558493;-83.558493;-83.558518;-83.558518;-83.558545;-83.55857;-83.55857;-83.55857;-83.558595];
Y = [40.302548;40.302548;40.302577;40.302577;40.302577;40.302605;40.302605;40.302633;40.302633;40.30266;40.30266;40.30266;40.30269;40.30269;40.302718;40.302718;40.302747;40.302775;40.302775;40.302775;40.302803;40.302803;40.302832;40.302832;40.30286;40.30286;40.30286;40.302888;40.302888;40.302917;40.302917;40.302945;40.302945;40.302973;40.302973;40.302973;40.303002;40.303002;40.30303;40.30303;40.303058;40.303058;40.303058;40.303087;40.303087;40.303115;40.303115;40.303143;40.303143;40.303172;40.303172;40.303172;40.3032;40.3032;40.303228;40.303228;40.303257;40.303257;40.303257;40.303285;40.303285;40.303313;40.303313;40.303342;40.303342;40.303372;40.303372;40.303372;40.303398;40.303398;40.303428;40.303428;40.303457;40.303457;40.303457;40.303485;40.303485;40.303513;40.303513;40.303542;40.303542;40.303542;40.30357;40.30357;40.3036;40.3036;40.303628;40.303628;40.303657;40.303657;40.303657;40.303683;40.303683;40.303713;40.303713;40.30374;40.30374;40.30374;40.303768;40.303768;40.303797;40.303797;40.303825;40.303825;40.303853;40.303853;40.303853;40.303882;40.303882;40.30391;40.30391;40.303938;40.303938;40.303938;40.303967;40.303967;40.303997;40.303997;40.304025;40.304025;40.304055;40.304055;40.304083;40.304083;40.304083;40.304113;40.304113;40.304142;40.30417;40.30417;40.30417;40.304198];
D = sqrt((X(2:end)-X(1:end-1)).^2 + (Y(2:end)-Y(1:end-1)).^2)
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Dylan Mecca
Dylan Mecca 2018 年 2 月 20 日
Thank you!!

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