Help with splitting data point pairs

조회 수: 14 (최근 30일)
BaconSwordfish
BaconSwordfish 2015년 4월 28일
댓글: BaconSwordfish 2015년 4월 30일
So I was tasked with defining my x and y with x= rcos(theta) and y=rsin(theta). Both r and theta are data arrays. My next step is to do the following: Utilize conditional statement(s) in order to split data point pairs into separate arrays depending on which quadrant they would fall into if plotted “y versus x”
I am unsure exactly what this is asking me to do and unsure how to go about it. Thank you!

채택된 답변

GGT
GGT 2015년 4월 28일
BaconSwordfish. After many hours I finally got it. You need two new counters in a for loop with an if end statement like so assuming you already have x and y saved into arrays.
for l=1:length(x)
if x(l)>0
xQ14=x(x>0); %x is positive in quadrants 1 & 4
end
if x(l)<0
xQ23=x(x<0);
end
end
  댓글 수: 2
GGT
GGT 2015년 4월 28일
By the way that is l as in lima in the parentheses; looks like the number one.
BaconSwordfish
BaconSwordfish 2015년 4월 30일
After trying to use this there seems to be a slight problem. Although this does a great job splitting the data into their positive and negative terms, I need to create arrays of the four quadrants. However, I can not do this because not all of the arrays are of equal length. So I can not take the positive xQ14 and yQ12 and say this is quadrant 1. If that makes any sense.

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추가 답변 (1개)

the cyclist
the cyclist 2015년 4월 28일
I think it is asking you about this type of quadrant.
  댓글 수: 4
BaconSwordfish
BaconSwordfish 2015년 4월 28일
I'm supposed to find what quadrant each data pair falls in. I am not sure how to write a code to do this. x =
4.0000
0.5532
-3.8499
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3.3788
2.6851
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1.2613
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3.8334
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4.1960
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2.9717
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3.8949
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y =
0
3.9628
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0.4664
4.1021
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1.0905
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1.9778
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2.8379
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2.5366
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0.8765
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1.3799
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3.6775
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3.9363
1.2811
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3.1116
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3.6102
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3.6870
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2.0381
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3.4495
3.7357
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3.1470
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3.8793
1.9521
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2.1781
3.6573
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-3.9628
the cyclist
the cyclist 2015년 4월 28일

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