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!

 採用された回答

GGT
GGT 2015 年 4 月 28 日

1 投票

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

2 件のコメント

GGT
GGT 2015 年 4 月 28 日
By the way that is l as in lima in the parentheses; looks like the number one.
BaconSwordfish
BaconSwordfish 2015 年 4 月 30 日
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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その他の回答 (1 件)

the cyclist
the cyclist 2015 年 4 月 28 日

0 投票

I think it is asking you about this type of quadrant.

4 件のコメント

BaconSwordfish
BaconSwordfish 2015 年 4 月 28 日
I think you are right. However, I am unsure how to separate the data based off of what quadrant they fall in. Do you know how to?
the cyclist
the cyclist 2015 年 4 月 28 日
Do you have values of x and y? You need to figure out if x and y are positive or negative.
BaconSwordfish
BaconSwordfish 2015 年 4 月 28 日
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 =
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y =
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3.3321
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1.4327
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0.3120
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2.4132
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4.1119
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2.3087
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3.9363
1.2811
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3.1116
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4.0580
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3.6102
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4.1780
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2.4053
5.1047
4.9043
1.7986
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3.6870
4.8552
2.0554
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2.0381
4.6074
2.3332
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3.4495
3.7357
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3.1470
3.4943
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3.8793
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2.1781
3.6573
-1.1037
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the cyclist
the cyclist 2015 年 4 月 28 日

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