problem with condtitions in program

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Krystian Sliwka
Krystian Sliwka on 15 Nov 2022
Commented: Krystian Sliwka on 17 Nov 2022
Hello I have the following problem. I would like my program to subtract data from t1 from t. The purpose of my program is to assume that the values will be in a given range from 0.1900 to 0.2600. If the program subtracts the value of t1-t it receives a value in the given range then it is supposed to write this value into a variable. If, on the other hand, it does not receive a given value because, for example, there is a value of 68 the next one is only 88 and you can see that there is a lack of samples here. The program has to subtract more data from the vector until it receives the required value. The problem I have is that when I enter only one boundary value the program works, but when I want to enter a second one unfortunately the program does not work properly.
t=[0.0840000000000000,0.935000000000000,1.83900000000000,2.85400000000000,3.84500000000000,4.73200000000000,5.71000000000000,6.77900000000000,7.75000000000000,8.62400000000000,9.40900000000000,10.1720000000000,10.9430000000000,11.8060000000000,12.6770000000000,13.8250000000000,14.9320000000000,16.0030000000000,17.0310000000000,17.9190000000000,18.6890000000000,19.4270000000000,20.2120000000000,21.1040000000000,22.0370000000000,22.9860000000000,23.9090000000000,24.7900000000000,25.6760000000000,26.5840000000000,27.4870000000000,28.4060000000000,29.3130000000000,30.1520000000000,30.9750000000000,31.8760000000000,32.8330000000000,33.7180000000000,34.6020000000000,35.5970000000000,36.6070000000000,37.5170000000000,38.4710000000000,39.4750000000000,40.3770000000000,41.2570000000000,42.2520000000000,43.2800000000000,44.2160000000000,45.1490000000000,46.1790000000000,47.2260000000000,48.1870000000000,49.1360000000000,50.2220000000000,51.3240000000000,52.3400000000000,53.2800000000000,54.2750000000000,55.4050000000000,56.5300000000000,57.5530000000000,58.5850000000000,59.6640000000000,60.6640000000000,61.6140000000000,62.7610000000000,63.8230000000000,64.8040000000000,65.9080000000000,67.0560000000000,68.0810000000000,69.1380000000000,70.2060000000000,71.1900000000000,72.2370000000000,73.3250000000000,74.3050000000000,75.3310000000000,76.4380000000000,77.4460000000000,78.4250000000000,79.4900000000000,80.5880000000000,81.5760000000000,82.6540000000000,83.7080000000000,84.6390000000000,85.6360000000000,86.7050000000000,87.7210000000000,88.6310000000000,89.6890000000000,90.7540000000000,91.7460000000000,92.7570000000000,93.8380000000000,94.8490000000000,95.7690000000000,96.8210000000000]
t1=[0.196000000000000,1.13900000000000,2.03600000000000,3.04600000000000,4.04900000000000,4.93400000000000,5.90100000000000,6.97500000000000,7.95300000000000,8.83400000000000,9.61600000000000,10.3740000000000,11.1410000000000,12.0020000000000,12.8700000000000,14.0140000000000,15.1210000000000,16.1940000000000,17.2230000000000,18.1150000000000,18.8850000000000,19.6220000000000,20.4060000000000,21.2920000000000,22.2280000000000,23.1800000000000,24.1070000000000,24.9870000000000,25.8710000000000,26.7790000000000,27.6820000000000,28.6010000000000,29.5120000000000,30.3530000000000,31.1680000000000,32.0640000000000,33.0270000000000,33.9180000000000,34.7930000000000,35.7840000000000,36.8060000000000,37.7130000000000,38.6610000000000,39.6720000000000,40.5790000000000,41.4500000000000,42.4420000000000,43.4790000000000,44.4190000000000,45.3440000000000,46.3700000000000,47.4230000000000,48.3910000000000,49.3320000000000,50.4140000000000,51.5180000000000,52.5440000000000,53.4820000000000,54.4680000000000,55.5970000000000,56.7320000000000,57.7540000000000,58.7790000000000,59.8600000000000,60.8690000000000,61.8090000000000,62.9540000000000,64.0280000000000,65.0040000000000,66.1040000000000,67.2660000000000,68.3030000000000,NaN,NaN,88.0120000000000,88.9140000000000,89.9530000000000,91.0270000000000,92.0150000000000,93.0150000000000,94.0890000000000,95.1110000000000,96.0140000000000,97.0600000000000]
indexA = 1;
indexB =1;
for indexB = 1:length(t1)
c=t1(indexB)-t(indexA);
if (c<0.2600)&&(c<0.1900)
disp(t(indexA))
disp(t1(indexB))
d(indexB)=c;
disp('----------')
else
indexA=indexA+1;
c=t1(indexB)-t(indexA);
while (c>0.2600)&&(c>0.1900)
indexA=indexA+1;
c=t1(indexB)-t(indexA);
d(indexB)=c;
end
disp(t(indexA))
disp(t1(indexB))
d(indexB)=c;
disp('----------')
end
end
  2 Comments
Chandler Hall
Chandler Hall on 15 Nov 2022
What is the source of t and t1? They are different lengths. Would it not be more appropriate to ensure they are the same length and then check the value of t1(i) - t(i)? What is the ultimate purpose of d?
Krystian Sliwka
Krystian Sliwka on 17 Nov 2022
any idea to make them the same size ?
The purpose of d is to display the results on a graph. They should be within a given range

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Answers (1)

David Hill
David Hill on 15 Nov 2022
Edited: David Hill on 15 Nov 2022
Hard to understand what you are trying to do. Looks like you are good to about 72.
t=[0.0840000000000000,0.935000000000000,1.83900000000000,2.85400000000000,3.84500000000000,4.73200000000000,5.71000000000000,6.77900000000000,7.75000000000000,8.62400000000000,9.40900000000000,10.1720000000000,10.9430000000000,11.8060000000000,12.6770000000000,13.8250000000000,14.9320000000000,16.0030000000000,17.0310000000000,17.9190000000000,18.6890000000000,19.4270000000000,20.2120000000000,21.1040000000000,22.0370000000000,22.9860000000000,23.9090000000000,24.7900000000000,25.6760000000000,26.5840000000000,27.4870000000000,28.4060000000000,29.3130000000000,30.1520000000000,30.9750000000000,31.8760000000000,32.8330000000000,33.7180000000000,34.6020000000000,35.5970000000000,36.6070000000000,37.5170000000000,38.4710000000000,39.4750000000000,40.3770000000000,41.2570000000000,42.2520000000000,43.2800000000000,44.2160000000000,45.1490000000000,46.1790000000000,47.2260000000000,48.1870000000000,49.1360000000000,50.2220000000000,51.3240000000000,52.3400000000000,53.2800000000000,54.2750000000000,55.4050000000000,56.5300000000000,57.5530000000000,58.5850000000000,59.6640000000000,60.6640000000000,61.6140000000000,62.7610000000000,63.8230000000000,64.8040000000000,65.9080000000000,67.0560000000000,68.0810000000000,69.1380000000000,70.2060000000000,71.1900000000000,72.2370000000000,73.3250000000000,74.3050000000000,75.3310000000000,76.4380000000000,77.4460000000000,78.4250000000000,79.4900000000000,80.5880000000000,81.5760000000000,82.6540000000000,83.7080000000000,84.6390000000000,85.6360000000000,86.7050000000000,87.7210000000000,88.6310000000000,89.6890000000000,90.7540000000000,91.7460000000000,92.7570000000000,93.8380000000000,94.8490000000000,95.7690000000000,96.8210000000000];
t1=[0.196000000000000,1.13900000000000,2.03600000000000,3.04600000000000,4.04900000000000,4.93400000000000,5.90100000000000,6.97500000000000,7.95300000000000,8.83400000000000,9.61600000000000,10.3740000000000,11.1410000000000,12.0020000000000,12.8700000000000,14.0140000000000,15.1210000000000,16.1940000000000,17.2230000000000,18.1150000000000,18.8850000000000,19.6220000000000,20.4060000000000,21.2920000000000,22.2280000000000,23.1800000000000,24.1070000000000,24.9870000000000,25.8710000000000,26.7790000000000,27.6820000000000,28.6010000000000,29.5120000000000,30.3530000000000,31.1680000000000,32.0640000000000,33.0270000000000,33.9180000000000,34.7930000000000,35.7840000000000,36.8060000000000,37.7130000000000,38.6610000000000,39.6720000000000,40.5790000000000,41.4500000000000,42.4420000000000,43.4790000000000,44.4190000000000,45.3440000000000,46.3700000000000,47.4230000000000,48.3910000000000,49.3320000000000,50.4140000000000,51.5180000000000,52.5440000000000,53.4820000000000,54.4680000000000,55.5970000000000,56.7320000000000,57.7540000000000,58.7790000000000,59.8600000000000,60.8690000000000,61.8090000000000,62.9540000000000,64.0280000000000,65.0040000000000,66.1040000000000,67.2660000000000,68.3030000000000,NaN,NaN,88.0120000000000,88.9140000000000,89.9530000000000,91.0270000000000,92.0150000000000,93.0150000000000,94.0890000000000,95.1110000000000,96.0140000000000,97.0600000000000];
a=t1(1:72)-t(1:72)
a = 1×72
0.1120 0.2040 0.1970 0.1920 0.2040 0.2020 0.1910 0.1960 0.2030 0.2100 0.2070 0.2020 0.1980 0.1960 0.1930 0.1890 0.1890 0.1910 0.1920 0.1960 0.1960 0.1950 0.1940 0.1880 0.1910 0.1940 0.1980 0.1970 0.1950 0.1950
b=a(a>=.19&a<=.26)%a few are not in the range, do not know what you want to do if they are not in the range
b = 1×65
0.2040 0.1970 0.1920 0.2040 0.2020 0.1910 0.1960 0.2030 0.2100 0.2070 0.2020 0.1980 0.1960 0.1930 0.1910 0.1920 0.1960 0.1960 0.1950 0.1940 0.1910 0.1940 0.1980 0.1970 0.1950 0.1950 0.1950 0.1950 0.1990 0.2010
  1 Comment
Krystian Sliwka
Krystian Sliwka on 16 Nov 2022
Look at the vectors t. You can see there the value up to 68 further we have NaN and only the values 88. I want the program to subtract t1-t. If the value in the range doesn't come out we take another sample from the t vector. E.g.
68.303-68.081=0.222
Then we have NaN and then for 88.012 but that corresponds to 71.19.
So that the value is in the range:88.12-87.721=2.91.
Here I noticed that I should increase the range, however.
I want the program to keep adding samples until it gets a value in range.
That is, the while loop probably has the wrong condition
Translated with www.DeepL.com/Translator (free version)

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