Extract the index and find the value in other array that have the same extracted index?

I have many arrays but in my question I'll minimize it to be understood for you.

if I have

t1=[ 1.0000 1.1000 1.2000 1.3000 1.4000 1.5000]

the corresponsing x for t1 is

x1=[ 0.1690 0.6491 0.7317 0.6477 0.4509 0.5470]

I have other iteration with different length than t1,x1 as showing below:

t2=[

1.0000 1.0100 1.0200 1.0300 1.0400 1.0500 1.0600

1.0700 1.0800 1.0900 1.1000 1.1100 1.1200 1.1300

1.1400 1.1500 1.1600 1.1700 1.1800 1.1900 1.2000

1.2100 1.2200 1.2300 1.2400 1.2500 1.2600 1.2700

1.2800 1.2900 1.3000 1.3100 1.3200 1.3300 1.3400

1.3500 1.3600 1.3700 1.3800 1.3900 1.4000 1.4100

1.4200 1.4300 1.4400 1.4500 1.4600 1.4700 1.4800

1.4900 1.5000

]

the corresonsing x2 for t2 is

x2=[

0.4427 0.1067 0.9619 0.0046 0.7749 0.8173 0.8687 0.0844 0.3998 0.2599

0.8001 0.4314 0.9106 0.1818 0.2638 0.1455 0.1361 0.8693 0.5797 0.5499

0.1450 0.8530 0.6221 0.3510 0.5132 0.4018 0.0760 0.2399 0.1233 0.1839

0.2400 0.4173 0.0497 0.9027 0.9448 0.4909 0.4893 0.3377 0.9001 0.3692

0.1112 0.7803 0.3897 0.2417 0.4039 0.0965 0.1320 0.9421 0.9561 0.5752

0.0598

]

I need for loop that can extarct the index for t2 that has same value in t1, then after finding the index for same value in t2 with t1 , find the coressponding x2 for each t2 that have same value in t1.

the output will be new x which is extracted from x2 and have same length of x1

[Sargondjani]

Have a look at 'find'. there are some examples that will help you further. You can for example get row and column index and use a for loop over t1, with

[row,col]= find(t2 == t1(it1))

, something like that.

[Willim]

It is wrong , it asks to be same size

Answer 1

[Bob Thompson]

You shouldn't actually need a for loop to do this, just some logic indexing.

goal = x2(ismember(t2,t1));

[Stephen23]

Keep in mind that floating point error means that checking for exact equivalence of decimal values is not recommended, and can easily lead to unexpected outputs. For the data shown in the question it would be much more robust to use ismembertol.