How can I add data to an existing double at specific points?

Hello,

I have this 14x1 double, I need to add 3 more data points at specific locations, effectively making it a 17x1 double. It looks like this named mean_RT

765
413
309.006493506494
288.863636363636
562.833333333333
408
925.333333333333
605
201.800000000000
530
431.666666666667
640.659574468085
625.500000000000
851.666666666667

advars is only applicable on tables, I tried expanding it with undefined values as

mean_RT = nan(17,1)

but that just creates a double of 17 Nans. I tried indexing a 0 at the points I needed (these are test subject averages that need to be the same location for statistical purposes)

mean_RT([2,13,14]) = 0 

and while it does insert a zero at the technical point, instead of adding a value, it replaces the value so my double is still 14x1. It I try for separate points such as

mean_RT([2,1],[13,1],[14,1])= 0 

I wind up with a 14x13x14 double. What am I doing wrong?

Answer 1

[Dave B]

The easy to think about solution might feel a little messy:

% insert the numbers 13 and 18 between a(2)/a(3) and a(4)/a(5)
a=1:5

a = 1×5

1 2 3 4 5

b=[a(1:2) 13 a(3:4) 18 a(5)]

b = 1×7

1 2 13 3 4 18 5

How about taking this strategy, make the new array all NaN, assign the new values, then assign the old array to the remaining NaNs?

a = 1:5;
b = nan(1,7);
b([3 6]) = [13 18];
b(isnan(b)) = a

b = 1×7

1 2 13 3 4 18 5

[Dave B]

mean_RT([2,1],[13,1],[14,1])= 0

is saying assign a 0 to first dimension [2 1], second dimension [13 1], third dimension [14 1], here's an example with 1s instead of zeros (and smaller numbers) so you can see what's being indexed:

mean_RT([5,1],[4,1],[3,1])= 1

mean_RT = 
mean_RT(:,:,1) =

     1     0     0     1
     0     0     0     0
     0     0     0     0
     0     0     0     0
     1     0     0     1


mean_RT(:,:,2) =

     0     0     0     0
     0     0     0     0
     0     0     0     0
     0     0     0     0
     0     0     0     0


mean_RT(:,:,3) =

     1     0     0     1
     0     0     0     0
     0     0     0     0
     0     0     0     0
     1     0     0     1

[Ruth Ronalter]

Yeah the easier solution doesn't work for me, it runs into a horzcat error: Dimensions of arrays being concatenated are not consistent. I'm also on r2018a, so that might also be an issue.

Trying the second method, I'm understand the first 2 steps

mean_inserts = nan(17,1) %it's supposed to be the projected size, right?
mean_inserts([2 13 14])= nan nan nan; 
mean_inserts(isnan(mean_inserts))= mean_RT;

but when I'm overlaying the first array to the Nan array, it just comes out all Nans for the variable. What am I doing wrong? Appreciate the explanation about the third dimension output, I was confused why there were multiple matrices for 1 equation

[Dave B]

@Ruth Ronalter

WRT the first approach, I used a row vector as an example, but your data is in a column vector. That just means you need a ; between each value instead of a space. In the second approahc, I was explictly relying on the fact that the inserted data is not NaN, so when you inserted three NaN values I'd expect it to error.

Because we can now run code right in ML Answers, I'll just run my suggestions with the data you included below. To make matters just a tiny bit more confusing, I'll display the results using a row vector (because otherwise the display will just show ... after a few entries).

mean_RT=[765;413;309.006493506494;288.863636363636;562.833333333333;408;925.333333333333;605;201.800000000000;530;431.666666666667;640.659574468085;625.500000000000;851.666666666667];

First approach (simple):

new_mean_RT = [mean_RT(1); 0; mean_RT(2:11); 0; 0; mean_RT(12:end)];
disp(new_mean_RT')

  765.0000         0  413.0000  309.0065  288.8636  562.8333  408.0000  925.3333  605.0000  201.8000  530.0000  431.6667         0         0  640.6596  625.5000  851.6667

Second approach (as a tip, to understand exactly how this works take off the semi-colons and run each line separately. I broke out into an extra line just so that it's clear what's happening.

new_mean_RT = nan(17,1);
new_mean_RT([2 13 14])=0;
ind=isnan(new_mean_RT);
new_mean_RT(ind)=mean_RT;
disp(new_mean_RT')

  765.0000         0  413.0000  309.0065  288.8636  562.8333  408.0000  925.3333  605.0000  201.8000  530.0000  431.6667         0         0  640.6596  625.5000  851.6667

[Ruth Ronalter]

Sorry, realized I never accepted your answer. I got both methods to work, thank you for taking the time to explain it to me and making it work.