what can we use when we dont have value in a matrix instead of 0?

i have a matrix of my data and i dont want to put 0 for the items that i dont have ...is there any other way?

[Jonas]

what about NaN or missing ?

[arian hoseini]

% relay awayBus closeBus Line Backup1 Backup2

Data= [1 1 2 1-6,1-3 14 nan

2 3 1 3-4 3 nan

3 4 3 4-5 4 nan

4 5 4 5-6 5 nan

5 6 5 6-1,6-2 7 6

6 2 6 2-1 1 nan

7 1 6 1-3,1-2 2 8

8 2 1 2-6 13 nan

9 1 3 1-6,1-2 14 8

10 3 4 3-1 9 nan

11 4 5 4-3 10 nan

12 5 6 5-4 11 nan

13 6 2 6-1,6-5 7 12

14 6 1 6-2,6-5 6 12];

Dimensions of matrices being concatenated are not consistent.

Error in ProtectionTask4_ArianHoseini (line 16)

Data= [1 1 2 1-6,1-3 14 NaN

i get this error is it because of 4th column

Answer 1

[Cris LaPierre]

use nan?

data = rand(3);
data(2,3) = nan

data = 3×3

0.7049 0.8908 0.2471 0.2524 0.5020 NaN 0.4604 0.1822 0.3545

You could also use missing, especially if you are usure of the data type. The result is the same as nan for doubles.

data(3,2) = missing

data = 3×3

0.7049 0.8908 0.2471 0.2524 0.5020 NaN 0.4604 NaN 0.3545

c = ["a","b","c"];
c(1) = missing

c = 1×3 string array
    <missing>    "b"    "c"

[Cris LaPierre]

You can also create the vector with nan.

x = [1 nan 3]

x = 1×3

1 NaN 3

There are other related functions to be aware of. See here:

• https://www.mathworks.com/help/matlab/referencelist.html?type=function&category=preprocessing-data&s_tid=CRUX_topnav

[arian hoseini]

thank u sir but see my matrix(Data(:,4)) there is another problem....Error using vertcat....that column is just for explanation and i dont want to use it but i need it to be there also i cant make it comment

[Cris LaPierre]

The issue is that column 4 cannot be stored as a double. It probably needs to be string. I would consider using a table instead of a matrix. Then you could capture all the data.

Assuming your columns are separate variables, you could do the following (first 5 rows)

relay = (1:5)';
awayBus = [1;3;4;5;6];
closeBus = [2;1;3;4;5];
Line = ["1-6,1-3";"3-4";"4-5";"5-6";"6-1,6-2"];
Backup1 = [14;3;4;5;7];
Backup2 =[nan;nan;nan;nan;6];

data = table(relay, awayBus, closeBus, Line, Backup1, Backup2)

data = 5×6 table
    relay    awayBus    closeBus      Line       Backup1    Backup2
    _____    _______    ________    _________    _______    _______

      1         1          2        "1-6,1-3"      14         NaN  
      2         3          1        "3-4"           3         NaN  
      3         4          3        "4-5"           4         NaN  
      4         5          4        "5-6"           5         NaN  
      5         6          5        "6-1,6-2"       7           6  

Otherwise, could you explain more about your data? Perhaps share the code you use to create Data?

[arian hoseini]

for example for relay 1 i need to sumZ(1)+min(Z2(14),Z2(2))....the indexes u see are from my Data

[Cris LaPierre]

Sorry, but I don't see how Z relates to Data.

Also, it appears your outer for loop in Zone 1 is unnecessary (for k = 1 : 2). It runs the same code without any changes. You'd get the same results without it.