grouping values that as need

1 Ansicht (letzte 30 Tage)
Nicle Davidson
Nicle Davidson am 5 Nov. 2021
Kommentiert: Nicle Davidson am 5 Nov. 2021
I hava a csv file with totally random numbers all in one column, that I read from using
whoeCulomn = readtable('test2.csv');
this table have 60 values in one column,
I would like to splitt these 60 values into 10 groups in which each of these have 6 of the values. for example the frist group have from 1 ot 6 the second group have from 7 to 12 etc
How can I do that?
*the groups of my numbers should be presented such as:
x1=[the first group of six numbers]
x2=[the second group]
x3=[...];
x4=[...];
x5=[...];
x6=[...];

Akzeptierte Antwort

Sudharsana Iyengar
Sudharsana Iyengar am 5 Nov. 2021
Bearbeitet: Sudharsana Iyengar am 5 Nov. 2021
An example:
x=linspace(1,60,60);
k=1;
for i =1:6:length(x)
B(k,1:6)=x(i:i+5) %; add this semicolon if you dont want this to be printed.
k=k+1;
end
B = 1×6
1 2 3 4 5 6
B = 2×6
1 2 3 4 5 6 7 8 9 10 11 12
B = 3×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
B = 4×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
B = 5×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
B = 6×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
B = 7×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42
B = 8×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48
B = 9×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
B = 10×6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
A = 1:60;
B = reshape(A,[10,6]) %more easier way
  10 Kommentare
Sudharsana Iyengar
Sudharsana Iyengar am 5 Nov. 2021
May be this explanation is more clearer:
Instead of having has X1,X2...X10 you have X with 10 rows and 6 coloumns. Each row corresponds to each of X1,X2... So you can access them by calling the row index.
X(i,:) % will call the ith row and all the columns. So if i is 1 you are acessing X1 if it is 3
% you are acessing X3 and so on.
Nicle Davidson
Nicle Davidson am 5 Nov. 2021
I appreciate your time and effort and I accept this answer. Thank you

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Resizing and Reshaping Matrices finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by