CODEWITHSUNDEEP
matlabrowsequence
user2100039
user2100039

Reputation: 1366

Extract Sequence of Rows in Column (MATLAB)

I am trying to extract a sequence of rows from a column of data in MATLAB. For example, my data looks like this in one column:

10 
20 
30 
40 
50 
60 
70 
80 
90 
100 
110 
120 
130

I want to select a sequence of rows and store these rows in a column vector A that looks like this after extracting the first 3 rows skipping 2 rows then selecting the next 3 rows and skipping 2 rows and selecting the next 3 to the end of the column.

Finally, the data should look like this with 9 rows and 1 column:

10 
20
30
60 
70 
80
110 
120
130

Thank you!

Upvotes: 0

Views: 213

Answers (3)

user2100039
user2100039

Reputation: 1366

I just figured out another way to compliment the nice solutions using find function.

If my data is: 

x = [10; 20; 30; 40; 50; 60; 70; 80; 90; 100; 110; 120; 130]

and I want to extract a repeating sequence of rows then:
Develop an index of the repeating pattern or row sequence of keeps (3) and skips (2) rows. My index would be then: 

index = [1:5 1:5 1:5]

or 

index = 123451234512345

Then use the find function to logically index the "keep" sequence. 

a = find(index<4)

which gives: 

a = 1 2 3 6 7 8 11 12 13

then: 

b = x(a)

produces 

b = 10 20 30 60 70 80 110 120 130

Upvotes: 0

Luis Mendo
Luis Mendo

Reputation: 112749

Let

x = [10 20 30 40 50 60 70 80 90 100 110 120 130].'; %'// data
m = 3; %// keep
n = 2; %// skip

Then

y = x(mod(0:end-1,m+n)<=m-1);

or equivalently

y = x(mod(0:numel(x)-1,m+n)<=m-1);

How this works: generate the sequence 0,1,2,... numel(x)-1. Each number corresponds to a position in x. To keep m and skip n cyclically, you apply mod(...,m+n) to that sequence. For example, for m=3, n=2 the result is sequence 0,1,2,3,4,0,1,2,.... You then select those numbers (that is, the corresponding entries of x) that are <=m-1. This results in the periodic pattern "keep, keep, keep, skip, skip".

To first skip and then keep (reverse from above): just reverse the inequality:

y = x(mod(0:end-1,m+n)>=m-1);

or

y = x(mod(0:numel(x)-1,m+n)>=m-1);

Upvotes: 2

Divakar
Divakar

Reputation: 221614

Try this, assuming a is your input column vector -

r_accept = 3; %// number of rows to accept
r_reject = 2; %// number of rows to reject

ind1 = bsxfun(@plus,1:r_accept+r_reject:numel(a),[0:r_accept-1]') %//'
out = a(ind1(ind1(:)<=numel(a))) %// desired output

Upvotes: 1

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