Hi, I have a single column array of some positive and negative values. I am trying to use a loop to segregate them and to store the result in a separate vector. Unable to do ot so far. can anyone help please? This is my code:
Diff=MP_25-MP_60
y=nan(117,1)
for i=Diff(1:117) if i>0 y(i) = ('Upward') else y(i) = ('Downward') end end

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David Fletcher
David Fletcher am 18 Mär. 2018
Bearbeitet: David Fletcher am 18 Mär. 2018

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y=Diff>=0
will return a logical array of all elements in Diff that are greater than or equal to 0. You can either just use this as a marker, or if you wish index out all the positive values (and by extension also all the negative values) into separate vectors: -
positives=Diff(y)
negatives=Diff(~y)

6 Kommentare

Ali Ajaz
Ali Ajaz am 18 Mär. 2018
Thanks for answering to this question. My problem is I am trying to store the output of the loop as a string into the vector y. But I am unable to do it. Could you please guide me on how to store the output of this for loop in the vector?
Ali Ajaz
Ali Ajaz am 18 Mär. 2018
David Fletcher
David Fletcher am 18 Mär. 2018
Bearbeitet: David Fletcher am 18 Mär. 2018
for a start you won't be able to store 'Upward' and 'Downward' in a matrix without padding them to the same length (and that length will be a minimum of 8 columns ('Downward' has eight letters) not the one you have pre-allocated in y). Would 'U' and 'D' be enough? If you don't want to pad, or contract them down to a single letter, you can allocate to a cell array instead
Ali Ajaz
Ali Ajaz am 18 Mär. 2018
tried, it gives following error: Subscript indices must either be real positive integers or logicals.
David Fletcher
David Fletcher am 18 Mär. 2018
Tried what?
Ali Ajaz
Ali Ajaz am 18 Mär. 2018
y=nan(117,1)
for i=Diff(1:117,:) if i>0 y = ('U') else y = ('D') end end

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Weitere Antworten (1)

Ali Ajaz
Ali Ajaz am 18 Mär. 2018

0 Stimmen

y=nan(117,1)
for i=Diff(1:117,:) if i>0 y = ('U') else y = ('D') end end

3 Kommentare

David Fletcher
David Fletcher am 18 Mär. 2018
Bearbeitet: David Fletcher am 18 Mär. 2018
for iter=1:length(Diff)
if (Diff(iter)>=0)
y(iter)='U'
else
y(iter)='D'
end
end
You may find that because you pre-allocated y as NaN, when you add 'U' or 'D' you get their numeric codes instead of the letter.
you could pre-allocate y to a character vector instead
y=repmat('U',117,1)
Actually, if you pre-allocate y to 'D' you can lose the else clause in the condition block (since the array is already set to 'D')
y=repmat('D',117,1)
for iter=1:length(Diff)
if (Diff(iter)>=0)
y(iter)='U'
end
end

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