# How can I delete a row based on the value of the previous row?

조회 수: 5 (최근 30일)
Kellie Wilson 2022년 7월 31일
편집: Bruno Luong 2022년 7월 31일
I have a file containing one column and a large number of rows. I want to delete any row whose value is less than 2 greater than the previous row (ie if row 1 is 10, and row 2 is 11, then delete row 2. If row 1 is 10, and row 2 is 12.1, then don't delete it). How can I do this? I know how to delete specificied rows with something like
rowstodelete = xxxxxx
A(rowstodelete,:) = [ ];
But I don't know how to figure out which rows need to be deleted (ie what do I need to put in for xxxxx.) I was thinking maybe using logical indexing or if-else statments, but not really sure how. Open to other options too! Thanks in advance!
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Walter Roberson 2022년 7월 31일
Suppose you have three rows, with values 10, 11.5, 13.3 . The second row is not at least 2 greater than the first. The third row is not at least 2 greater than the second. Do I understand correctly that you would delete the second and the third row both?
Or should the deletion in a sense be carried out "immediately", so that after the 11.5 is deleted, the 13.3 would be compared to the now-previous 10 ?
Kellie Wilson 2022년 7월 31일
Good question - I want the latter. So I would want to compare the 13.3 to the now previous 10 and thus keep it.

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### 답변 (5개)

Bruno Luong 2022년 7월 31일
편집: Bruno Luong 2022년 7월 31일
This avoids deletetion which like growsing array wouls kill the performance.
A=[10, 11.5, 13.3 13.5 13.7 17];
keep = false(size(A));
b = -Inf;
for i=1:length(A)
if A(i) >= b
keep(i) = true;
b = A(i) + 2;
end
end
A = A(keep)
A = 1×3
10.0000 13.3000 17.0000
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Matt J 2022년 7월 31일
rowstodelete = diff([inf;A])<2;
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Kellie Wilson 2022년 7월 31일
This works well for the first scenario that Walter laid out, where if I had 3 rows: 10, 11.5, 13.3, then it would delete 11.5 and 13.3. But I just want to delete the 11.5, and then keep 13.3 since it would be more than 2 away from the 10 that is now right before it.

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Walter Roberson 2022년 7월 31일
편집: Walter Roberson 2022년 7월 31일
Not a full algorithm, but something to think about
rng(1234)
data = randi(99, 1, 20) %row vector
data = 1×20
19 62 44 78 78 27 28 80 95 87 36 50 68 71 37 56 50 2 77 88
G = tril(data.'-2 >= data)
G = 20×20 logical array
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 0
pos = sum(cumprod(~G,1),1)+1
pos = 1×20
2 4 4 8 8 8 8 9 21 21 12 13 14 19 16 19 19 19 20 21
find the first 1 in the first column of G; it is at row 2. Entry 2 is the first entry in data that is at least 2 greater than entry 1. Cross check: 62 >= 19+2. Let C = 2 (column 2). pos(1) = 2 -- the required value of the first 1
Look in column C (#2) for the first 1; it is at entry 4. Entry 4 is the first entry in data that is at least 2 greater than entry 2. Cross-check: 78 >= 62+2. Let C = 4 (column 4). pos(2) = 4 -- the required value of the first 1
Look in column C (#4) for the first 1; it is at entry 8. Entry 8 is the first entry in data that is at least 2 greater than entry 4. Cross-check: 80 >= 78+2. Let C = 8 (column 8). pos(4) = 8 -- the required value of the first 1
Look in column C (#8) for the first 1. pos(8) = 9, entry 9 is the first entry, 95 >= 80+2
and so on. The G matrix is the details, the pos vector is the condensed information, the position you need. So you can just read the positions out of pos. pos(1) gives the next index to look at in pos, pos() at that gives the next index, and so on.
When you get an index that is 1 greater than the number of elements in data then you have reached the end, there are no more positions that satisfy the condition.
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Matt J 2022년 7월 31일
It'll have to be done with a loop,
A=[10, 11.5, 13.3 13.5 13.7 17];
i=2;
while i<=numel(A)
if A(i)<A(i-1)+2
A(i)=[];
else
i=i+1;
end
end
A
A = 1×3
10.0000 13.3000 17.0000
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Bruno Luong 2022년 7월 31일
편집: Bruno Luong 2022년 7월 31일
If your data is sorted you could try this
A = 1:0.1:10
A = 1×91
1.0000 1.1000 1.2000 1.3000 1.4000 1.5000 1.6000 1.7000 1.8000 1.9000 2.0000 2.1000 2.2000 2.3000 2.4000 2.5000 2.6000 2.7000 2.8000 2.9000 3.0000 3.1000 3.2000 3.3000 3.4000 3.5000 3.6000 3.7000 3.8000 3.9000
A = uniquetol(A,2*(1-eps),'DataScale',1)
A = 1×5
1 3 5 7 9

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