delete rows depend on repeated value from 1 col

조회 수: 1 (최근 30일)
bero
bero 2016년 2월 16일
댓글: Walter Roberson 2016년 2월 18일
Hi I have a 20000x60 cell matrix I want to delete rows depend on repeated values from the first columns.
s = {
1 '2013-10-01' '05:35:00' '01:05:00'
2 '2013-10-01' '10:20:00' '00:00:00'
2 '2013-10-01' '10:20:00' '00:00:00'
3 '2013-10-01' '11:00:00' '00:40:00'
3 '2013-10-01' '11:00:00' '00:40:00'
3 '2013-10-01' '11:00:00' '00:40:00'
3 '2013-10-01' '11:00:00' '00:40:00'
5 '2013-10-01' '14:50:00' '00:50:00'
5 '2013-10-01' '14:50:00' '00:50:00'
6 '2013-10-01' '15:15:00' '01:15:00'
7 '2013-10-01' '15:55:00' '00:05:00'
7 '2013-10-01' '15:55:00' '00:05:00'
7 '2013-10-01' '15:55:00' '00:05:00'
7 '2013-10-01' '15:55:00' '00:05:00'}
the result should like:
s1 =
1 '2013-10-01' '05:35:00' '01:05:00'
2 '2013-10-01' '10:20:00' '00:00:00'
3 '2013-10-01' '11:00:00' '00:40:00'
5 '2013-10-01' '14:50:00' '00:50:00'
6 '2013-10-01' '15:15:00' '01:15:00'
7 '2013-10-01' '15:55:00' '00:05:00'
8 '2013-10-01' '19:00:00' '00:05:00'
9 '2013-10-01' '19:10:00' '00:10:00'
I use unique for it but because the different type of data there is Error using cell/unique (line 85) Input A must be a cell array of strings."

채택된 답변

Star Strider
Star Strider 2016년 2월 16일
If the number in the first column is the same for all repeated values in the rest of the row, just use it.
Using your example data:
s = { 1 '2013-10-01' '05:35:00' '01:05:00'
2 '2013-10-01' '10:20:00' '00:00:00'
2 '2013-10-01' '10:20:00' '00:00:00'
3 '2013-10-01' '11:00:00' '00:40:00'
3 '2013-10-01' '11:00:00' '00:40:00'
3 '2013-10-01' '11:00:00' '00:40:00'
3 '2013-10-01' '11:00:00' '00:40:00'
5 '2013-10-01' '14:50:00' '00:50:00'
5 '2013-10-01' '14:50:00' '00:50:00'
6 '2013-10-01' '15:15:00' '01:15:00'
7 '2013-10-01' '15:55:00' '00:05:00'
7 '2013-10-01' '15:55:00' '00:05:00'
7 '2013-10-01' '15:55:00' '00:05:00'
7 '2013-10-01' '15:55:00' '00:05:00'};
sc1 = cellfun(@(x)num2str(x, '%.0f'), s(:,1));
[s1u, ia] = unique(sc1);
s1 = s(ia,:)
s1 =
[1.0000e+000] '2013-10-01' '05:35:00' '01:05:00'
[2.0000e+000] '2013-10-01' '10:20:00' '00:00:00'
[3.0000e+000] '2013-10-01' '11:00:00' '00:40:00'
[5.0000e+000] '2013-10-01' '14:50:00' '00:50:00'
[6.0000e+000] '2013-10-01' '15:15:00' '01:15:00'
[7.0000e+000] '2013-10-01' '15:55:00' '00:05:00'
You may have to modify this slightly if your actual cell array is different, but it works with the array you posted, and as I interpreted it.
  댓글 수: 8
Star Strider
Star Strider 2016년 2월 17일
The '%.0f' is one of a number of format descriptors (see the link Stephen provided for a full discussion of all of them), this one telling MATLAB to produce a string representing a floating-point number with no digits to the right of the decimal. So using that format descriptor, pi=3.1415... would print as 3 with no trailing decimal point. When you read the documentation, I leave it for you to determine the reason I chose this format rather than, for example, '%d'.
Walter Roberson
Walter Roberson 2016년 2월 18일
bero commented "good"

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추가 답변 (1개)

MHN
MHN 2016년 2월 16일
편집: MHN 2016년 2월 16일
Consider this example (it is not the fastest and easiest way, but it solves your problem)
s = {3 '2013-10-01' '11:00:00' '00:40:00'; 3 '2013-10-01' '11:00:00' '00:40:00'; 3 '2013-10-01' '11:00:00' '00:40:00'; 5 '2013-10-01' '14:50:00' '00:50:00'; 5 '2013-10-01' '14:50:00' '00:50:00'; 6 '2013-10-01' '15:15:00' '01:15:00'; 7 '2013-10-01' '15:55:00' '00:05:00'};
col1 = cell2mat(s(:,1));
uniqcol1 = union(col1,col1);
% change all the repeated rows to '0' (or some number which is unused in the first column) except the first one and then remove them.
for i=1:length(uniqcol1)
for j = 1:size(s,1)
if uniqcol1(i)==s{j,1}
for k = j+1: size(s,1)
if uniqcol1(i)==s{k,1}
s{k,1} = 0;
end
end
end
end
end
col1 = cell2mat(s(:,1));
s(col1==0,:)=[];
  댓글 수: 1
bero
bero 2016년 2월 17일
It is so slowly for my my big matrix, I need more efficient and faster process...

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