delete rows depend on repeated value from 1 col

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bero
bero 2016-2-16
评论： Walter Roberson 2016-2-18
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."

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采纳的回答

Star Strider
Star Strider 2016-2-16
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.
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Star Strider
Star Strider 2016-2-17
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'.

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更多回答（1 个）

MHN
MHN 2016-2-16
编辑：MHN 2016-2-16
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,:)=[];
  1 个评论
bero
bero 2016-2-17
It is so slowly for my my big matrix, I need more efficient and faster process...

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