How to use cumsum for data in a table which is not NaN?
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Hi. I have a question about how to use cumsum function for data in a table which is not NaN.
'02-Nov-2017 10:50:49' NaN
'02-Nov-2017 10:50:53' NaN
'03-Nov-2017 00:00:00' NaN
'03-Nov-2017 08:00:00' NaN
'04-Nov-2017 00:00:00' NaN
'04-Nov-2017 08:00:00' NaN
'05-Nov-2017 00:00:00' NaN
'05-Nov-2017 08:00:00' NaN
'05-Nov-2017 14:00:41' 0.100000000000000
'05-Nov-2017 14:04:08' 0.100000000000000
'05-Nov-2017 14:06:40' 0.100000000000000
'05-Nov-2017 14:10:00' NaN
'05-Nov-2017 14:12:58' NaN
'05-Nov-2017 14:13:24' NaN
'05-Nov-2017 14:14:00' NaN
'05-Nov-2017 14:15:58' 0.100000000000000
'05-Nov-2017 14:16:24' 0.100000000000000
For example I have this above data. I want the cumsum function only calculate the value which is not NaN. The result should be like this:
'02-Nov-2017 10:50:49' NaN
'02-Nov-2017 10:50:53' NaN
'03-Nov-2017 00:00:00' NaN
'03-Nov-2017 08:00:00' NaN
'04-Nov-2017 00:00:00' NaN
'04-Nov-2017 08:00:00' NaN
'05-Nov-2017 00:00:00' NaN
'05-Nov-2017 08:00:00' NaN
'05-Nov-2017 14:00:41' 0.100000000000000
'05-Nov-2017 14:04:08' 0.200000000000000
'05-Nov-2017 14:06:40' 0.300000000000000
'05-Nov-2017 14:10:00' NaN
'05-Nov-2017 14:12:58' NaN
'05-Nov-2017 14:13:24' NaN
'05-Nov-2017 14:14:00' NaN
'05-Nov-2017 14:15:58' 0.100000000000000
'05-Nov-2017 14:16:24' 0.200000000000000
I try to find the example for cumsum, but I found nothing. Could you please hep me? Thank you very much for your help.
0 个评论
采纳的回答
Jan
2017-12-2
编辑:Jan
2017-12-3
idx = ~isnan(Val);
Val2 = Val;
Val2(idx) = cumsum(Val(idx));
Or:
Val2 = cumsum(Val, 'omitnan');
Val2(isnan(Val)) = NaN;
[EDITED] And with a reset of the sum after each NaN block:
Data = [NaN, NaN, 0.5, 0.1, 0.4, NaN, 0.1, 0.2];
X = Data;
idx = isnan(X);
new = strfind(idx(:).', [true, false]); % (:).' because STRFIND needs a row
Y = cumsum(X, 'omitnan');
X(new) = [0; -diff(Y(new))]; % [EDITED 2] ROUNDING PROBLEMS!!!
R = cumsum(X, 'omitnan');
R(idx) = NaN;
[EDITED 2] I've fixed the code, but it suffers from rounding problems. After a certain number of elements you get e.g. 0.0999999999997975 instead of 0.1 . So let's try a simply loop:
function R = cumsumResetNaN(R)
c = 0;
for k = 1:numel(R)
if isnan(R(k))
c = 0;
else
c = c + R(k);
R(k) = c;
end
end
end
This needs 0.00068 sec for input data with 64700 elements, which is faster than the vectorized method above (which has the severe rounding problem).
[EDITED 3] And for completeness a C-Mex function:
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
// Jan Simon, 2017, License: CC BY-SA 3.0
// Create cumulative sum of a double vector, which is reset to 0
// at every NaN.
double *x, *xf, c;
if (!mxIsDouble(prhs[0]) || mxIsSparse(prhs[0]) || mxIsComplex(prhs[0]) ||
(mxGetM(prhs[0]) != 1 && mxGetM(prhs[0]) != 1)) {
mexErrMsgIdAndTxt("JSimon:cumsumRestartNaN:BadInput",
"Input must be a real full double vector.");
}
plhs[0] = mxDuplicateArray(prhs[0]);
x = mxGetPr(plhs[0]);
xf = x + mxGetNumberOfElements(plhs[0]);
c = 0;
while (x < xf) {
if (*x == *x) { // NaN==NaN is FALSE by defintion
c += *x;
*x++ = c;
} else {
c = 0;
x++;
}
}
}
This needs 0.26 sec compared to 0.91 sec for the M-version [EDITED 2] (input data: [64700] elements, 1000 iterations, MSCV2012, Win7/64, Matlab 2016b)
6 个评论
更多回答(2 个)
Akira Agata
2017-12-2
How about keeping position of NaN, applying cumsum with 'omitnan' option and finally putting NaN for the position, like:
Val = [NaN NaN 0.1 0.1 0.1 NaN]';
idx = isnan(Val);
Val2 = cumsum(Val,'omitnan');
Val2(idx) = NaN;
3 个评论
Akira Agata
2017-12-3
If you have Image Processing Toolbox, you can simply do that by using bwlabel function. Here is my second try!
Val = [NaN NaN 0.1 0.1 0.1 NaN NaN 0.1 0.1 0.1 NaN]';
idx = isnan(Val);
group = bwlabel(~idx);
tmp = splitapply(@(x) {cumsum(x)},Val(~idx), group(~idx));
Val2 = nan(size(Val));
for kk = 1:max(group)
Val2(group == kk) = tmp{kk};
end
Pai-Feng Teng
2018-10-5
How to find the sum of every single number in a table? I searched every board I can find and all they had is the sum of rows or columns.
1 个评论
Jan
2018-10-6
Please do not attach a new question in the section for answers of another question. Such thread-hijacking produces confusions, because it is not clear, to which question an answer belongs. Create a new question in your own thread and delete this "answer". Thanks.
By the way: "The sum of each single number" is not clear and should be elaborated.
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