Replacing missing data with the previous data in a column

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Olu B
Olu B 2019-8-9
评论： Olu B 2019-8-9
Hi Guys,
1.How can I replace missing data points with previous data in a column?
For example I have a set of data called 'newcycl' that has 3 columns and 6 rows each but row 3 is missing in column (:,1), I want to replace row 3 in column (:,1) with the previous value in column (:,1) , However I will like to replace missing data points all through a large pool of data this same way. lets say a 70000 by 1 matrix
newcycl(:,1) = { 0, 4, ,5,6,7 } newcycl(:,2) = { 1, 2, 3 ,4,0, 6 } newcycl(:,3) = { 2, 4, 6 ,5,6,7 }
2. Also how can I replace numbers of a set range lets say between 0 and 1 with the next higher number in a column
3. If I have a column of 9 rows, how can i divide the column into 2 columns of 4 rows each and remove the tail or pad it with the previous number
e.g. A = { 1 2 4 5 6 7 2 4 6 }
split into two columns below and remove or add to the last value so it can be divided evenly:
B = {1 2 4 5 }, C = {6,7,2,4}
Thanks guys
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the cyclist
the cyclist 2019-8-9
I'll ask the same question again. In what data type are your data currently stored? If it is in a numeric type, it literally cannot have a missing value.
Messages Image(3526219204).png
Olu B
Olu B 2019-8-9
It is a numeric type and was stored in a csv format from a picolog file.
OK.. is there a way you can replace all the zeros in a column with the previous number non zero number in the same column?

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回答（1 个）

Neuropragmatist
Neuropragmatist 2019-8-9
You are making cell arrays in your question by using curly {} brackets. I'm going to assume this is a mistake because you call these things matrices and not cell arrays and because the syntax you use makes me think its a mistake.
1) It sounds like you might be looking for fillmissing:
https://uk.mathworks.com/help/matlab/ref/fillmissing.html
Using the 'previous' method and DIM set to columns.
2) This should work:
data; % is your data matrix
range_to_test = [0 1];
first_greater = min(data-range_to_test(2),[],2);
idx = data>range_to_test(1) & data<range_to_test(2);
new_data = idx.*first_greater;
data(idx) = new_data;
3) This should work:
n = floor(numel(A)/2);
B = A(1:n);
C = A(n+1:end);
min_length = min([numel(B) numel(C)]);
B = B(1:min_length)
C = C(1:min_length)
Hope this helps,
M.

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