# How to break data in to groups using while loop?

조회 수: 3 (최근 30일)
Julia 2023년 1월 31일
댓글: dpb 2023년 2월 3일
matrix = [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 NaN NaN]
matrix = 7×6
1 50 60 70 50 40 2 NaN 10 20 10 10 3 NaN 20 NaN NaN NaN 1 NaN 60 30 40 50 2 10 20 10 20 NaN 1 30 20 40 NaN 50 2 NaN 50 50 NaN NaN
The first column indicates to which group the rows belong to. "1, 2, 3" (row 1, 2, 3) are Group 1, "1, 2" (row 4, 5 ) are Group 2, the next "1, 2" ( row 6, 7) are Group 3.
I am trying to find the number of NaN in each group. The desired result would be:
ans =
5
2
4
Is it possible to do this using a while loop?
eg. In Column 1, while "the current element" is larger than "the element in the previous row", statement
I am sorry if this description is confusing.

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### 채택된 답변

dpb 2023년 1월 31일
편집: dpb 2023년 1월 31일
Answer the Q? actually asked -- I'll post this as a separate answer for convenience but not remove the first.
OK, as @Stephen23 pointed out earlier, I didn't read the Q? carefully enough and just assumed the first column values were the group IDs. As the comment below says, I'd first build the grouping variable from the definition of the group; it'll bound to be useful later, anyway. (And, it illustrates another technique worth knowing...)
M= [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 NaN NaN];
g=nan(size(M,1),1); % first build a grouping variable of right size; missing values
g(M(:,1)==1)=unique(M(:,1)); % populate the first of each group location with its ID/group
g=fillmissing(g,'previous'); % and fill in the rest
nNaN=sum(groupsummary(M,g,@(x)sum(isnan(x))),2)
nNaN = 3×1
5 2 4
NOTA BENE: the whole M array can be passed here; the first column is immaterial to the count...
@the cyclist's use of splitapply will also work here as well
nNaN=splitapply(@(x)sum(isnan(x),'all'),M(:,2:end),g)
nNaN = 3×1
5 2 4
Oh. NOTA BENE SECOND:
The above needs modification to build the grouping variable in general -- it's probably just coincidence there are three groups and a maximum of three rows in any one group. So, unique isn't the generic answer for the RHS of the assignment of the initial group indices to the grouping variable -- it would be more like
g=nan(size(M,1),1); % first build a grouping variable of right size; missing values
ix=find(M(:,1)==1); % the locations of each group start
g(ix)=1:numel(ix); % populate the first of each group location with its ID/group
This will count the number of times the first index value occurs and where and generate that many groups irrespective of the number of records/group.
##### 댓글 수: 8이전 댓글 6개 표시이전 댓글 6개 숨기기
Julia 2023년 2월 3일
If now, I want to find the number of elements that are NaN in one column but are non-NaN in its previous column, how should I modify the code?
I tried MM=isfinite(circshift(M,1,2))&isnan(M)
however, this code compares Column 2 (the 1st meaningful column) against Column 1 (the group indicator column), which does not give me my desired answer.
M= [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 NaN NaN];
MM=isfinite(circshift(M,1,2))&isnan(M)
MM = 7×6 logical array
0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0
Instead of the result given by this line of code (above), would you please teach me how to obtain this result I want (below)?
0 0 0 0 0 0
0 0 0 0 0 0
0 0 1 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 1
0 0 1 0 0 0
dpb 2023년 2월 3일
"...elements that are NaN in one column but are non-NaN in its previous column"
That's precisely the result of MM above; that the first column is in there is immaterial; it simply is a placeholder to match the size of M.
That's why the solution above uses either MM(:,2:end) or cleans out MM(:,1) when done; it is known that the first column isn't of interest. But, that doesn't affect the remaining columns.
Oh. Brain freeze -- it's the first two columns of the result that are invalid, not just the first. So, use
M= [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 NaN NaN];
MM=isfinite(circshift(M,1,2))&isnan(M);
MM(:,1:2)=false
MM = 7×6 logical array
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 1 0 0 0 0 1 0 0 0 0 0 1 0

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### 추가 답변 (6개)

the cyclist 2023년 1월 31일
Here is one way:
matrix = [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 NaN NaN];
g = findgroups(matrix(:,1));
out = splitapply(@(x)sum(isnan(x(:))),matrix(:,2:end),g)
out = 3×1
2 5 4
##### 댓글 수: 2없음 표시없음 숨기기
dpb 2023년 1월 31일
@the cyclist, this one suffers the same fate of my first solution below w/ groupsummary. findgroups considers each numeric value a group; one must build a new grouping variable here based on the occurrence of the first element in the subsequent sequence...
the cyclist 2023년 1월 31일
Ah ... I misread OP's question (apparently in the exact same way you did). Thanks for pointing that out.

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dpb 2023년 1월 31일
편집: dpb 2023년 1월 31일
No explicit looping construct needed; let MATLAB do it for you...I shortened your variable name to M...
M= [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 NaN NaN];
nNaN=sum(groupsummary(M(:,2:end),M(:,1),@(x)sum(isnan(x))),2)
nNaN = 3×1
2 5 4
See <groupsummary> for the details on using it; in short, the first argument is the array to compute the summary over (in your case everything except the first column) while the second is the grouping variable (your first column). Then the function to apply can be a builtin such as 'mean' or as shown above, whatever you want it to be.
groupsummary applies the function to each column of the array by grouping variable so above we first add up how many True values are returned for each column by isnan, that returns an array of counts by column; The total for each group then is simply the sum of those by row (the second, optional argument, 2) applied to the resulting array.
OK, as @Stephen23 points out below, I didn't read the Q? carefully enough and just assumed the first column values were the group IDs. As the comment below says, I'd first build the grouping variable from the definition of the group; it'll bound to be useful later, anyway. (And, it illustrates another technique worth knowing...)
g=nan(size(M,1),1); % first build a grouping variable of right size; missing values
g(M(:,1)==1)=unique(M(:,1)); % populate the first of each group location with its ID/group
g=fillmissing(g,'previous'); % and fill in the rest
nNaN=sum(groupsummary(M(:,2:end),g,@(x)sum(isnan(x))),2)
nNaN = 3×1
5 2 4
Same solution still works, just use the new grouping variable in place of the column values...
The above can/will work generically for other numbering sequences but one would need to save the result of unique and explicitly code for whatever was the indicator value for starting the new sequence other than 1.
*PS. I wondered why so many people were getting the wrong answer first time... :)
##### 댓글 수: 2없음 표시없음 숨기기
Stephen23 2023년 1월 31일
@dpb: I really like this solution, but note that the groups given in the matrix are not duplcaites of the same number, but are given in increasing runs of integers. A bit trickier.
dpb 2023년 1월 31일
Oh. In skimming over the Q? text, I missed that. I'd probably make a new grouping variable, then, first...it'll probably be wanted/needed for subsequent analyses on the dataset anyway...

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Benjamin Thompson 2023년 1월 31일
편집: Jan 2023년 2월 1일
Even better, use the vectorization features in Matlab to count NaNs in each row all at once.
nanCounts = sum(isnan(matrix),1); % This will count NaNs in each row.
groupCounts = zeros(size(unique(matrix),1),); % Creates a zero vector whose length is equal to number of groups
for i = 1:length(groupCounts)
I = nanCounts(matrix(:,1) == i; % Creates an index vector for group i counts
groupCounts(i) = sum(I);
end
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Voss 2023년 1월 31일
matrix = [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 NaN NaN];
group_start_idx = find(matrix(:,1) == 1);
group_end_idx = find(diff([matrix(:,1); 1]) <= 0);
n_groups = numel(group_start_idx);
n_nans = zeros(n_groups,1);
for ii = 1:n_groups
n_nans(ii) = nnz(isnan(matrix(group_start_idx(ii):group_end_idx(ii),:)));
end
disp(n_nans);
5 2 4
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Voss 2023년 1월 31일
"Is it possible to do this using a while loop?"
matrix = [1 50 60 70 50 40
2 NaN 10 20 10 10
3 NaN 20 NaN NaN NaN
1 NaN 60 30 40 50
2 10 20 10 20 NaN
1 30 20 40 NaN 50
2 NaN 50 50 NaN NaN];
n_nans = [];
row = 1;
n_rows = size(matrix,1);
while row <= n_rows
group_start_row = row;
while row < n_rows && matrix(row+1,1) > matrix(row,1)
row = row+1;
end
group_end_row = row;
n_nans(end+1,1) = nnz(isnan(matrix(group_start_row:group_end_row,:)));
row = row+1;
end
disp(n_nans);
5 2 4
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Jan 2023년 2월 1일
matrix = [1 50 60 70 50 40; ...
2 NaN 10 20 10 10; ...
3 NaN 20 NaN NaN NaN; ...
1 NaN 60 30 40 50; ...
2 10 20 10 20 NaN; ...
1 30 20 40 NaN 50; ...
2 NaN 50 50 NaN NaN];
group = cumsum([1; diff(matrix(:, 1)) < 0]); % [1 1 1 2 2 3 3].'
data = sum(isnan(matrix), 2);
result = accumarray(group, data)
result = 3×1
5 2 4
##### 댓글 수: 1이전 댓글 -1개 표시이전 댓글 -1개 숨기기
dpb 2023년 2월 1일
@Jan, cumsum is very clever for grouping variable computation. +1
I first thought of using the <0 for the breakpoint, but didn't have the flash on cumsum so got bogged down...

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