This is just a very simple example to show my problem.
a=ones(5)
How can i insert NaN after every two rows like:
I know the way to do this simple example is:
b(:,1:5)=NaN
[a(1:2,:);b;a(3:4,:);b;a(end,:)]
But the problem is if the martrix is 60000-by-200 (may be more larger), so how can i insert 'NaN' after every two rows.
Thanks so much.
a = ones(5); %// example data
n = 2; %// number of rows
N = floor(size(a,1)*(1+1/n)); %// final number of rows
ind = mod(1:N, n+1) ~= 0; %// logical index for non-NaN rows
b = NaN(N,size(a,2)); %// initiallize result to NaN
b(ind,:) = a; %// fill in non-NaN rows
I can't think of an easy, one-line type solution. It can be done in a pretty tight loop though.
a = ones(5);
a_with_nans = nan(floor(size(a,1)*(3/2)), size(a,2)); %Start with all nans in a larger matrix
for ix = 1:2:size(a,1)
a_with_nans(ix*3/2-(1/2),:) = a(ix,:);
if ix+1<=size(a,1)
a_with_nans(ix*3/2-(1/2)+1,:) = a(ix+1,:);
end
end
Then:
a_with_nans =
1 1 1 1 1
1 1 1 1 1
NaN NaN NaN NaN NaN
1 1 1 1 1
1 1 1 1 1
NaN NaN NaN NaN NaN
1 1 1 1 1
You can do it like this:
>> a= [ 1 2 3 4 5 6 7 8 9]
a =
1 2 3 4 5 6 7 8 9
>> b = nan(floor(length(a)/2),1)'
b =
NaN NaN NaN NaN
>> a_new = zeros(1, length(a)+length(b))
a_new =
0 0 0 0 0 0 0 0 0 0 0 0 0
>> b_i = 3:2:length(a)
b_i =
3 5 7 9
>> a_new(b_i+(0:length(b_i)-1)) = b
a_new =
0 0 NaN 0 0 NaN 0 0 NaN 0 0 NaN 0
>> a_new(~isnan(a_new))=a
a_new =
1 2 NaN 3 4 NaN 5 6 NaN 7 8 NaN 9
Related
I have a matrix like this:
A(a,b)=
NaN 1 1
NaN 2 NaN
NaN 3 1
NaN 4 2
NaN 5 3
NaN 6 4
NaN 7 5
NaN 8 6
NaN 9 NaN
NaN 10 NaN
NaN NaN NaN
NaN NaN 1
NaN NaN 2
NaN NaN 3
NaN NaN NaN
and I would like to obtain a matrix (logical or numerical) where a 1 is assigned to all the values there are not NaN and whenever a certain value, lets say 5, is reached. But (here it comes the difficult part) the 1 has to be assigned also to the numbers before of the 5 (from 1 to 5), again if and only if the 5 is reached.
Therefore, my expected result should be a matrix like the following:
B(a,b) =
0 1 0
0 1 0
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 0
0 1 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
Some ideas please?
Additional info following the comments:
I am considering the order column wise. Every run of numbers in each column will start always from 1 and continue in the form 1,2,3,...
A NaN will stop this sequence and when it will restart will restart from 1.
If a sequence between two NaN contains the specified value, then all entries have to be true (value=1) between the prior and posterior NaN (see 3rd column as example)
Let your data matrix be defined as
A = [NaN 1 1
NaN 2 NaN
NaN 3 1
NaN 4 2
NaN 5 3
NaN 6 4
NaN 7 5
NaN 8 6
NaN 9 NaN
NaN 10 NaN
NaN NaN NaN
NaN NaN 1
NaN NaN 2
NaN NaN 3
NaN NaN NaN];
Here's one approach using regular expressions to detect runs of at least 5 numeric values:
B = mat2cell(char(~isnan(A).'+'0'),ones(size(A,2),1)); %'// make a string from each column
C = regexprep(B, '(1{5}1*)', '${repmat(''+'',1,numel($0))}'); %// change that 5 as needed
result = vertcat(C{:}).'=='+';
This gives
result =
0 1 0
0 1 0
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 0
0 1 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
The following code is not very efficient but does the job. If you need an efficient code, something else might be more suitable. However, this does the trick:
function stackoverflow
A =[
NaN 1 1
NaN 2 NaN
NaN 3 1
NaN 4 2
NaN 5 3
NaN 6 4
NaN 7 5
NaN 8 6
NaN 9 NaN
NaN 10 NaN
NaN NaN NaN
NaN NaN 1
NaN NaN 2
NaN NaN 3
NaN NaN NaN];
[n,m] = size(A);
k= 5;
B =(A>=k);
for j=1:m % for all rows
for i=1:n % for all columns
if B(i,j)==1 % if the value is >=5, all 4 prior values also have to be 1
B(i-(k-1):i,j) =1;
end
end
end
disp(A)
disp(B)
But I'm pretty sure, that there is a way to get rid of the loop(s), which I can't figure out right now.
for i = 1 : numel(T);
j = 1 : numel(T(i).n);
P(i,j) = (T(i).n);
G(i) = (T(i).lastPulse)-1100;
Y = P(1,G(1):length(T(1).n));
S = P(2,G(2):length(T(2).n));
end
I have the preceeding code. P is a (191x10000) matrix. I want to take out a specific portion of each row as I showed in S and Y and then concatenate S and Y and other row matrices corresponding to other rows of P to create matrix A(191x[max length of (S,Y,...)]). BUT the tricky part is that I cannot make S and Y aligned.
EXAMPLE:
P = [1 2 1 3 1 1 1 0 3 1 0]
[3 0 2 0 1 1 4 1 1 2 0];
S = P(1,1:7) = [1 2 1 3 1 1 1];
Y = P(2,5:10) = [1 1 4 1 1 2];
% A = concatenated S and Y aligned to original P.
A = [ 1 2 1 3 1 1 1 nan nan nan nan]
[nan nan nan nan 1 1 4 1 1 2 nan];
Preferably I would like to use a loop instead of separated matrices such as S and Y since I have many rows.
Suggested Answer:
I have the idea that probably I have to use indices corresponding to P and use them to concatenate Y and S, I just don't know how to execute this thought especially in a loop.
If I got the question correctly in my head, it seems bsxfun could be used here for creating a mask and then keep the masked elements from P and thus have an aligned output. Here's an implementation to go along those lines -
%// Random input array
P = randi(9,5,11)
%// Define the start and stop(end) indices as vectors
start_idx = [1 5 3 4 11]
stop_idx = [7 10 3 6 11]
%// Get the size of P and initialize output array
[M,N] = size(P);
P_out = NaN(M,N);
%// Create the mask for extracting specific elements from P
mask = bsxfun(#le,start_idx(:),1:N) & bsxfun(#ge,stop_idx(:),1:N);
%// Put masked elements from P into output array
P_out(mask) = P(mask)
Another way to get the output without initializing it, would be like this -
P_out = P.*mask;
P_out(~mask) = NaN;
So, to correlate with the variables used in the question, start_idx would be G and stop_idx would be [length(T(1).n),length(T(2).n).length(T(3).n),...].
Sample run -
P =
1 6 8 8 8 1 9 1 2 4 2
8 8 6 3 7 6 7 2 5 1 2
6 8 9 5 6 6 6 8 6 5 2
9 9 5 9 3 7 9 5 1 2 1
7 1 5 6 6 9 6 8 6 2 6
start_idx =
1 5 3 4 11
stop_idx =
7 10 3 6 11
P_out =
1 6 8 8 8 1 9 NaN NaN NaN NaN
NaN NaN NaN NaN 7 6 7 2 5 1 NaN
NaN NaN 9 NaN NaN NaN NaN NaN NaN NaN NaN
NaN NaN NaN 9 3 7 NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN 6
I have a 3D matrix with NaN padded to obtain equal number of rows in each 2D matrix i.e each (:,:,ind). Now i need to find the number of actual non-NaN rows in each (:,:,ind).
A simple example of what I need:
% Input:
A(:,:,1) = [ 1 1;
2 2;
NaN NaN];
A(:,:,2) = [ 2 2;
NaN NaN;
NaN NaN];
% Function call:
B = callingfunction(A);
% Output:
B = [2 1] % Number of Non-NaN rows in each 2D Matrix
Approach #1
B = squeeze(sum(all(~isnan(A),2),1))
Here's the build-up process to get a hang of it -
Start>>> Given A:
>> A
A(:,:,1) =
1 1
2 2
NaN NaN
A(:,:,2) =
2 2
NaN NaN
NaN NaN
1) Detect all non-NaN positions:
>> ~isnan(A)
ans(:,:,1) =
1 1
1 1
0 0
ans(:,:,2) =
1 1
0 0
0 0
2) Find rows with all non-Nan elements:
>> all(~isnan(A),2)
ans(:,:,1) =
1
1
0
ans(:,:,2) =
1
0
0
3) Sum up the number of all such rows:
>> sum(all(~isnan(A),2),1)
ans(:,:,1) =
2
ans(:,:,2) =
1
4) Get the result as a 1D array:
>> squeeze(sum(all(~isnan(A),2),1))
ans =
2
1
Approach #2
B = squeeze(sum(~any(isnan(A),2),1))
Use the same break-up-my-code-into-pieces process as listed earlier here and in all your future MATLAB codes and all past MATLAB codes that didn't make sense to do so now!
I have an incomplete dataset,
N = [NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN]'
I wish to identify a cluster of Nans, that is, if the subsequent number of them exceeds 2. how do i do that?
You could do something like this:
aux = diff([0; isnan(N); 0]);
clusters = [find(aux == 1) find(aux == -1) - 1];
Then clusters will be a Nx2 matrix, where N is the number of NaN clusters (all of them), and each row gives you the start and end index of the cluster.
In this example, that would be:
clusters =
1 1
5 5
8 9
15 19
It means you have 4 NaN clusters, and cluster one ranges from index 1 to index 1, cluster two ranges from 5 to 5, cluster three ranges from 8 to 9 and cluster four ranges from 15 to 19.
If you want only the clusters with at least K NaNs, you could do it like this (for example, with K = 2):
K = 2;
clusters(clusters(:,2) - clusters(:,1) + 1 >= K, :)
That would give you this:
ans =
8 9
15 19
That is, clusters 8-9 and 15-19 have 2 or more NaNs.
Explanation:
Finding the clusters
isnan(N) gives you a logical vector containing the NaNs as ones:
N --------> NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN
isnan(N) -> 1 0 0 0 1 0 0 1 1 0 0 0 0 0 1 1 1 1 1
We want to know where each sequence of ones start, so we use diff, which calculates each value minus the previous one, and gives us this:
aux = diff(isnan(N));
N ----> NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN
aux --> -1 0 0 1 -1 0 1 0 -1 0 0 0 0 1 0 0 0 0
Where a 1 indicates the group start and a -1 indicates a group end. But it misses the first group start and the last group end, because the first 1 element is absent (it doesn't have a previous on N because it is the first) and the last -1 is absent too (because there is nothing after the last 1 on N). A common fix is to add a zero before and after the array, which gives us this:
aux = diff([0; isnan(N); 0]);
N ----> NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN
aux --> 1 -1 0 0 1 -1 0 1 0 -1 0 0 0 0 1 0 0 0 0 -1
Notice two things:
If the diff at index i is 1, N(i) is the start of the NaN block.
If the diff at index i is -1, N(i - 1) is the end of the NaN block.
To get the start and end, we use find to get the indexes where aux == 1 and aux == -1. Hence, we call find twice, and concatenate both calls using [ and ]:
aux = diff([0; isnan(N); 0]);
clusters = [find(aux == 1) find(aux == -1) - 1];
Filtering the clusters whick have K or more elements
The last step is to find clusters which have K or more elements. To do that, we first take the cluster matrix and subtract the first column from the first, and add 1, like this:
clusters(:,2) - clusters(:,1) + 1
ans =
1
1
2
5
It means clusters 1 and 2 have 1 NaN, cluster 3 have 3 NaNs and cluster 4 have 5 NaNs. If we ask which values are greather than or equal K, we get this:
clusters(:,2) - clusters(:,1) + 1 >= K
ans =
0
0
1
1
It's a logical array. We can use that to index only the 1 (true) rows of the cluster matrix, like this:
clusters(clusters(:,2) - clusters(:,1) + 1 >= K, :)
ans =
8 9
15 19
It's like asking: give us only the clusters where the rows match the ones on this logical vector, and give us all columns (denoted by the :).
Here is a modular solution:
% the number of NaN you consider as a cluster
num = 3;
% moving average filter
Z = filter(ones(num,1),1,isnan(N));
x = arrayfun(#(x) find(Z == num) - num + x, 1:num,'uni',0)
y = unique(cell2mat(x))
(UPDATE: faster version below)
gives for num = 1:
y = 1 5 8 9 15 16 17 18 19
for num = 2:
y = 8 9 15 16 17 18 19
for num = 3, num = 4 and num = 5:
y = 15 16 17 18 19
and finally for num = 6 ... and more
y = Empty matrix: 1-by-0
Explanation
isnan(N) returns a logical array with ones at the positions of NaN.
Z = filter(ones(num,1),1,isnan(N)); is a implementation for a moving average filter with a filter window of ones(num,1) = [1 1 1] (for num = 3). So the filter of size 3 glides of the array and just reaches the value num = 3 when there are 3 NaN in a row.
So it basicall looks like:
%// N isnan(N) Z
NaN 1 1
1 0 1
2 0 1
3 0 0
NaN 1 1
5 0 1
6 0 1
NaN 1 1
NaN 1 2
7 0 2
8 0 1
10 0 0
12 0 0
20 0 0
NaN 1 1
NaN 1 2
NaN 1 3
NaN 1 3
NaN 1 3
Now it is easy to find all elements which are 3: find(Z == num) - but you also need all 2 right before: find(Z == num) - num + 2 and all 1 right before: find(Z == num) - num + 1. Instead of a loop arrayfun is used, which is basically the same. As result you get a matrix with a lot of indices, lot of them mulitple, but you just need the unique ones. I hope everything is clear now.
Actually it would be much faster to get find out of arrayfun, which can then even be substituted by bsxfun and you can get rid of cell2mat, which leads to the following form:
Faster:
Z = find( filter(ones(num,1),1,isnan(N)) == num ) - num;
y = unique( bsxfun(#plus, Z,1:num) );
or faster the obligatory fancy one-liner:
y = unique(bsxfun(#plus,find(filter(ones(num,1),1,isnan(N))==num)-num,1:num));
STRFIND Approach
I. Fancy One Liner:
%%// Given input N
N = [NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN]
out = [strfind(num2str(isnan([ 0 N 0]),'%1d'),'011');strfind(num2str(isnan([ 0 N 0]),'%1d'),'110')]'
Output
out =
8 9
15 19
II. Detailed one with explanation:
Basically you are trying to do sliding window checks, for which there is no direct method when working with double arrays, but after converting to strings, one can use strfind. This trick is used here.
I would suggest following the comments used in the code and the output numbers to understand it. Please note that for this particular case a cluster means a group of two or more consecutive NaNs
Code
%%// Given input N
N = [NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN]
%%// Set the locations where NaNs are present and then
%%// append at the start and end with zeros
N2 = isnan([ 0 N 0])
%%// Find the start indices of all NaN clusters
start_ind = strfind(num2str(N2,'%1d'),'011')
%%// Find the stop indices of all NaN clusters
stop_ind = [strfind(num2str(N2,'%1d'),'110')]
%%// Put start and stop indices into a Mx2 matrix
out = [start_ind' stop_ind']
Output
N =
NaN 1 2 3 NaN 5 6 NaN NaN 7 8 10 12 20 NaN NaN NaN NaN NaN
N2 =
0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 1 1 1 1 1 0
start_ind =
8 15
stop_ind =
9 19
out =
8 9
15 19
This uses diff, as Rafael Monteiro's answer, but seems to be simpler:
ind = diff([0; isnan(N(:))]);
result = find(ind(1:end-1)==1 & ind(2:end)==0);
In your example, this gives [8 15].
How it works: ind takes the values:
1 where a run of (one or more) NaN values starts;
0 where there is no change between NaN and numeric with respect to the previous value;
-1 where a run of (one o more) numeric values starts.
The second line selects the positions at which a run of NaN starts and such that the next position is also NaN. Thus it gives the start of each run of more than one NaN, as desired.
I have a tensor T
T=ones(2,2,2)
T(:,:,1) =
1 1
1 1
T(:,:,2) =
1 1
1 1
Now I want to add an element by doing
T(3,3,3)=100
and I get the following result
T(:,:,1) =
1 1 0
1 1 0
0 0 0
T(:,:,2) =
1 1 0
1 1 0
0 0 0
T(:,:,3) =
0 0 0
0 0 0
0 0 100
As you can see matlab automatically inserts 0 for the new row and column elements.
I know that I can convert the zeros using T(T==0)=NaN. But I'm looking for a way where NaN is inserted immediately so I won't have to do the additional conversion.
Is there a way that matlab automatically inserts NaN instead of 0 for those elements?
Desired result:
T(:,:,1) =
1 1 NaN
1 1 NaN
NaN NaN NaN
T(:,:,2) =
1 1 NaN
1 1 NaN
NaN NaN NaN
T(:,:,3) =
NaN NaN NaN
NaN NaN NaN
NaN NaN 100
Appreciate your help.
Code
T=ones(2,2,2)
T(3,3,3)=100
T(T==0)=NaN
%%// T(~T)=NaN would work too, but not a good practice as T is not logical
Or
T=ones(2,2,2)
T1 = NaN(3,3,3)
T(1:2,1:2,1:2) = T;
T1(3,3,3)=100
Or
T1 = NaN(3,3,3)
T1(1:2,1:2,1:2)=1;
T1(3,3,3)=100