Quick way to sum array elements based on flag in another array?

Say I have 2 arrays: one with "clean" data, and one that tells me which data gets lost, as indicated by zeros. If an element is lost, it should be added to the next observed element. Anybody got a quick one-or-two-liner to come up with the summed array?
Example:
clean = [1 2 3 4 5 6]
lost = [1 0 3 0 0 6]
desired = [1 5 15]

 Risposta accettata

Stephen23
Stephen23 il 14 Lug 2015
Modificato: Stephen23 il 14 Lug 2015
Try this:
tmp = (1:numel(lost))-[0,cumsum(lost(1:end-1)==0)];
desired = accumarray(tmp.',clean).'
Tested:
>> clean = [1 2 3 4 5 6];
>> lost = [1 0 3 0 0 6];
>> tmp = (1:numel(lost))-[0,cumsum(lost(1:end-1)==0)];
>> desired = accumarray(tmp.',clean).'
desired =
1 5 15
>> clean = [40 10 50 30 40 10 50 30 40 10 50 30];
>> lost = [40 0 50 30 0 10 0 0 40 10 50 30];
>> tmp = (1:numel(lost))-[0,cumsum(lost(1:end-1)==0)];
>> desired = accumarray(tmp.',clean).'
desired =
40 60 30 50 120 10 50 30

Più risposte (2)

desired = flip(accumarray(cumsum(lost(end:-1:1)~=0)',clean(end:-1:1)),1)';
or
desired = accumarray(cumsum(xor([0,diff(lost > 0)],lost))',clean)';

2 Commenti

Thanks for the help. Try:
clean = [40 10 50 30 40 10 50 30 40 10 50 30];
lost = [40 0 50 30 0 10 0 0 40 10 50 30];
It should give:
desired = [40 60 30 50 120 10 50 30];
Your solution does not.

Accedi per commentare.

accumarray(interp1(find(lost),1:nnz(lost),1:numel(lost),'next')',clean)

4 Commenti

Thanks for the help. Try:
clean = [40 10 50 30 40 10 50 30 40 10 50 30];
lost = [40 0 50 30 0 10 0 0 40 10 50 30];
It should give:
desired = [40 60 30 50 120 10 50 30];
Your solution does not.
Yes it does! Well it's transposed but the numbers are correct
clean = [40 10 50 30 40 10 50 30 40 10 50 30];
lost = [40 0 50 30 0 10 0 0 40 10 50 30];
accumarray(interp1(find(lost),1:nnz(lost),1:numel(lost),'next')',clean)
ans =
40
60
30
50
120
10
50
30
You could always transpose the answer with '
Sorry Sean, I must have fat-fingered the test!

Accedi per commentare.

Richiesto:

Ted
il 14 Lug 2015

Commentato:

Ted
il 15 Lug 2015

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