Consecutive count of values based on multiple conditions

I have a very big table consisting of approximately 3 million rows. The table consists of four columns [ID period amount pbehind], see the following screenshot including the output column I want.

For every ID, I want to count the consecutive amount of times of behind payment (pbehind). But there are a few conditions to this:

1. when pbehind <= 0.5, then output = 0
2. when pbehind > 1.5, but value in previous period is 0, then output = 0;
3. when pbehind > 0.5, but amount in next period (which is last period) is zero, then output=0.

There are a couple of things to note:

1. period does not have to start at 1 and increase by 1 (see ID = 2). However it is ordered by ID and period using sortrows(table,[1,2]).
2. An ID can make multiple payments for a certain period, however pbehind will always be equal to for that period (see ID=3).

I know for the first condition that I can just do the following:

output = pbehind;
output(output<=0.5)=0;

But I cannot figure out how to efficiently implement the other conditions. Does anyone know how to do this?

Answer 1

[Peter Perkins]

Put your data in a table, and use rowfun with ID as the grouping variable. Write a function that does all the logic you need to enforce. Your function will be applied to one group of rows at a time, and should return a column vector of output values. Here's a simple example:

>> t = table([1;1;1;2;2],rand(5,1),rand(5,1),'VariableNames',{'ID' 'X' 'Y'})
t =
  5×3 table
    ID       X           Y    
    __    ________    ________
    1      0.82202     0.26854
    1     0.041591     0.63908
    1      0.91635    0.031734
    2      0.17678     0.33395
    2      0.92236      0.8908
>> t2 = rowfun(@(x,y) x - mean(y),t,'GroupingVariable','ID')
t2 =
  5×3 table
    ID    GroupCount      Var3  
    __    __________    ________
    1         3          0.50891
    1         3         -0.27152
    1         3          0.60323
    2         2          -0.4356
    2         2          0.30998
>> t.Output = t2.Var3
t =
  5×4 table
    ID       X           Y         Output 
    __    ________    ________    ________
    1      0.82202     0.26854     0.50891
    1     0.041591     0.63908    -0.27152
    1      0.91635    0.031734     0.60323
    2      0.17678     0.33395     -0.4356
    2      0.92236      0.8908     0.30998

Obviously you will need to write your own function.

[mtango]

The problem is that I cannot figure out how to create code to count the consecutive amount of times of behind payment (pbehind) for each ID...

Answer 2

[Bastian Arturo Romero Mardones]

you should try with a series of if..else. For example:

for i = length(output)
if(pbehind(i) > 1.5 && pbehind(i-1) == 0)
    output(i) =0;
elseif(condition)
    ...
end