Identify the peers within a firm

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Long Hong 2020-2-29

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https://ww2.mathworks.cn/matlabcentral/answers/508223-identify-the-peers-within-a-firm

评论： Image Analyst 2020-5-18

Dear all:

I am working on a panel dataset, which contains worker ID, firm ID, and year. Each worker works for one firm only in each year - so it is a worker-year panel. I hope to create a matrix that shows whether worker j is worker i's coworker. Ideally, I hope to create a matrix as follows:

where

and 1 indicates worker i's coworker's worker index in year t.

I am wondering if anyone knows how to have a quick algorithm instead looping over the worker's index (because there are thousands of workers in the data).

Thank you very much and I look forward to hearing from you!

Best,

Long

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KALYAN ACHARJYA 2020-2-29

Can you elaborate more (Maths)?

Long Hong 2020-2-29

Hi Kalyan, thanks for asking! An easy example is as follows (two firms and one period).

IDw IDf

1 1

2 1

3 1

4 2

5 2

Then

and

. Hope it helps.

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Answer 1

Image Analyst 2020-2-29

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在 MATLAB Online 中打开

You can simply use the firm vector and put it into pdist2(). Any distance that is zero means those two workers are in the same firm

firm2 = [firms(:), firms(:)];
distances = pdist2(firm2, firm2);

For example, if there is a zero at row 30 and column 142 of distances, it means that worker #30 and worker #142 work in the same firm.

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Image Analyst 2020-2-29

在 MATLAB Online 中打开

Does this help:

numEmployees = 10;
numFirms = 3;
firms = randi(numFirms, numEmployees, 1)
firm2 = [firms(:), firms(:)]
distances = pdist2(firm2, firm2) == 0
% Print out what employees work in what firm together
for k = 1 : numEmployees
	thisRow = distances(k, :);
	thisRow(1:k) = 0;
	otherEmployees = find(thisRow);
	if isempty(otherEmployees)
		continue;
	end
	for k2 = 1 : length(otherEmployees)
		fprintf('Employee %d works with employee %d in firm %d.\n',...
			k, otherEmployees(k2), firms(k));
	end
end

firms =

3

2

1

2

1

3

1

firm2 =

3 3

2 2

1 1

2 2

1 1

3 3

1 1

distances =

10×10 logical array

1 1 1 0 0 0 0 0 1 0

0 0 0 1 0 1 0 0 0 0

0 0 0 0 1 0 1 1 0 1

0 0 0 1 0 1 0 0 0 0

0 0 0 0 1 0 1 1 0 1

1 1 1 0 0 0 0 0 1 0

0 0 0 0 1 0 1 1 0 1

Employee 1 works with employee 2 in firm 3.

Employee 1 works with employee 3 in firm 3.

Employee 1 works with employee 9 in firm 3.

Employee 2 works with employee 3 in firm 3.

Employee 2 works with employee 9 in firm 3.

Employee 3 works with employee 9 in firm 3.

Employee 4 works with employee 6 in firm 2.

Employee 5 works with employee 7 in firm 1.

Employee 5 works with employee 8 in firm 1.

Employee 5 works with employee 10 in firm 1.

Employee 7 works with employee 8 in firm 1.

Employee 7 works with employee 10 in firm 1.

Employee 8 works with employee 10 in firm 1.

Long Hong 2020-2-29

在 MATLAB Online 中打开

Hi Image Analyst:

Thank you for the helpful insight. I hope I did not bother you too much. I am wondering if how you would apply this case into a panel data.

In the following example, I have created five different workers (idw), who are observed in both periods (year 1 and year 2) - therefore in total there are 10 observations.

idw = [1  1  2  2  3  3  4  4  5  5]';
idf = [1  1  1  2  1  2  1  1  2  2]';
idft = [11 12 11 22 11 22 11 12 21 22]'; %firm*year index
idft2 = [idft(:), idft(:)];
C = pdist2(idft2, idft2) == 0;

C =

10×10 logical array

1 0 1 0 1 0 1 0 0 0

0 1 0 0 0 0 0 1 0 0

1 0 1 0 1 0 1 0 0 0

0 0 0 1 0 1 0 0 0 1

1 0 1 0 1 0 1 0 0 0

0 0 0 1 0 1 0 0 0 1

1 0 1 0 1 0 1 0 0 0

0 1 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 0 1 0

0 0 0 1 0 1 0 0 0 1

However, some of the columns are actually placebo - they are by construction 0. For example, C(1,2) - I am not in the same firm in year 2 with myself in year 1.

I am wondering if there is a quick way to deal with the panel data setting. So, instead of generating a 10x10 matrix, I will have a 10x5 matrix. This will also free up a lot of space while running this program since I have thousands of workers.

Thank you very much and I look forward to hearing from you!

Best, Long

Long Hong 2020-5-17

Dear @Image Analyst,

I am writing to ask a very quick follow-up question on the use of pdist2,

Using my current (huge) data, pdist2 requires 0.7 million x 0.7 million matrix , and my computer cannot handle such a big matrix. Is there any way to solve this problem (say, adding a sparse matrix option since there are basically 0 and 1 in the matrix)?

Thank you, and I look forward to hearing from you!

Best,

Long

Image Analyst 2020-5-18

No idea. Give it a try.

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