How to establish overlap of participant IDs when using the tab command

[Scott Forrester]

Dear Statalist,

I'm trying to find out how much overlap exists in the different responses to a question.
For example, 91% do not perform an action when it is rational, while only 34% perform an action when it is rational to do so.

How can I find out if the same participants (by their unique IDs) belong to both 91% and 34%?

Thank you very much in advance.

[Nick Cox]

I would appreciate a data example. I can't guess at your data structure here or your variable names.

[Scott Forrester]

Hi Nick,

Thank you very much for your reply. Here is the data example:

tab statusquo rational, column

+-------------------+
| Key               |
|-------------------|
|     frequency     |
| column percentage |
+-------------------+

           |       rational
 statusquo |         0          1 |     Total
-----------+----------------------+----------
         0 |        33        466 |       499
           |     40.74      42.99 |     42.83
-----------+----------------------+----------
         1 |        48        618 |       666
           |     59.26      57.01 |     57.17
-----------+----------------------+----------
     Total |        81      1,084 |     1,165
           |    100.00     100.00 |    100.00

This table shows rationality (a binary variable) and status-quo (also a binary variable) (ie when participants confirmed their decision instead of rethinking their answer): Perhaps, as an illustration of how to read this table: When it rational to leave the decision as is (second column of rational), 57.01% behave rationally and remain in the status quo. When it is not rational to leave the decision as is (rational = 0), 40.74% behave rationally and do not leave the decision as is (ie status-quo =0). Now my question is, how many of the same participants, as identified by their unique ID, are both in 0,0 and 1,1, or how many are both in 0,1 and 1,0?

* Example generated by -dataex-. For more info, type help dataex
clear
input int CASE float(statusquo rational)
 888 1 1
 981 0 0
 553 0 1
 560 0 1
 471 1 1
 513 1 1
 563 1 1
1341 1 0
 573 1 0
 524 0 1
 906 0 1
 526 0 1
1291 0 1
 541 0 1
1086 1 1
1161 1 1
 667 0 1
 797 1 1
 594 0 1
 240 0 1
 915 1 1
 615 0 1
 741 0 1
1351 0 1
 666 1 1
 244 1 1
1295 1 1
 234 0 1
 482 0 1
1272 1 1
 689 0 1
 462 1 1
1165 1 0
 385 1 1
 446 0 1
 187 1 1
 395 1 1
 570 0 1
 760 1 1
 227 0 1
 168 1 1
 224 1 1
 631 1 1
1254 1 1
1173 1 1
 171 0 1
 970 0 1
 780 0 1
 649 0 1
 470 1 1
 586 1 1
 534 1 1
1223 0 1
 375 1 1
 242 0 1
1258 0 1
 776 1 1
 788 1 1
 740 1 1
 713 0 1
 974 0 1
 507 1 1
 907 1 1
 785 1 1
 232 1 1
 929 1 1
 228 1 1
 988 0 1
1217 1 1
 500 0 0
 635 0 1
 583 1 1
1328 1 1
 664 1 1
1163 1 1
 318 1 1
 710 1 1
1241 1 1
 723 0 1
 536 1 1
 196 0 1
1207 0 1
 499 1 1
 489 1 1
 612 0 1
1347 0 1
 670 1 1
 958 1 1
 475 0 0
 323 0 1
 724 1 1
1175 0 1
1255 0 1
 702 1 0
1232 0 1
 540 0 1
 578 0 1
1097 1 1
1235 1 1
 365 1 1
end

Thank you very much for your assistance!

Kind regards,
Scott

[Nick Cox]

both in 0,0 and 1,1

both in 0,1 and 1,0

I interpret as answered by

count if statusquo == rational 

count if statusquo != rational

which should yield 651 and 514 respectively, although you'll get more if there are any missing values. Adding the condition

& !missing(statusquo, rational)

will ignore missings.

So, those counts are equivalent to adding cells on each diagonal of the table.

[Scott Forrester]

Thank you, Nick! Are you saying that the 514 from statusquo != rational are reflected in the 651 from statusquo = rational?

If not, may I ask how to relate this to the potential overlap of participants, i.e., how can I identify participants that are part of more than one of the four (1,1; 0,0; 1,0; 0,1) groups?

I was hoping to get a statement like 20% (as an example) of the same individuals are in 0,1 as well as 1,0.

Thank you so much for your assistance!

[Nick Cox]

Sorry, but I don't understand your questions. If two values are not equal, they can't also be equal. Also, there is no overlap between the subsets (0,0), (0,1), (1,0), (1,1). So, there is no information about frequencies that isn't in the two-way table.

[Scott Forrester]

Dear Nick, Thank you for your reply. I'm sorry I did not clarify this: each participant faces four decisions, which I have laid out in the long format. Each decision is one row of data, so I have four rows per participant and, subsequently, four decisions to stay in the status-quo or not.
Hence, the question from my supervisor to analyze whether some participants are irrational (stay when irrational to do so or leave when actually it's rational to stay) in some decisions but rational in others (stay when rational or leave when irrational to stay).

I would be most grateful for your insights.
Thank you in advance!

[Nick Cox]

So, how is that repetition manifested in your data structure? The only other variable in #3 is CASE and in that data example a check with

isid CASE

shows that each occurs just once, and there is no repetition.

[Scott Forrester]

Hi Nick, Thank you for your reply. I don't know what happened with my dataex example, but when I closed and reloaded the data, this is what I got:

CASE is the unique participant identifier, round is not always 4 (as initially there were four decisions per individual) but some decisions were dropped as part of data cleaning.

* Example generated by -dataex-. For more info, type help dataex
clear
input int CASE byte round float(rational statusquo)
 79 1 1 0
 79 2 1 0
 79 3 1 0
 79 4 1 0
 80 1 1 1
 80 2 1 1
 97 1 0 0
 97 3 1 0
 97 4 0 0
104 1 1 0
104 2 0 1
104 3 1 0
104 4 1 0
109 1 1 0
109 4 1 0
128 1 1 0
128 2 1 0
128 3 1 0
128 4 1 0
129 1 1 0
129 2 0 0
129 3 0 0
129 4 0 0
140 1 1 0
140 2 1 0
140 3 1 0
140 4 1 0
147 2 1 0
147 3 0 0
149 1 1 0
149 3 1 1
149 4 1 1
151 2 1 0
151 4 1 0
155 1 1 0
155 2 1 0
155 3 0 0
155 4 1 0
160 1 0 0
160 2 1 0
160 3 1 0
160 4 1 0
162 1 1 0
162 2 1 0
162 3 1 0
165 1 0 0
165 2 1 0
165 4 1 0
171 1 1 0
171 2 0 0
171 3 1 0
171 4 0 0
172 1 1 0
172 2 1 0
172 3 1 0
172 4 0 0
174 1 1 0
174 2 1 0
174 3 0 0
176 1 1 0
176 2 0 0
176 3 1 0
176 4 1 0
177 1 1 0
177 3 1 0
177 4 1 0
179 1 1 0
179 2 0 0
179 3 0 0
180 1 1 0
180 2 1 0
180 3 1 1
180 4 1 0
181 1 1 0
181 2 1 0
181 3 1 0
181 4 1 0
187 1 1 0
187 2 1 0
187 4 1 0
189 1 1 0
189 2 0 0
189 3 1 0
189 4 1 0
191 1 0 0
191 2 1 0
191 3 1 0
191 4 1 0
193 2 0 0
193 3 0 0
193 4 1 0
194 1 1 0
194 2 1 0
194 3 1 0
194 4 1 0
198 1 1 0
198 2 1 0
198 3 1 0
198 4 1 0
201 1 1 0
end

I'm very grateful for your assistance!
Kind regards,
Scott

[Nick Cox]

Thanks for the detail. There are at least three possibilities.

1. Each case's history can be represented concisely as a string profile so case 79 is 10101010 (or 11110000 if you prefer). That raises the slightly alarming prospect of 2^8 = 256 distinct profiles, although they may not all occur, and they are unlikely to be equally frequent either.

Constructing such a profile is routine manipulation, but the details are spelled out at https://journals.sagepub.com/doi/pdf...36867X20909698

2. We can count how many combinations each case chose, as a measure of inconstency.

3. We can reduce the latter to staying the same or changing ever.

As you'll realise, there needs to be some care over what is done with cases with just 1, 2 or 3 rounds AND over whether you're counting observations or cases.

Some technique follows:

* Example generated by -dataex-. For more info, type help dataex
clear
input int CASE byte round float(rational statusquo)
 79 1 1 0
 79 2 1 0
 79 3 1 0
 79 4 1 0
 80 1 1 1
 80 2 1 1
 97 1 0 0
 97 3 1 0
 97 4 0 0
104 1 1 0
104 2 0 1
104 3 1 0
104 4 1 0
109 1 1 0
109 4 1 0
128 1 1 0
128 2 1 0
128 3 1 0
128 4 1 0
129 1 1 0
129 2 0 0
129 3 0 0
129 4 0 0
140 1 1 0
140 2 1 0
140 3 1 0
140 4 1 0
147 2 1 0
147 3 0 0
149 1 1 0
149 3 1 1
149 4 1 1
151 2 1 0
151 4 1 0
155 1 1 0
155 2 1 0
155 3 0 0
155 4 1 0
160 1 0 0
160 2 1 0
160 3 1 0
160 4 1 0
162 1 1 0
162 2 1 0
162 3 1 0
165 1 0 0
165 2 1 0
165 4 1 0
171 1 1 0
171 2 0 0
171 3 1 0
171 4 0 0
172 1 1 0
172 2 1 0
172 3 1 0
172 4 0 0
174 1 1 0
174 2 1 0
174 3 0 0
176 1 1 0
176 2 0 0
176 3 1 0
176 4 1 0
177 1 1 0
177 3 1 0
177 4 1 0
179 1 1 0
179 2 0 0
179 3 0 0
180 1 1 0
180 2 1 0
180 3 1 1
180 4 1 0
181 1 1 0
181 2 1 0
181 3 1 0
181 4 1 0
187 1 1 0
187 2 1 0
187 4 1 0
189 1 1 0
189 2 0 0
189 3 1 0
189 4 1 0
191 1 0 0
191 2 1 0
191 3 1 0
191 4 1 0
193 2 0 0
193 3 0 0
193 4 1 0
194 1 1 0
194 2 1 0
194 3 1 0
194 4 1 0
198 1 1 0
198 2 1 0
198 3 1 0
198 4 1 0
end

gen profile = strofreal(rational) + strofreal(statusquo) if round == 1 
bysort CASE (round) : replace profile = profile[_n-1] + strofreal(rational) + strofreal(statusquo) if round > 1
by CASE : replace profile = profile[_N]

bysort CASE : gen nrounds = _N 
bysort profile : egen pcount = total(nrounds == 4 & round == 4)

graph hbar pcount if nrounds == 4 & round == 4, over(profile, sort(1) descending) ysc(alt) ytitle(Instances of profile)

egen group = group(rational statusquo) 
egen tag = tag(CASE group)
bysort CASE: egen nchoices = total(tag)
gen changed = nchoices > 1 

tabdisp CASE if nrounds == 4 & round == 4, c(profile nchoices changed) 

----------------------------------------------
     CASE |    profile    nchoices     changed
----------+-----------------------------------
       79 |   10101010           1           0
      104 |   10011010           2           1
      128 |   10101010           1           0
      129 |   10000000           2           1
      140 |   10101010           1           0
      155 |   10100010           2           1
      160 |   00101010           2           1
      171 |   10001000           2           1
      172 |   10101000           2           1
      176 |   10001010           2           1
      180 |   10101110           2           1
      181 |   10101010           1           0
      189 |   10001010           2           1
      191 |   00101010           2           1
      194 |   10101010           1           0
      198 |   10101010           1           0
----------------------------------------------

[Scott Forrester]

Thank you very much! I appreciate the detailed code!