Interactions in control variables (Z)

[MartinHaus1]

Hi there,

first of all, thank you very much for this amazing package!!

I was wondering whether there is a way to use interactions as control variables Z, i.e., something like A*B.

example:
interflex(estimator="kernel", Y="outcome", D="treatment", X="moderator", Z=c("A*B", "other_control"), FE=c("year", "state"),data=dataset)
This currently leads to
A*B is not in the data

Is there any way to do this at the moment?

Thank you!

Best,
Martin

[xuyiqing]

What about create a new variable?

…

On Thu, Jun 10, 2021, 03:43 MartinHaus1 ***@***.***> wrote: Hi there, first of all, thank you very much for this amazing package!! I was wondering whether there is a way to use interactions as control variables Z, i.e., something like A*B. example: interflex(estimator="kernel", Y="outcome", D="treatment", X="moderator", Z=c("A*B", "other_control"), FE=c("year", "state"),data=dataset) This currently leads to A*B is not in the data Is there any way to do this at the moment? Thank you! Best, Martin — You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub <#9>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AB2PKGGZ4MSQXNJNEH257L3TSCJMXANCNFSM46OFTO4Q> .

[MartinHaus1]

Thank you for your response!

Yes, this is a feasible work-around. I was just wondering whether I just missed how to feed interaction terms into interflex, in case there is a direct way.

Thanks!

[MartinHaus1]

Just a quick follow-up:

I tried to create a new variable (as a factor) and included it in the Z-vector to control for it.

It throws the following error:

Cross-validating bandwidth ... 
Parallel computing with 8 cores...
Optimal bw=0.0609.
Error in names(result) <- c("x0", "(Intercept)", use.variable[2:length(use.variable)]) : 
  'names' attribute [119] must be the same length as the vector [118]

If I remove the new categorical variable from the control vector, it works. Using the same variable for creating a model with plm does not lead to any issues.

Am I doing something wrong? For context, I use the following command options:

interflex(estimator="kernel", Y="y", D="d", X="x", Z=c("new_control_factor","other_controls"), 
          IV="iv", FE=c("year", "state"),data=dataset, cores=8, na.rm = T, vcov.type="cluster", cl="state")

Thank you!

[xuyiqing]

Yes, interflex does not work well with factorial controls. Ziyi, do you have any ideas? Best, Yiqing

…

On Thu, Jun 10, 2021 at 9:48 AM MartinHaus1 ***@***.***> wrote: Just a quick follow-up: I tried to create a new variable (as a factor) and included it in the Z-vector to control for it. It throws the following error: Cross-validating bandwidth ... Parallel computing with 8 cores... Optimal bw=0.0609. Error in names(result) <- c("x0", "(Intercept)", use.variable[2:length(use.variable)]) : 'names' attribute [119] must be the same length as the vector [118] If I remove the new categorical variable from the control vector, it works. Using the same variable for creating a model with *plm* does not lead to any issues. Am I doing something wrong? For context, I use the following command options: interflex(estimator="kernel", Y="y", D="d", X="x", Z=c("new_control_factor","other_controls"), IV="iv", FE=c("year", "state"),data=dataset, cores=8, na.rm = T, vcov.type="cluster", cl="state") Thank you! — You are receiving this because you commented. Reply to this email directly, view it on GitHub <#9 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AB2PKGEI5ZX4MO35BMXY23DTSDUFFANCNFSM46OFTO4Q> .

-- Yiqing Xu Assistant Professor Department of Political Science University of California, San Diego http://yiqingxu.org/

[lzy318]

Can you send a sample dataset to [email protected] for debugging as this code works well on my sample dataset? Thanks a lot!

[MartinHaus1]

Hi,

thanks both for your responses. I mailed a sample dataset that should reproduce the error.

Best,
Martin