Regression table for outputs from 10 countries - regression

How can I combine the regression results for 10 countries in Stata?
I use just one do-file to run the regressions for each country. After running the do-file every time, I can save the results with estout command to a word document. In this case, I have 10 regression outcomes saved separately for each country. Do you know a nice way of combining these regressions in one or two tables at least?

You definitely could've just googled this (as I did) and got a quicker answer but here goes:
The basic approach to do what you want is: run a model, say reg y x, save the estimates into an object using the eststo command (which is shorthand for estimates store), repeat this for as many models you want, and then export the saved estimates into a table using the estout command.
For example:
use https://stats.idre.ucla.edu/stat/stata/notes/hsb2, clear
regress read female write
estimates store m1, title(Model 1)
regress read female write math
estimates store m2, title(Model 2)
regress read female write math science socst
estimates store m3, title(Model 3)
estout m1 m2 m3
---------------------------------------------------
m1 m2 m3
b b b
---------------------------------------------------
female -4.532084 -2.739657 -1.328513
write .7067537 .3924361 .1503085
math .4753659 .2934723
science .2508791
socst .2694578
_cons 17.40106 7.986659 2.44264
---------------------------------------------------
You can add significance stars and standard errors in parentheses below the coefficients with these options:
estout m1 m2 m3, cells(b(star fmt(3)) se(par fmt(2)))
------------------------------------------------------------
m1 m2 m3
b/se b/se b/se
------------------------------------------------------------
female -4.532*** -2.740* -1.329
(1.17) (1.09) (1.05)
write 0.707*** 0.392*** 0.150
(0.06) (0.07) (0.08)
math 0.475*** 0.293***
(0.07) (0.07)
science 0.251***
(0.07)
socst 0.269***
(0.06)
_cons 17.401*** 7.987* 2.443
(3.20) (3.23) (3.11)
------------------------------------------------------------
And if you want to export to a Word document you need only specify a file name and the correct extension. E.g.
estout m1 m2 m3 using "~/Desktop/Regression_results.rtf", cells(b(star fmt(3)) se(par fmt(2)))
And then open it with MS Word.
These examples come from this UCLA resource which you can review for more info

You could create a new dataset with estimation results appending from each country, like this:
use https://stats.idre.ucla.edu/stat/stata/notes/hsb2, clear
regress read female write
estimates store m1, title(Model 1)
regress read female write math
estimates store m2, title(Model 2)
regress read female write math science socst
estimates store m3, title(Model 3)
estout m1 m2 m3
keep _est_m1 _est_m2 _est_m3
save data1.dta, replace
use https://stats.idre.ucla.edu/stat/stata/notes/hsb2, clear
regress read female write
estimates store m4, title(Model 1)
regress read female write math
estimates store m5, title(Model 2)
regress read female write math science socst
estimates store m6, title(Model 3)
keep _est_m4 _est_m5 _est_m6
save data2.dta, replace
use data1.dta, clear
append using data2.dta
estout m1 m2 m3 m4 m5 m6
estout m1 m2 m3 m4 m5 m6 using "~/Desktop/Regression_results.rtf", cells(b(star fmt(3)) se(par fmt(2)))

Related

How to calculate Moran's I and GWR when given duplicated factor data

I'm trying to do the statistical analysis by using the Moran's I, but it leads me into a serious problem.
Suppose the data look like:
y
indep1
indep2
coord_x
coord_y
District
y1
indep1 1
indep2 1
coord_x 1
coord_y 1
A
y2
indep1 2
indep2 2
coord_x 1
coord_y 1
A
y3
indep1 3
indep2 3
coord_x 1
coord_y 1
A
y4
indep1 4
indep2 4
coord_x 2
coord_y 2
B
y5
indep1 5
indep2 5
coord_x 2
coord_y 2
B
Note that the data is given as form of *.shp.
This dataframe has 2 unique districts, but total 5 rows of data. If we want to calculate the Moran's I, then the weighted matrix W is 2 by 2.
But If I run the code
ols <- lm(y~indep1+indep2, data=dataset)
We cannot calculate the
lm.morantest(ols, w) #w:weighted matrix
# returns "different length"
How can we solve that problem? IF the total number of data and the number of unique districts are different, then how can we check the spatial auto-correlation between the districts and how can we apply GWR(geographically weighted regression?) Any reference papers or your advises could be helpful.
Thank you for your help in advance.
I tried to calculate the moran's I by making the number our dependent variables same as the number of unique districts. It can be possible when aggregating the sum of dependent variables with respect to the districts.
library(reshape2)
y_agg <- dcast(shp#data, district~., value.var="dependent variable", sum)
y_agg <- y_agg$.
moran.test(y_agg, W)
But I think it is not the right way to analyze the spatial regression since all the other independent variables are ignored. How can I solve that problem? Is there any way for solving that problem without making the aggregated independent variables of my data?
Thank you.

Panel data or time-series data and xt regression

Need help observing simple regression as well as xt-regression for panel data.
The dataset consists of 16 participants in which daily observations were made.
I would like to observe the difference between pre-test (from the first date on which observations were taken) and post-test (the last date on which observations were made) across different variables.
also I was advised to do xtregress, re
what is this re? and its significance?
If the goal is to fit some xt model at the end, you will need the data in long form. I would use:
bysort id (year): keep if inlist(_n,1,_N)
For each id, this puts the data in ascending chronological order, and keeps the first and last observation for each id.
The RE part of your question is off-topic here. Try Statalist or CV SE site, but do augment your questions with details of the data and what you hope to accomplish. These may also reveal that getting rid of the intermediate data is a bad idea.
Edit:
Add this after the part above:
bysort id (year): gen t= _n
reshape wide x year, i(id) j(t)
order id x1 x2 year1 year2
Perhaps this sample code will set you in the direction you seek.
clear
input id year x
1 2001 11
1 2002 12
1 2003 13
1 2004 14
2 2001 21
2 2002 22
2 2003 23
3 1005 35
end
xtset id year
bysort id (year): generate firstx = x[1]
bysort id (year): generate lastx = x[_N]
list, sepby(id)
With regard to xterg, re, that fits a random effects model. See help xtreg for more details, as well as the documentation for xtreg in the Stata Longitudinal-Data/Panel-Data Reference Manual included in your Stata documentation.

SQL Commands for Statistical Modelling

I have a set of Hockey stats imported into a data frame in R. I'm having trouble finding the right queries for the set of data I want.
All records are kept in 1 table called skaters
Name Team Opp G
AAAAA PHI BOS 2
BBBBB NYR OTT 7
AAAAA PHI BOS 9
DDDDD BOS PHI 3
EEEEE BOS PHI 1
FFFFF PHI BOS 2
GGGGG OTT NYR 3
I'd like to find a way to take a sum of G per team. Thoughts were to you use a query like
SELECT DISTINCT(Team), SUM(G) FROM skaters but this didn't give me what I was anticipating.
Then I found the GROUP BY function, which found the correct data. Then I wanted to up the ante once more.
I want to perform a query that will perform the sum I was looking for, but for both the home and away teams and compare them.
Hope this what you are looking for.
SELECT Team, SUM(G) 'Sum' FROM skaters group by Team
library(data.table)
setDT(df) # convert data frame to data table
df[,list(sum_G = sum(G)), by=Team]
# Team sum_G
# 1: PHI 13
# 2: NYR 10
# 3: BOS 4
# Or
library(dplyr)
df %>% group_by(Team) %>% summarise(sum_G = sum(G))
Since you state that you have the stats in a data frame:
aggregate(skaters$G, list(skaters$Team), sum)
Select Team ,Sum(G) as Total from skaters Group by Team

COUNTIF for rows which contain a given value in another column

My table lists every character from all 5 of George R. R. Martin's currently published A Song of Ice and Fire novels. Each row contains a record indicating which book in the series the character is from (numbered 1-5) and a single letter indicating the character's gender (M/F). For example:
A B C
1 Character Book Gender
------------------------------
2 Arya Stark - 1 - F
3 Eddard Stark - 1 - M
4 Davos Seaworth - 2 - M
5 Lynesse Hightower - 2 - F
6 Xaro Xhoan Daxos - 2 - M
7 Elinor Tyrell - 3 - F
I can use COUNTIF to find out that there are three females and three males in this table, but I want to know, for example, how many males there are in book 2. How could I write a formula that would make this count? Here is a pseudocode of what I'm trying to achieve:
=COUNTIF(C2:C7, Column B = '2' AND Column C = 'M')
This would output 2.
I'm aware that this task is far better suited to databases and a SELECT query, but I'd like to know how to solve this problem within the constraints of a LibreOffice Calc spreadsheet, without using a macro. Excel-based solutions are fine, so long as they also work in Calc. If there's no solution that uses COUNTIF, it doesn't matter, so long as it works.
I worked it out, thanks to a prompt by assylias. The COUNTIFS formula produces the result I want by counting multiple search criteria. For example, this formula works out how many male characters are in Book 1 (A Game of Thrones).
=COUNTIFS($A$2:$A$2102, "=1", $L$2:$L$2102, "=M")

Ranking algorithm using likes / dislikes and average views per day

I'm currently ranking videos on a website using a bayesian ranking algorithm, each video has:
likes
dislikes
views
upload_date
Anyone can like or dislike a video, a video is always views + 1 when viewed and all videos have a unique upload_date.
Data Structure
The data is in the following format:
| id | title | likes | dislikes | views | upload_date |
|------|-----------|---------|------------|---------|---------------|
| 1 | Funny Cat | 9 | 2 | 18 | 2014-04-01 |
| 2 | Silly Dog | 9 | 2 | 500 | 2014-04-06 |
| 3 | Epic Fail | 100 | 0 | 200 | 2014-04-07 |
| 4 | Duck Song | 0 | 10000 | 10000 | 2014-04-08 |
| 5 | Trololool | 25 | 30 | 5000 | 2014-04-09 |
Current Weighted Ranking
The following weighted ratio algorithm is used to rank and sort the videos so that the best rated are shown first.
This algorithm takes into account the bayesian average to give a better overall ranking.
Weighted Rating (WR) = ((AV * AR) + (V * R))) / (AV + V)
AV = Average number of total votes
AR = Average rating
V = This items number of combined (likes + dislikes)
R = This items current rating (likes - dislikes)
Example current MySQL Query
SELECT id, title, (((avg_vote * avg_rating) + ((likes + dislikes) * (likes / dislikes)) ) / (avg_vote + (likes + dislikes))) AS score
FROM video
INNER JOIN (SELECT ((SUM(likes) + SUM(dislikes)) / COUNT(id)) AS avg_vote FROM video) AS t1
INNER JOIN (SELECT ((SUM(likes) - SUM(dislikes)) / COUNT(id)) AS avg_rating FROM video) AS t2
ORDER BY score DESC
LIMIT 10
Note: views and upload_date are not factored in.
The Issue
The ranking currently works well but it seems we are not making full use of all the data at our disposal.
Having likes, dislikes, views and an upload_date but only using two seems a waste because the views and upload_date are not factored in to account how much weight each like / dislike should have.
For example in the Data Structure table above, items 1 and 2 both have the same amount of likes / dislikes however item 2 was uploaded more recently so it's average daily views are higher.
Since item 2 has more likes and dislikes in a shorter time than those likes / dislikes should surely be weighted stronger?
New Algorithm Result
Ideally the new algorithm with views and upload_date factored in would sort the data into the following result:
Note: avg_views would equal (views / days_since_upload)
| id | title | likes | dislikes | views | upload_date | avg_views |
|------|-----------|---------|------------|---------|---------------|-------------|
| 3 | Epic Fail | 100 | 0 | 200 | 2014-04-07 | 67 |
| 2 | Silly Dog | 9 | 2 | 500 | 2014-04-06 | 125 |
| 1 | Funny Cat | 9 | 2 | 18 | 2014-04-01 | 2 |
| 5 | Trololool | 25 | 30 | 5000 | 2014-04-09 | 5000 |
| 4 | Duck Song | 0 | 10000 | 10000 | 2014-04-08 | 5000 |
The above is a simple representation, with more data it gets a lot more complex.
The question
So to summarise, my question is how can I factor views and upload_date into my current ranking algorithm in a style to improve the way that videos are ranked?
I think the above example by calculating the avg_views is a good way to go but where should I then add that into the ranking algorithm that I have?
It's possible that better ranking algorithms may exist, if this is the case then please provide an example of a different algorithm that I could use and state the benefits of using it.
Taking a straight percentage of views doesn't give an accurate representation of the item's popularity, either. Although 9 likes out of 18 is "stronger" than 9 likes out of 500, the fact that one video got 500 views and the other got only 18 is a much stronger indication of the video's popularity.
A video that gets a lot of views usually means that it's very popular across a wide range of viewers. That it only gets a small percentage of likes or dislikes is usually a secondary consideration. A video that gets a small number of views and a large number of likes is usually an indication of a video that's very narrowly targeted.
If you want to incorporate views in the equation, I would suggest multiplying the Bayesian average you get from the likes and dislikes by the logarithm of the number of views. That should sort things out pretty well.
Unless you want to go with multi-factor ranking, where likes, dislikes, and views are each counted separately and given individual weights. The math is more involved and it takes some tweaking, but it tends to give better results. Consider, for example, that people will often "like" a video that they find mildly amusing, but they'll only "dislike" if they find it objectionable. A dislike is a much stronger indication than a like.
I can point you to a non-parametric way to get the best ordering with respect to a weighted linear scoring system without knowing exactly what weights you want to use (just constraints on the weights). First though, note that average daily views might be misleading because movies are probably downloaded less in later years. So the first thing I would do is fit a polynomial model (degree 10 should be good enough) that predicts total number of views as a function of how many days the movie has been available. Then, once you have your fit, then for each date you get predicted total number of views, which is what you divide by to get "relative average number of views" which is a multiplier indicator which tells you how many times more likely (or less likely) the movie is to be watched compared to what you expect on average given the data. So 2 would mean the movie is watched twice as much, and 1/2 would mean the movie is watched half as much. If you want 2 and 1/2 to be "negatives" of each other which sort of makes sense from a scoring perspective, then take the log of the multiplier to get the score.
Now, there are several quantities you can compute to include in an overall score, like the (log) "relative average number of views" I mentioned above, and (likes/total views) and (dislikes / total views). US News and World Report ranks universities each year, and they just use a weighted sum of 7 different category scores to get an overall score for each university that they rank by. So using a weighted linear combination of category scores is definitely not a bad way to go. (Noting that you may want to do something like a log transform on some categories before taking the linear combination of scores). The problem is you might not know exactly what weights to use to give the "most desirable" ranking. The first thing to note is that if you want the weights on the same scale, then you should normalize each category score so that it has standard deviation equal to 1 across all movies. Then, e.g., if you use equal weights, then each category is truly weighted equally. So then the question is what kinds of weights you want to use. Clearly the weights for relative number of views and proportion of likes should be positive, and the weight for proportion of dislikes should be negative, so multiply the dislike score by -1 and then you can assume all weights are positive. If you believe each category should contribute at least 20%, then you get that each weight is at least 0.2 times the sum of weights. If you believe that dislikes are more important that likes, then you can say (dislike weight) >= c*(like weight) for some c > 1, or (dislike_weight) >= c*(sum of weights) + (like weight) for some c > 0. Similarly you can define other linear constraints on the weights that reflect your beliefs about what the weights should be, without picking exact values for the weights.
Now here comes the fun part, which is the main thrust of my post. If you have linear inequality constraints on the weights, all of the form that a linear combination of the weights is greater than or equal to 0, but you don't know what weights to use, then you can simply compute all possible top-10 or top-20 rankings of movies that you can get for any choice of weights that satisfy your constraints, and then choose the top-k ordering which is supported by the largest VOLUME of weights, where the volume of weights is the solid angle of the polyhedral cone of weights which results in the particular top-k ordering. Then, once you've chosen the "most supported" top-k ranking, you can restrict the scoring parameters to be in the cone that gives you that ranking, and remove the top k movies, and compute all possibilities for the next top-10 or top-20 ranking of the remaining movies when the weights are restricted to respect the original top-k movies' ranking. Computing all obtainale top-k rankings of movies for restricted weights can be done much, much faster than enumerating all n(n-1)...(n-k+1) top-k possible rankings and trying them all out. If you have two or three categories then using polytope construction methods the obtainable top-k rankings can be computed in linear time in terms of the output size, i.e. the number of obtainable top-k rankings. The polyhedral computation approach also gives the inequalities that define the cone of scoring weights that give each top-k ranking, also in linear time if you have two or three categories. Then to get the volume of weights that give each ranking, you triangulate the cone and intersect with the unit sphere and compute the areas of the spherical triangles that you get. (Again linear complexity if the number of categories is 2 or 3). Furthermore, if you scale your categories to be in a range like [0,50] and round to the nearest integer, then you can prove that the number of obtainable top-k rankings is actually quite small if the number of categories is like 5 or less. (Even if you have a lot of movies and k is high). And when you fix the ordering for the current top group of movies and restrict the parameters to be in the cone that yields the fixed top ordering, this will further restrict the output size for the obtainable next best top-k movies. The output size does depend (polynomially) on k which is why I recommended setting k=10 or 20 and computing top-k movies and choosing the best (largest volume) ordering and fixing it, and then computing the next best top-k movies that respect the ordering of the original top-k etc.
Anyway if this approach sounds appealing to you (iteratively finding successive choices of top-k rankings that are supported by the largest volume of weights that satisfy your weight constraints), let me know and I can produce and post a write-up on the polyhedral computations needed as well as a link to software that will allow you to do it with minimal extra coding on your part. In the meantime here is a paper http://arxiv.org/abs/0805.1026 I wrote on a similar study of 7-category university ranking data where the weights were simply restricted to all be non-negative (generalizing to arbitrary linear constraints on weights is straightforward).
A simple approach would be to come up with a suitable scale factor for each average - and then sum the "weights". The difficult part would be tweaking the scale factors to produce the desired ordering.
From your example data, a starting point might be something like:
Weighted Rating = (AV * (1 / 50)) + (AL * 3) - (AD * 6)
Key & Explanation
AV = Average views per day:
5000 is high so divide by 50 to bring the weight down to 100 in this case.
AL = Average likes per day:
100 in 3 days = 33.33 is high so multiply by 3 to bring the weight up to 100 in this case.
AD = Average dislikes per day:
10,000 seems an extreme value here - would agree with Jim Mischel's point that dislikes may be more significant than likes so am initially going with a negative scale factor of twice the size of the "likes" scale factor.
This gives the following results (see SQL Fiddle Demo):
ID TITLE SCORE
-----------------------------
3 Epic Fail 60.8
2 Silly Dog 4.166866
1 Funny Cat 1.396528
5 Trololool -1.666766
4 Duck Song -14950
[Am deliberately keeping this simple to present the idea of a starting point - but with real data you might find linear scaling isn't sufficient - in which case you could consider bandings or logarithmic scaling.]
Every video have:
likes
dislikes
views
upload_date
So we can deduct the following parameters from them:
like_rate = likes/views
dislike_rate = likes/views
view_rate = views/number_of_website_users
video_age = count_days(upload_date, today)
avg_views = views/upload_age
avg_likes = likes/upload_age
avg_dislikes = dislikes/upload_age
Before we can set the formula to be used, we need to specify how different videos popularity should work like, one way is to explain in points the property of a popular video:
A popular video is a recent one in most cases
The older a video gets, the higher avg_views it requires to become popular
A video with a like_rate over like_rate_threshold or a dislike_rate over dislike_rate_threshold, can compete by the difference from its threshold with how old it gets
A high view_rate of a video is a good indicator to suggest that video to a user who have not watched it before
If avg_likes or avg_dislikes make most of avg_views, the video is considered active in the meantime, in case of active videos we don't really need to check how old it's
Conclusion: I don't have a formula, but one can be constructed by converting one unit into another's axis, like cutting a video age by days based on a calculation made using avg_likes, avg_dislikes, and avg_views
Since no one has pointed it out yet (and I'm a bit surprised), I'll do it. The problem with any ranking algorithm we might come up with is that it's based on our point of view. What you're certainly looking for is an algorithm that accomodates the median user point of view.
This is no new idea. Netflix had it some time ago, only they personalized it, basing theirs on individual selections. We are looking - as I said - for the median user best ranking.
So how to achieve it? As others have suggested, you are looking for a function R(L,D,V,U) that returns a real number for the sort key. R() is likely to be quite non-linear.
This is a classical machine learning problem. The "training data" consists of user selections. When a user selects a movie, it's a statement about the goodness of the ranking: selecting a high-ranked one is a vote of confidence. A low-ranked selection is a rebuke. Function R() should revise itself accordingly. Initially, the current ranking system can be used to train the system to mirror its selections. From there it will adapt to user feedback.
There are several schemes and a huge research literature on machine learning for problems like this: regression modeling, neural networks, representation learning, etc. See for example the Wikipedia page for some pointers.
I could suggest some schemes, but won't unless there is interest in this approach. Say "yes" in comments if this is true.
Implementation will be non-trivial - certainly more than just tweaking your SELECT statement. But on the plus side you'll be able to claim your customers are getting what they're asking for in very good conscience!