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summary many regressions

4 messages · Gary Dong, arun, David Winsemius +1 more

#
Hi,
Try:
res <- do.call(rbind,lapply(split(data,data$city),function(z) {fit_city <- lm(y~x,data=z);data.frame(City=unique(z$city),Coefficient=coef(fit_city)[2],Adjusted_R_square= summary(fit_city)$adj.r.squared)}))

A.K.
On Monday, November 25, 2013 6:37 PM, Gary Dong <pdxgary163 at gmail.com> wrote:
Dear R users,

I have a large data set which includes data from 300 cities. I want to run
a biviriate regression for each city and record the coefficient and the
adjusted R square.

For example, in the following, I have 10 cities represented by numbers from
1 to 10:

x = cumsum(c(0, runif(999, -1, +1)))
y = cumsum(c(0, runif(999, -1, +1)))
city = rep(1:10,each=100)
data<-data.frame(cbind(x,y,city))

I can manually run regressions for each city:
fit_city1 <- lm(y ~ x,data=subset(data,data$city==1))
summary(fit_city1)

Obvious, it is very tedious to run 300 regressions. I wonder if there is a
quicker way to do this. Use for loop?? what I want to see is something like
this:

City? ? ? ? Coefficient? ? ?  Adjusted R square
1? ? ? ? ? ? ? -0.05? ? ? ? ? ? ? ? ? 0.36
2? ? ? ? ? ? ? -0.12? ? ? ? ? ? ? ? ? 0.20
3? ? ? ? ? ? ? -0.05? ? ? ? ? ? ? ? ? 0.32
.....

Any advice is appreciated!

Gary

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#
On Nov 25, 2013, at 3:35 PM, Gary Dong wrote:

            
The way to get the most rapid response from this list is to post a dataset that represents the complexity of the problem. Presumably this large dataset is either a dataframe with a column of city entries or a list of dataframes. Why not post dput() applied to an extract of three of the cities and include sufficient rows to allow a regression?

        
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#
Hi

It is work for split/lapply or sapply approach.

ff<-function(data) {ss<-lm(y~x, data); c(coef(ss), summary(ss)$adj.r.squared)}
lapply(split(data[,1:2], data$city), ff)

Regards
Petr