Lecture 12
========================================================
author: John S
date: October 21, 2015

More on *apply family
========================================================

- Continued...
- Often things that can be done in a loop can often be done with an "apply type" function.
- Sometimes this is faster.
- It can be easier to program and read.
- **split** the data - **apply** the function - **combine** the results
- how's midterm going?


Today
=====
- Extended example with strike data



Strike Data 
========================================================

From Bruce Western, Sociology, Harvard.

- Data frame of 8 columns: country, year, days on strike per 1000 workers, unemployment, inflation, left-wing share of gov’t, centralization of unions, union density
- 625 observations from 18 countries, 1951–1985
- Since 18 × 35 = 630 > 625, some years missing from some countries

Strike Data 
========================================================

```{r, tidy=T, cache=T}
strike <- read.csv("~/strikes.csv")
names(strike)
strike[1,]
```

Question
==========
Does having a friendlier government make labor action more or less likely?

More specific:  Is there a relationship between a country's ruling party alignment (left vs. right) and the volume of strikes?

One way to address: linear regression by country.

Covariate: left.parliament

Response: strike.volume




Why is it important to do it by country?
========

```{r,fig.width=14,tidy=T}
par(cex=2)
plot(strike$left.parliament,strike$strike.volume,
     xlab="% Left Parliament",ylab="Strike Volume")
```

First step: look at all the data!
========

```{r,echo=F,fig.width=14,tidy=T}
plot(strike$left.parliament[strike$country=="Australia"],strike$strike.volume[strike$country=="Australia"],
     xlab="% Left Parliament",ylab="Strike Volume",main="Australia")
```
Use par(mfrow=c(6,3)) to make for all 18 countries.

Could use a for loop:
=====
```{r,tidy=T,cache=T}
coefs <- data.frame(country=unique(strike$country),int=NA,slope=NA)
  for (i in 1:dim(coefs)[1]){
  temp <- strike[strike$country==coefs$country[i],]
  coefs[i,2:3] <- lm(strike.volume~left.parliament,data=temp)$coef
}  
head(coefs)
```

Another way: sapply()
===


- sapply(x,fun,...)
- write a function that takes one country's data, fits model, and return coefficients.
```{r,cache=T,tidy=T}
fit.for.one.country <- function(one.country.data){
  coef <- 
    lm(strike.volume~left.parliament,data=one.country.data)$coef
  return(coef)
}
```

sapply
=====
Make a list that has datasets for each country in each element
```{r,cache=T,tidy=T}
datasets <- split(strike,strike$country)
names(datasets)
```

sapply
=====
Make a list that has datasets for each country in each element
```{r,cache=T,tidy=T}
fits <- sapply(datasets,fit.for.one.country)
fits[1:2,1:3]
```

plot results: not too conlusive...
===
```{r,echo=F,fig.width=14}
par(cex=2)
plot(fits[2,],xaxt="n",xlab="",ylab="Regression coefficient", 
     main="Countrywise Labor Activity By Left-Wing Score")
axis(side=1,at=seq(along=colnames(fits)),labels=colnames(fits),
las=2,cex.axis=0.5)
abline(h=0,col="grey")
```

Other ways:
=====
* by() (Please see book.)
* lapply(): same as sapply, but it gives a list of outcomes

Summary: 
==========
* **split** - **apply**- **combine** can be done with a loop
* *apply family is another option
  + sometimes much more efficient
  + easier to code and read (after you learn it!)
  + but, output can have a format that is hard to control
* Next: plyr library will address that problem