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Created August 29, 2019 20:28
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That quadruple sapply()
# Hi Glen. Here is comes:
countries <- unique(country)
everything <- sapply(countries, function(cy) {
parties <- unique(party[country==cy]) # all parties
this <- sapply(parties, function(z) sapply(1990:2013, function(y) mav(as.numeric(sapply(seq(y-3,y+3), function(x) experts.raw[country==cy & party==z & year==x])))))
rownames(this) <- 1990:2013
colnames(this) <- parties
print(everything )
# >> mav() is a moving average function. I have wrapped it into an sapply() at the time to get the seven years in my dataset separately.
mav <- function(x,k=3){
n <- length(x)
return(sapply((k+1):(n-k), function(i) sum(x[(i-k):(i+k)],na.rm=TRUE)/(2*k+1-sum([(i-k):(i+k)])))))
# >> The dataset consists of several columns, of which "country", "year", "party", and "expert.raw" are relevant. In the "expert.raw" we have the party position from an expert survey, the other three columns identify the country, year, and party. There are plenty of missing values because we do not have expert estimates for all the elections.
# >> In the first sapply(), I basically ask R to run the code for each of the countries
# >> In the second sapply(), I ask R to run the code for each party in a country
# >> In the third sapply(), I ask R to run the code for each year (for a given party in a given country).
# >> The fourth sapply() feeds the existing expert estimates for the year examined +/- 3 years (i.e. 7 years) and calculates the moving average on that.
# >> After closing all the relevant brackets, I just label the column and row names
# >> The resulting object "everything" is a list of 8 (countries) with a column for each party in a country and the years 1990 to 2013. The cells give the moving average of the expert positions.
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