How to save data column of zoo object to matrix?

Question

I am downloading some data using R package tseries,

require('tseries')
tickers<- c('JPM','AAPL','MSFT','FB','GE');
prices = matrix(NA,nrow=40,ncol=6)
startdate<-'2015-02-02'
enddate<-'2015-03-30'# 40 rows dim()
for(i in 1:5){
  prices[,i]<-get.hist.quote(
    instrument=tickers[i],
    start=startdate,
    end=enddate,
    quote='AdjClose',
    provider='yahoo')
}
colnames(prices)<-c('JPM','AAPL','MSFT','FB','GE');

I want to construct a matrix saving the adjclose price and date information, but I don't know how to access the zoo date column, say when I construct a zoo object using get.hist.quote(), I can view the object like this

But when I save them to matrix, the date column is missing

Answer 1

Here Map applied to get.hist.quote will create a zoo object for each ticker. Then we use zoo's multiway merge.zoo to merge them all together creating a final zoo object prices :

prices <- do.call(merge, 
  Map(get.hist.quote, tickers,
    start=startdate,
    end=enddate,
    quote='AdjClose',
    provider='yahoo')
)

Answer 2

I would probably keep all the series in a zoo object. This can be done like in the following code, thereby also avoiding your for-loop etc. You can always convert this object to a matrix by as.matrix() afterwards.

prices <-lapply(tickers, get.hist.quote, start=startdate, end=enddate, quote='AdjClose')
prices <- Reduce(cbind, prices)
names(prices) <- tickers
prices <- as.matrix(prices)
head(prices)
             JPM   AAPL  MSFT    FB    GE
2015-02-02 55.10 118.16 40.99 74.99 23.99
2015-02-03 56.35 118.18 41.31 75.40 24.25
2015-02-04 56.01 119.09 41.54 75.63 23.94
2015-02-05 56.40 119.94 42.15 75.61 24.28
2015-02-06 57.51 118.93 42.11 74.47 24.30
2015-02-09 57.44 119.72 42.06 74.44 24.42