Displaying 1 result from an estimated 1 matches for "19e6".
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1996
2011 Feb 15
4
string parsing
...5B 116.00 166.25 4965150
2 19.1M 3.75 5.47 8521
3 226.6B 22.73 31.58 57127000
4 886.4M 30.80 74.54 226690
5 142.4B 3.21 5.15 541804992
6 276.4M 11.98 21.30 149656
7 55.823B 9.75 18.97 89369000
now I need to do this:
--> convert 55.823B to 55e9 and 19.1M to 19e6
parse.num <- function (s) { as.numeric(gsub("M$","e6",gsub("B$","e9",s))); }
data[1]<-lapply(data[1],parse.num);
seems like awfully inefficient (two regexp substitutions),
is there a better way?
--> iterate over stocks & data at the same time...