try this:
x - c(36, 40, 10, 4)
x.m - matrix(as.integer(intToBits(x)), byrow = TRUE, ncol = 32)[, 1:20]
x.m - data.frame(x.m) # convert to data.frame
x.m
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20
1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
On Fri, Jul 8, 2011 at 6:39 PM, Justin Haynes jto...@gmail.com wrote:
Happy weekend helpeRs!
As usual, I'm stumped by R...
My plan was to take an integer number, convert it to binary and wind
up with a data.frame where each column is either 1 or 0 so I can see
which bits are changing:
bb-function(i) ifelse(i, paste(bb(i %/% 2), i %% 2, sep=), )
my.dat-c(36,40,10,4)
my.binary.dat-bb(my.dat)
my.list-strsplit(my.binary.dat,'')
max.len-max(ldply(my.list,length))
len-length(my.list)
my.df-data.frame(two=rep(0,len),four=rep(0,len),eight=rep(0,len),sixteen=rep(0,len),thirtytwo=rep(0,len),sixtyfour=rep(0,len))
for(i in 1:length(my.list)){
for(j in 1:length(my.list[[i]])){
my.df[i,max.len-length(my.list[[i]])+j]-my.list[[i]][j]
}
}
But this isn't exactly feasable on a million+ rows where some binary
numbers are 20 digits... I know theres a way without loops I just
know it!
Ideally, I can do this to multiple columns of a data.frame and have
them named accordingly (V1.two,V1.four... V2.two,V2.four, etc.)
Thanks,
Justin
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Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
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and provide commented, minimal, self-contained, reproducible code.
