Hi,
I have a very large data set (aprox. 100,000 rows.)
The data comes from around 10,000 groups with about 10 entered per group.
The values are in one column, the group ID is an integer in the second column.
I want to normalize the values by group:
for(g in unique(groups){
Not tested but should work:
sums = tapply(x, group, sum);
sums.ext = sums[ match(group, names(sums))]
normalized = x/sums.ext
It may be that the tapply is just as slow as your loop though, I'm not sure.
HTH,
Peter
On Thu, Nov 29, 2012 at 10:55 AM, Noah Silverman noahsilver...@ucla.edu wrote:
Yes, type in:
?by
for example:
data - data.frame(fac=factor(c(A,A,B,B)), vec=c(1:4) )
by(data$vec,data$fac, FUN=sum)
Best,
MikoÅaj Hnatiuk
2012/11/29 Noah Silverman noahsilver...@ucla.edu
Hi,
I have a very large data set (aprox. 100,000 rows.)
The data comes from around 10,000 groups
Hello,
If yopu want one value per group use tapply(), if you want one value per
value of x use ave()
tapply(x, group, FUN = function(.x) .x/sum(.x))
ave(x, group, FUN = function(.x) .x/sum(.x))
Hope this helps,
Rui Barradas
Em 29-11-2012 18:55, Noah Silverman escreveu:
Hi,
I have a very
try the 'data.table' package. Takes about 0.1 seconds to normalize the data.
x - data.frame(id = sample(1, 10, TRUE), value = runif(10))
require(data.table)
Loading required package: data.table
data.table 1.8.2 For help type: help(data.table)
system.time({
+ x - data.table(x)
Close, but not quite what I need.
That very nicely gives me sums by group.
I need to take each value of X and divide it by the sum of the group it belongs
to.
With your example, I have 100,000 X and only 10,000 group. The by command
gives me 10,000 sums. I still have to loop over all
On 29-11-2012, at 19:55, Noah Silverman wrote:
Hi,
I have a very large data set (aprox. 100,000 rows.)
The data comes from around 10,000 groups with about 10 entered per group.
The values are in one column, the group ID is an integer in the second column.
I want to normalize the
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