Not a new approach, but some benchmark data (the perl=TRUE speeds up Jim's
suggestion):
> x <- c('18x.6','12x.9','302x.3')
> y <- rep(x,10)
> system.time(temp <- unlist(lapply(strsplit(y,".",fixed=TRUE),function(x)
>x[1])))
user system elapsed
1.203 0.018 1.222
> system.time(temp2 <-
> df$a[is.infinite(df$a) | is.nan(df$a) ] <- NA
> df
a
1 NA
2 NA
3 NA
4 1
5 2
6 3
On 5/26/11 3:18 PM, "Albert-Jan Roskam" wrote:
>Hi,
>
>I want to recode all Inf and NaN values to NA, but I;m surprised to see
>the
>result of the following code. Could anybody enlighten me about this?
>
>>
Everything looks OK. Does this help?
> test <-
>data.frame(alpha=as.factor(c("A","A","B","B","C")),number=c(1,2,3,4,5))
> mode(test)
[1] "list"
> class(test)
[1] "data.frame"
> sapply(test, mode)
alphanumber
"numeric" "numeric"
> sapply(test, class)
alphanumber
"factor" "numeric
3975"
>one.month "5" "30.894195075" "37.657271835"
>one.month "6" "29.27843098" "37.59689852"
>one.month "7" "27.5014142975" "34.36265367"
>one.month "8" "26.4055425"
Gregory: Would setting limit.list <- NULL at the start do the trick? See
example below:
Analogue of your example:
lower <- 0
upper <- 0
limit.list<-data.frame(lower,upper)
for(i in 1:12) {
some.data <- rnorm(2)
lower <- min(some.data)
upper <- max(some.data)
one.month <- data.frame(low
Geoffrey:
There may be something for this in one of the packages dealing with dates.
If not, here's one (incomplete) idea, based on something I used for a similar
issue a little while ago. Essentially, make a data frame that ranks each
weekday over a period in ascending order. This data frame
ofile and then use
Sys.getenv("DROPBOX_PATH")
to access the path. It seems from looking at forums for Dropbox that there
is no easily accessed environment variable that Dropbox sets for the path
to the Dropbox folder.
--
Ian Gow
Accounting Information and Management
Kellogg School of
Hi:
Reordering the dimensions, then doing a vectorized addition, then reordering
(back) again is faster, it seems.
> m <- 20; n <- 30; p <- 40; q <- 30
> a <- NA
> length(a) <- m * n * p * q
> dim(a) <- c(m, n, p, q)
> x <- 1:n
> a[1:m,,1:p,1:q] <- 0
> b <- a
>
> # Approach 1
> system.time({
+
That approach relies on df1 and df2 not having overlapping values in b.
Slight variation in df2 gives different results:
> df1 <- data.frame(a=c("A","A"),b=c("B","B"))
> df2 <- data.frame(b=c("B","B"),c=c("c","c"))
> merge(df1,df2,all=TRUE)
b a c
1 B A c
2 B A c
3 B A c
4 B A c
On 5/15/11 11:1
Hi:
This is a bit of a kluge, but works for your test case:
> df2[,setdiff(names(df1),names(df2))] <- NA
> df1[,setdiff(names(df2),names(df1))] <- NA
> df3 <- rbind(df1,df2)
> df3
a b c
1 A B
2 A B
3 b c
4 b c
-Ian
On 5/15/11 7:41 PM, "Jonathan Flowers" wrote:
>Hello,
>
>I would like to
t; df.2
> from to value
> 1 99 303 1
> 2 500 702 3
> 3 799 950 5
>
> what I want is:
> time value
> 1 1011
> 2 1991
> 3 3011
> 4 401NA
> 5 5013
> 6 6013
> 7 7003
> 8 8005
> 9 9005
If I assume that the third column in data.frame.2 is named "val" then in
SQL terms it _seems_ you want
SELECT a.time, b.val FROM data.frame.1 AS a LEFT JOIN data.frame.2 AS b ON
a.time BETWEEN b.start AND b.end;
Not sure how to do that elegantly using R subsetting/merge, but you might
try a packa
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