How to choose non-interruped numbers only?

问题

I have two vectors:

x<-c(0,1,0,2,3,0,1,1,0,2)
y<-c("00:01:00","00:02:00","00:03:00","00:04:00","00:05:00",
     "00:06:00","00:07:00","00:08:00","00:09:00","00:10:00")

I need to choose only those in y, where values of x is not interrupted by 0. As a result, I'd like to get a dataframe like this

We built a script like this, but with a big dataset it takes time. Is there a more elegant solution? And I wonder, why df<-rbind(bbb,df) returns inverted df?

aaa<-data.frame(y,x)
df<-NULL
for (i in 1:length(aaa$x)){
  bbb<-ifelse((aaa$x[i]*aaa$x[i+1])!=0, 
              aaa$x[i], 
              ifelse((aaa$x[i]*aaa$x[i-1])!=0, 
                     aaa$x[i], 
                     NA))
  df<-rbind(bbb,df)
}
df<-data.frame(rev(df))
aaa$x<-df$rev.df.
bbb<-na.omit(aaa)
bbb

I'm a newbie in R, so please, as much detail as you can :) Thank you!

回答1:

aaa <- data.frame(y,x)
rles <- rle(aaa$x == 0)
bbb <- aaa[rep(rles$values == FALSE & rles$lengths >= 2, rles$lengths),]

which gives

> bbb
         y x
4 00:04:00 2
5 00:05:00 3
7 00:07:00 1
8 00:08:00 1

The sub-question you had: df<-rbind(bbb,df) returns df reversed because you are adding the new row (bbb) before the rest (existing) rows; invert the order of the arguments and you won't need to reverse df.

Now to break down the answer, since it involves a lot of parts. First, rephrasing your criteria, you want stretches of aaa that don't have 0's for at least 2 rows. So the first criteria is finding the 0's

> aaa$x == 0
 [1]  TRUE FALSE  TRUE FALSE FALSE  TRUE FALSE FALSE  TRUE FALSE

Then you want to figure out the length of each of these stretches; rle does this.

> rle(aaa$x == 0)
Run Length Encoding
  lengths: int [1:8] 1 1 1 2 1 2 1 1
  values : logi [1:8] TRUE FALSE TRUE FALSE TRUE FALSE ...

This means there was 1 TRUE, then 1 FALSE, then 1 TRUE, then 2 FALSEs, etc. This result is assigned to rles. The parts you want are where the value is FALSE (not 0), and the length of that run is 2 or more.

> rles$values == FALSE & rles$lengths >= 2
[1] FALSE FALSE FALSE  TRUE FALSE  TRUE FALSE FALSE

This needs to be expanded back out to the length of aaa, and rep will do that, using the rles$lengths to replicate the appropriate entries.

> rep(rles$values == FALSE & rles$lengths >= 2, rles$lengths)
 [1] FALSE FALSE FALSE  TRUE  TRUE FALSE  TRUE  TRUE FALSE FALSE

This gives a logical vector appropriate for indexing aaa

> aaa[rep(rles$values == FALSE & rles$lengths >= 2, rles$lengths),]
         y x
4 00:04:00 2
5 00:05:00 3
7 00:07:00 1
8 00:08:00 1

来源：https://stackoverflow.com/questions/12790378/how-to-choose-non-interruped-numbers-only

标签

selection