在 mutate 中应用函数

Question

我有一个包含时间戳的 data.table 对象（以午夜后的秒数衡量）。我的目标是运行一个函数，该函数为每一行返回观察前最大 $k$ 秒发生的观察次数。

require(data.table, dplyr, dtplyr)
set.seed(123)
DF  <- data.frame(Secs=cumsum(rexp(10000,1)))
setDT(DF)  
> DF
               Secs
1: 8.434573e-01
2: 1.420068e+00
3: 2.749122e+00
4: 2.780700e+00
5: 2.836911e+00
---             
 9996: 1.003014e+04
 9997: 1.003382e+04
 9998: 1.003384e+04
 9999: 1.003414e+04
10000: 1.003781e+04

我想应用于每一行的功能是

nS<-function(Second,k=5) 
    max(1,nrow(DF%>%filter(Secs<Second & Secs>=Second-k)))

获得我想要的东西的一种方法是使用apply，这需要很长时间。

 system.time(val <- apply(DF,1,nS))
   User      System verstrichen 
  20.56        0.03       20.66 

#Not working 
DF%>%mutate(nS=nS(Secs,100))%>%head()

# Also not working
library(lazyeval)
f = function(col1, new_col_name) {
    mutate_call = lazyeval::interp(~ nS(a), a = as.name(col1))
    DF%>%mutate_(.dots=setNames(list(mutate_call),new_col_name))
}
head(f('Secs', 'nS'))

DF%>%mutate(minTime=Secs-k)%>%head()

难道不能通过使用 mutate 来实现这种方法吗？非常感谢你的帮助！

Answer 1

使用rowwise()对你有用吗？

DF %>% rowwise() %>% mutate(ns = nS(Secs), # default k = 5, equal to your apply
                            ns2 = nS(Secs, 100)) # second test case k = 100
Source: local data frame [10,000 x 3]
Groups: <by row>

# A tibble: 10,000 × 3
        Secs    ns   ns2
       <dbl> <dbl> <dbl>
1  0.1757671     1     1
2  1.1956531     1     1
3  1.6594676     2     2
4  2.6988685     3     3
5  2.8845783     4     4
6  3.1012975     5     5
7  4.1258548     6     6
8  4.1584318     7     7
9  4.2346702     8     8
10 6.0375495     8     9
# ... with 9,990 more rows

它只比apply在我的机器上快一点...

system.time(DF %>% rowwise() %>% mutate(ns = nS(Secs)))
   user  system elapsed 
 13.934   1.060  15.280 

system.time(apply(DF, 1, nS))
   user  system elapsed 
 14.938   1.101  16.438