有效计数data.table中的非NA元素 [英] Efficiently counting non-NA elements in data.table

问题描述

有时候我需要计算 data.table NA 个元素的数量>。 data.table 量身定制的最佳方法是什么？

Sometimes I need to count the number of non-NA elements in one or another column in my data.table. What is the best data.table-tailored way to do so?

具体来说，让我们一起工作：

For concreteness, let's work with this:

DT <- data.table(id = sample(100, size = 1e6, replace = TRUE),
                 var = sample(c(1, 0, NA), size = 1e6, replace = TRUE), key = "id")

我想到的第一件事就是这样：

The first thing that comes to my mind works like this:

DT[!is.na(var), N := .N, by = id]

但是这有一个不幸的缺点是 N 不会分配给缺少 var 的任何行，即 DT [is。 na（var），N] = NA 。

But this has the unfortunate shortcoming that N does not get assigned to any row where var is missing, i.e. DT[is.na(var), N] = NA.

所以我通过添加以下内容来解决此问题：

So I work around this by appending:

DT[!is.na(var), N:= .N, by = id][ , N := max(N, na.rm = TRUE), by = id] #OPTION 1

但是，我不确定这是最好的方法；我想到的另一种选择，也是由 data.frame 的这个问题将是：

However, I'm not sure this is the best approach; another option I thought of and one suggested by the analog to this question for data.frames would be:

DT[ , N := length(var[!is.na(var)]), by = id] # OPTION 2

和

DT[ , N := sum(!is.na(var)), by = id] # OPTION 3

比较这些计算时间（平均100多次试验） ，最后一个似乎是最快的：

Comparing computation time of these (average over 100 trials), the last seems to be the fastest:

OPTION 1 | OPTION 2 | OPTION 3
  .075   |   .065   |   .043

有人知道 data.table ？

推荐答案

是的，选项3rd似乎是最好的选择。我添加了另一个仅当您考虑将data.table的键从 id 更改为 var ，但选项3仍然是您数据上最快的。

Yes the option 3rd seems to be the best one. I've added another one which is valid only if you consider to change the key of your data.table from id to var, but still option 3 is the fastest on your data.

library(microbenchmark)
library(data.table)

dt<-data.table(id=(1:100)[sample(10,size=1e6,replace=T)],var=c(1,0,NA)[sample(3,size=1e6,replace=T)],key=c("var"))

dt1 <- copy(dt)
dt2 <- copy(dt)
dt3 <- copy(dt)
dt4 <- copy(dt)

microbenchmark(times=10L,
               dt1[!is.na(var),.N,by=id][,max(N,na.rm=T),by=id],
               dt2[,length(var[!is.na(var)]),by=id],
               dt3[,sum(!is.na(var)),by=id],
               dt4[.(c(1,0)),.N,id,nomatch=0L])
# Unit: milliseconds
#                                                         expr      min       lq      mean    median        uq       max neval
#  dt1[!is.na(var), .N, by = id][, max(N, na.rm = T), by = id] 95.14981 95.79291 105.18515 100.16742 112.02088 131.87403    10
#                     dt2[, length(var[!is.na(var)]), by = id] 83.17203 85.91365  88.54663  86.93693  89.56223 100.57788    10
#                             dt3[, sum(!is.na(var)), by = id] 45.99405 47.81774  50.65637  49.60966  51.77160  61.92701    10
#                        dt4[.(c(1, 0)), .N, id, nomatch = 0L] 78.50544 80.95087  89.09415  89.47084  96.22914 100.55434    10

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