nfmcclure - 1 year ago 91

R Question

The following code/outcome baffles me as to why data.table returns NA for the mean functions and not the sd function.

`library(data.table)`

test <- data.frame('id'=c(1,2,3,4,5),

'A'=seq(2,9,length=5),

'B'=seq(3,9,length=5),

'C'=seq(4,9,length=5),

'D'=seq(5,9,length=5))

test <- as.data.table(test)

test[,`:=`(mean_test = mean(.SD), sd_test = sd(.SD)),by=id,.SDcols=c('A','B','C','D')]

> test

id A B C D mean_test sd_test

1: 1 2.00 3.0 4.00 5 NA 1.2909944

2: 2 3.75 4.5 5.25 6 NA 0.9682458

3: 3 5.50 6.0 6.50 7 NA 0.6454972

4: 4 7.25 7.5 7.75 8 NA 0.3227486

5: 5 9.00 9.0 9.00 9 NA 0.0000000

I've learned quite a bit searching around, going through the DT tutorials/examples. This question is very similar to what I was hoping to do.

Why does the standard deviation function work and the mean function return NA?

`test[,`:=`(mean_test = apply(.SD, 1, mean), sd_test = apply(.SD, 1, sd),by=id,.SDcols=c('A','B','C','D')]`

> test

id A B C D mean_test sd_test

1: 1 2.00 3.0 4.00 5 3.500 1.2909944

2: 2 3.75 4.5 5.25 6 4.875 0.9682458

3: 3 5.50 6.0 6.50 7 6.250 0.6454972

4: 4 7.25 7.5 7.75 8 7.625 0.3227486

5: 5 9.00 9.0 9.00 9 9.000 0.0000000

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

`.SD`

is itself a `data.table`

Thus, when you take `mean(.SD)`

you are (attempting) to take the mean of an entire data.table

The function `mean()`

does not know what to do with the data.table and returns `NA`

```
## the .SD in your question is the same as
test[, c('A','B','C','D'), with=FALSE]
## try taking its mean
mean(test[, c('A','B','C','D'), with=FALSE])
# Warning in mean.default(test[, c("A", "B", "C", "D"), with = FALSE]) :
# argument is not numeric or logical: returning NA
# [1] NA
```

use `lapply(.SD, mean)`

for column-wise
or `apply(.SD, 1, mean)`

for row-wise

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**