You could try something like this in `data.table`

```
data <- data.table(yourdataframe)
bar <- data[,.N,by=y]
foo <- data[x==1 & z==1,list(mean.t=mean(t,na.rm=T),median.t=median(t,na.rm=T)),by=y]
merge(bar[,list(y)],foo,by="y",all.x=T)
y mean.t median.t
1: 1 12.5 12.5
2: 2 NA NA
3: 3 NA NA
4: 4 NA NA
```

You probably could do the same in `aggregate`

, but I am not sure you can do it in one easy step.

An answer to to an additional request in the comments…

```
bar <- data.table(expand.grid(y=unique(data$y),z=unique(data[z %in% c(1,2,3,4),z])))
foo <- data[x==1 & z %in% c(1,2,3,4),list(
mean.t=mean(t,na.rm=T),
median.t=median(t,na.rm=T),
Q25.t=quantile(t,0.25,na.rm=T),
Q75.t=quantile(t,0.75,na.rm=T)
),by=list(y,z)]
merge(bar[,list(y,z)],foo,by=c("y","z"),all.x=T)
y z mean.t median.t Q25.t Q75.t
1: 1 1 12.5 12.5 11.25 13.75
2: 1 2 NA NA NA NA
3: 1 3 NA NA NA NA
4: 1 4 NA NA NA NA
5: 2 1 NA NA NA NA
6: 2 2 NA NA NA NA
7: 2 3 NA NA NA NA
8: 2 4 NA NA NA NA
9: 3 1 NA NA NA NA
10: 3 2 NA NA NA NA
11: 3 3 NA NA NA NA
12: 3 4 NA NA NA NA
13: 4 1 NA NA NA NA
14: 4 2 18.0 18.0 18.00 18.00
15: 4 3 45.0 45.0 45.00 45.00
16: 4 4 NA NA NA NA
```

11

solved Multiple aggregation in R with 4 parameters