I would like to set the first and the last value in a group to NA. Here is an example:
我想将组中的第一个和最后一个值设置为NA。这是一个例子:
DT <- data.table(v = rnorm(12), class=rep(1:3, each=4))
DT[, v[c(1,.N)] := NA , by=class]
But this is not working. How can I do it?
但这不起作用。我该怎么做?
4 个解决方案
#1
9
At the moment, the way to go about this would be to first extract the indices, and then do one assignment by reference.
目前,解决这个问题的方法是首先提取索引,然后通过引用进行一次分配。
idx = DT[, .(idx = .I[c(1L, .N)]), by=class]$idx
DT[idx, v := NA]
I'll try and add this example to the Reference semantics vignette.
我将尝试将此示例添加到Reference语义晕影中。
#2
3
This may not be a one-liner, but it does have 'first' and 'last' in the code :)
这可能不是一个单行,但它在代码中确实有“第一个”和“最后一个”:)
> DT <- data.table(v = rnorm(12), class=rep(1:3, each=4))
> setkey(DT, class)
> classes = DT[, .(unique(class))]
> DT[classes, v := NA, mult='first']
> DT[classes, v := NA, mult='last']
> DT
v class
1: NA 1
2: -1.8191 1
3: -0.6355 1
4: NA 1
5: NA 2
6: -1.1771 2
7: -0.8125 2
8: NA 2
9: NA 3
10: 0.2357 3
11: 0.3416 3
12: NA 3
>
Order is also preserved for the non-key columns. I think that is a documented (committed to) feature.
还为非键列保留顺序。我认为这是一个记录(承诺)功能。
#3
1
With a helper function it's easy
使用辅助功能很容易
set.na = function(x,y) {x[y] = NA; x}
DT[, set.na(v,c(1,.N)) , by=class]
#4
0
The canonical way to modify subsets of the data is to use i
to define the subset. You cannot use [
together with :=
. Either create a temporary i
as suggested by @David Arenburg or you can create the outcome vector yourself using a c(NA, v[-c(1, .N)], NA)
construction.
修改数据子集的规范方法是使用i来定义子集。你不能使用[with with:=。根据@David Arenburg的建议创建临时i,或者您可以使用c(NA,v [-c(1,.N)],NA)构造自己创建结果向量。
DT[, v := c(NA, v[-c(1, .N)], NA)[1:.N], by = class]
However, you should also note that the row order can change when you e.g. set a new key or use any number of functions. So you should be very careful with this operation.
但是,您还应该注意,例如,您可以更改行顺序。设置新密钥或使用任意数量的功能。所以你应该非常小心这个操作。
#1
9
At the moment, the way to go about this would be to first extract the indices, and then do one assignment by reference.
目前,解决这个问题的方法是首先提取索引,然后通过引用进行一次分配。
idx = DT[, .(idx = .I[c(1L, .N)]), by=class]$idx
DT[idx, v := NA]
I'll try and add this example to the Reference semantics vignette.
我将尝试将此示例添加到Reference语义晕影中。
#2
3
This may not be a one-liner, but it does have 'first' and 'last' in the code :)
这可能不是一个单行,但它在代码中确实有“第一个”和“最后一个”:)
> DT <- data.table(v = rnorm(12), class=rep(1:3, each=4))
> setkey(DT, class)
> classes = DT[, .(unique(class))]
> DT[classes, v := NA, mult='first']
> DT[classes, v := NA, mult='last']
> DT
v class
1: NA 1
2: -1.8191 1
3: -0.6355 1
4: NA 1
5: NA 2
6: -1.1771 2
7: -0.8125 2
8: NA 2
9: NA 3
10: 0.2357 3
11: 0.3416 3
12: NA 3
>
Order is also preserved for the non-key columns. I think that is a documented (committed to) feature.
还为非键列保留顺序。我认为这是一个记录(承诺)功能。
#3
1
With a helper function it's easy
使用辅助功能很容易
set.na = function(x,y) {x[y] = NA; x}
DT[, set.na(v,c(1,.N)) , by=class]
#4
0
The canonical way to modify subsets of the data is to use i
to define the subset. You cannot use [
together with :=
. Either create a temporary i
as suggested by @David Arenburg or you can create the outcome vector yourself using a c(NA, v[-c(1, .N)], NA)
construction.
修改数据子集的规范方法是使用i来定义子集。你不能使用[with with:=。根据@David Arenburg的建议创建临时i,或者您可以使用c(NA,v [-c(1,.N)],NA)构造自己创建结果向量。
DT[, v := c(NA, v[-c(1, .N)], NA)[1:.N], by = class]
However, you should also note that the row order can change when you e.g. set a new key or use any number of functions. So you should be very careful with this operation.
但是,您还应该注意,例如,您可以更改行顺序。设置新密钥或使用任意数量的功能。所以你应该非常小心这个操作。