mth_mad - 11 months ago 51

R Question

Here is an example that is showing a clear difference between "zoo" and "xts".

`library(xts)`

mydf = as.data.frame(replicate(6, sample(c(1:10), 10, rep = T)))

myzoo = zoo(mydf, order.by = Sys.Date() + 1:10)

resultzoo = sapply(myzoo, function(x) x+1)

Although I am losing the date (which is a behaviour already commented here), the code above works fine. However, the code below gives error

`myxts = xts(mydf, order.by = Sys.Date() + 1:10)`

resultxts = sapply(myxts, function(x) x+1)

# Error in array(r, dim = d, dimnames = if (!(is.null(n1 <- names(x[[1L]])) & :

# length of 'dimnames' [1] not equal to array extent

I cannot find any explanation for this weird behaviour. Any idea is welcome.

Answer Source

I think you have raised a very good question. Before making my answer, I would like to comment that you can use

```
sapply(myzoo, "+", 1)
sapply(myxts, "+", 1)
```

instead of

```
sapply(myzoo, function (x) x + 1)
sapply(myxts, function (x) x + 1)
```

This is because `"+"`

is already a function. Try `1 + 2`

and `"+"(1, 2)`

.

`sapply`

takes two stages. The first stage is an ordinary call to `lapply`

; the second stage is a call to `simplify2array`

for result simplification. The error message you get announces that something wrong happens in the second stage. Indeed, if we try:

```
x1 <- lapply(myzoo, "+", 1)
x2 <- lapply(myxts, "+", 1)
```

we get no error at all!

However, if we compare `x1`

and `x2`

, we will see the difference. For neatness I will just extract the first list element:

```
x1[[1]]
#2016-09-30 2016-10-01 2016-10-02 2016-10-03 2016-10-04 2016-10-05 2016-10-06
# 3 4 5 7 2 2 4
#2016-10-07 2016-10-08 2016-10-09
# 3 5 3
x2[[1]]
# V1
#2016-09-30 3
#2016-10-01 4
#2016-10-02 5
#2016-10-03 7
#2016-10-04 2
#2016-10-05 2
#2016-10-06 4
#2016-10-07 3
#2016-10-08 5
#2016-10-09 3
```

Ah, for "zoo" object, dimension is dropped so we get a vector; while for "xts" object, dimension is not dropped hence we get a single column matrix!

It is exactly for this reason that `sapply`

fails. By default, the simplification option for `sapply`

is `simplify = TRUE`

which always tries to simplify to a 1D vector or a 2D matrix. For `x1`

, this is no problem; but for `x2`

, this is impossible.

If we use a milder setting: `simplify = "array"`

, we will get appropriate behaviour:

`sapply(myzoo, "+", 1, simplify = "array")`

gives a 2D array (i.e., a matrix you see);`sapply(myxts, "+", 1, simplify = "array")`

gives a**3D array**.

From this example, we can see that `sapply`

is not always desirable. Why not use the following:

```
y1 <- do.call(cbind, x1)
y2 <- do.call(cbind, x2)
# V1 V2 V3 V4 V5 V6
#2016-09-30 3 8 6 4 11 3
#2016-10-01 4 3 9 2 5 7
#2016-10-02 5 7 9 7 7 10
#2016-10-03 7 2 5 3 5 3
#2016-10-04 2 6 7 2 4 5
#2016-10-05 2 2 11 2 4 7
#2016-10-06 4 3 10 10 8 2
#2016-10-07 3 6 4 5 9 4
#2016-10-08 5 4 10 10 3 8
#2016-10-09 3 3 11 8 11 7
```

**They give the same output, and you get dates as row names! What is more, the original object class is respected!**

```
class(y1)
# [1] "zoo"
class(y2)
# [1] "xts" "zoo"
```