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My data table looks like:

head(data)
    Date         AI   AGI      ADI   ASI   ARI   ERI   NVRI  SRI   FRI  IRI
1: 1991-09-06    NA 2094.19    NA    NA    NA    NA    NA    NA    NA    NA
2: 1991-09-13    NA 2204.94    NA    NA    NA    NA    NA    NA    NA    NA
3: 1991-09-20    NA 2339.10    NA    NA    NA    NA    NA    NA    NA    NA
4: 1991-09-27    NA 2387.81    NA    NA    NA    NA    NA    NA    NA    NA
5: 1991-10-04    NA 2459.94    NA    NA    NA    NA    NA    NA    NA    NA
6: 1991-10-11    NA 2571.07    NA    NA    NA    NA    NA    NA    NA    NA

Don't worry about the NAs. What I want to do is make a "percentage change" column for each of the columns apart from date.

What I've done so far is:

names_no_date <- unique(names(data))[!unique(names(data)) %in% "Date"]

for (i in names_no_date){
      data_ch <- data[, paste0(i, "ch") := i/shift(i, n = 1, type = "lag")-1]}

I get the error: 

Error in i/shift(i, n = 1, type = "lag") : 
  non-numeric argument to binary operator

I'm wondering how I get around this error?

Jaap
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Gin_Salmon
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  • please be more clear, what do you mean by "percentage change" -you mean a growth function? – Cyrus Mohammadian Sep 14 '16 at 02:21
  • You 'get around it' by only passing 'numerics' into a 'binary operator'. Your `i / shift(i, ...)` is expecting numeric values as both the numerator & denominator. Maybe you need to worry about the `NA`s? And make sure you're not `lag`ging to a non-existent row. – SymbolixAU Sep 14 '16 at 02:28
  • If you're using data.table, use data.table syntax. To make a new data.table, `dt[, lapply(.SD, function(i){i / shift(i)}), .SDcols = -1]`. To add on to the existing one; `dt[, paste0(names(dt)[-1], 'ch') := lapply(.SD, function(i){i / shift(i)}), .SDcols = -1]`. And if you must use `for` loops, preallocate an object of the appropriate size. – alistaire Sep 14 '16 at 04:16

1 Answers1

4

i is a string, so you are trying to divide a string in i/shift(i, n = 1, type = "lag"):

> "AI"/NA
Error in "AI"/NA : non-numeric argument to binary operator

Instead, do

for (i in names_no_date){
      data[, paste0(i, "ch") := get(i)/shift(get(i), n = 1, type = "lag")-1]
}

Also see Referring to data.table columns by names saved in variables.

Edit: @Frank writes in the comments that a more concise way to produce OP's output is

data[, paste0(names_no_date, "_pch") := .SD/shift(.SD) - 1, .SDcols=names_no_date]
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Weihuang Wong
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  • Thanks, that's excellent! I read the ?get help file, but i couldn't understand what it does. Do you mind explaining what get() does? – Gin_Salmon Sep 14 '16 at 03:34
  • `get("x")` returns the object with the name "x" in the environment where `get` was called. For example, try `x <- rnorm(5); get("x")`. – Weihuang Wong Sep 14 '16 at 03:40
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    ...that's how you `get` around this error `*ducks*` – Weihuang Wong Sep 14 '16 at 03:43
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    You can also do it in one step, I guess: `data[, paste0(names_no_date, "_pch") := .SD/shift(.SD) - 1, .SDcols=names_no_date]` – Frank Sep 14 '16 at 11:03
  • Hi @Frank, I'm curious, what does the .SD mean? – Gin_Salmon Sep 14 '16 at 23:16
  • Sure, it refers to the Subset of Data used in `j` of `DT[i, j]`. You can find details like this in the introductory materials for the package: https://github.com/Rdatatable/data.table/wiki/Getting-started @Gin_Salmon – Frank Sep 15 '16 at 00:06