R Refer to (part of) data frame using string in R

Question

I have a large data set in which I have to search for specific codes depending on what i want. For example, chemotherapy is coded by ~40 codes, that can appear in any of 40 columns called (diag1, diag2, etc).

I am in the process of writing a function that produces plots depending on what I want to show. I thought it would be good to specify what I want to plot in a input data frame. Thus, for example, in case I only want to plot chemotherapy events for patients, I would have a data frame like this:

Dataframe name: Style
 Name   SearchIn                                codes         PlotAs  PlotColour
 Chemo  data[substr(names(data),1,4)=="diag"]  1,2,3,4,5,6     |       red

I already have a function that searches for codes in specific parts of the data frame and flags the events of interest. What i cannot do, and need your help with, is referring to a data frame (Style$SearchIn[1]) using codes in a data frame as above.

> Style$SearchIn[1]
[1] data[substr(names(data),1,4)=="diag"]
Levels:  data[substr(names(data),1,4)=="diag"]

I thought perhaps get() would work, but I cant get it to work:

> get(Style$SearchIn[1])
Error in get(vars$SearchIn[1]) : invalid first argument
enter code here

or

> get(as.character(Style$SearchIn[1]))
Error in get(as.character(Style$SearchIn[1])) :   
object 'data[substr(names(data),1,5)=="TDIAG"]' not found

Obviously, running data[substr(names(data),1,5)=="TDIAG"] works.

Example:

  library(survival)
  ex <- data.frame(SearchIn="lung[substr(names(lung),1,2) == 'ph']")

  lung[substr(names(lung),1,2) == 'ph'] #works
  get(ex$SearchIn[1]) # does not work

@TimBiegeleisen I disagree, I don't think this question is poorly written. The OP is just wrestling with a design issue, and trying to use data.frames and dynamic logic in inadvisable ways. I think the overall intention of the question is clear. — bgoldst
– bgoldst, Commented May 26, 2015 at 2:42

bgoldst · Accepted Answer · 2015-05-26 02:33:50Z

3

It is not a good idea to store R code in strings and then try to eval them when needed; there are nearly always better solutions for dynamic logic, such as lambdas.

I would recommend using a list to store the plot specification, rather than a data.frame. This would allow you to include a function as one of the list's components which could take the input data and return a subset of it for plotting.

For example:

library(survival);

plotFromSpec <- function(data,spec) {
    filteredData <- spec$filter(data);
    ## ... draw a plot from filteredData and other stuff in spec ...
};

spec <- list(
    Name='Chemo',
    filter=function(data) data[,substr(names(data),1,2)=='ph'],
    Codes=c(1,2,3,4,5,6),
    PlotAs='|',
    PlotColour='red'
);
plotFromSpec(lung,spec);

If you want to store multiple specifications, you could create a list of lists.

answered May 26, 2015 at 2:33

bgoldst

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2 Comments

Luc Over a year ago

This is an interesting idea... The potential problem with this is that I can have dozens of things to include, and it might vary from 1 project to another. I guess i can have a list of each of the items, and combine lists in order to make the complete specs list...

Luc Over a year ago

EDIT (10 mins later) mmmm I am struggling how to use this approach with multiple specs (e.g chemo, surgery, radiotherapy)

score 2 · Accepted Answer · 2015-05-26 03:42:05Z

2

Have you tried using quote()

I'm not entirely sure what you want but maybe you could store the things you're trying to get() like

quote(data[substr(names(data),1,4)=="diag"])

and then use eval()

eval(quote(data[substr(names(data),1,4)=="diag"]), list(data=data))

For example,

dat <- data.frame("diag1"=1:10, "diag2"=1:10, "other"=1:10)
Style <- list(SearchIn=c(quote(data[substr(names(data),1,4)=="diag"]), quote("Other stuff")))

> head(eval(Style$SearchIn[[1]], list(data=dat)))
  diag1 diag2
1     1     1
2     2     2
3     3     3
4     4     4
5     5     5
6     6     6

edited May 26, 2015 at 3:42

answered May 26, 2015 at 2:27

user4938532

4 Comments

Luc Over a year ago

Its not 'data[substr(names(data),1,4)=="diag"]' that doesnt work, it is Style$SearchIn[1], that doesnt work. quote(Style$SearchIn[1]) doesn't help me.Thnx

user4938532 Over a year ago

@Luc You would need to quote the data[...] and eval() the Style$SearchIn[1]

Luc Over a year ago

Sorry, I don't fully understand. Can you write it out?

user4938532 Over a year ago

@Luc I added an example

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