In R: pass column name as argument and use it in f

This question links to this SO answer except that here I want to use the variable specified as function arg in a mutate_(). It works if I don't make any "calculations" in the mutate_():

data <- 
  data.frame(v1=c(1,2),
             v2=c(3,4))



func1 <- function(df, varname){
    res <-
      df %>%
      mutate_(v3=varname)
    return(res)
  }
func1(data, "v1")

This give the expected:

  v1 v2 v3
1  1  3  1
2  2  4  2

But if I do anything like this, it seems that I have not specified "v3" correctly:

func2 <- function(df, varname){
  res <-
    df %>%
    mutate_(v3=sum(varname))
  return(res)
}
func2(data, "v1")

Does not work; how come it is not equivalent to this outside a function ?:

data %>%
  mutate(v3=sum(v1))

Gives:

  v1 v2 v3
1  1  3  3
2  2  4  3

UPDATE (after @docendo discimus 's solution): The solution about using lazyeval::interp() works. But it seems that Im getting a lot of typing if one have a little more complex function. Eg. I wanted a function that could return score and fisher's 2x2 pvalue for all combinations of N-P in a data frame of counts, c.

require(plyr)
require(dplyr)
require(lazyeval)
set.seed(8)
df <- 
  data.frame(
    N = sample(c("n1","n2","n3","n4"),20, replace=T),
    P = sample(c("p1","p2","p3","p4"),20, replace=T),
    c = round(runif(20,0,10),0)) %>%
  distinct()

So I started making a function test.df using a lot of lines with group_byand mutate. Without lazyeval it does NOT work (of cause), but would look something like this:

test.df <- function(df=NULL, N=NULL, P=NULL, count=NULL, ...){
  require(plyr)
  require(dplyr)


  test <- function(a,b,c,d){
    data <- matrix(c(a,b,c,d),ncol=2)
    c(p = fisher.test(data)$p.value,
      OR = fisher.test(data)$estimate)
  }

  df %>%
    ungroup() %>%
    mutate(n.total = sum(count)) %>% 
    group_by(N) %>%
    mutate(n.N=sum(count)) %>%
    group_by(P) %>%
    mutate(n.P = sum(count)) %>%
    rowwise() %>%
    mutate(score(count/n.N)/(n.P/n.total), #simple enrichment score
           p=test(count,n.N-count,n.P-count,n.total-n.N-n.P+2*count)[[1]], #p values
           OR=test(count,n.N-count,n.P-count,n.total-n.N-n.P+2*count)[[2]]) #Odds ratio
    ungroup() %>%
    mutate(p_adj=p.adjust(p, method="BH"))

}

Then I turned to the lazyval-way, and it works!:

test.df <- function(df=NULL, N=NULL, P=NULL, count=NULL, ...){
  require(plyr)
  require(dplyr)
  require(lazyeval)

  test <- function(a,b,c,d){
    data <- matrix(c(a,b,c,d),ncol=2)
    c(p = fisher.test(data)$p.value,
      OR = fisher.test(data)$estimate)
  }

  df %>%
    ungroup() %>%
    mutate_(n.total = interp(~sum(count), count=as.name(count))) %>% 
    group_by_(interp(~N, N=as.name(N))) %>%
    mutate_(n.N = interp(~sum(count), count=as.name(count))) %>%
    group_by_(interp(~P, P=as.name(P))) %>%
    mutate_(n.P = interp(~sum(count), count=as.name(count))) %>%
    rowwise() %>%
    mutate_(score=interp(~(count/n.N)/(n.P/n.total), 
                       .values=list(count=as.name(count),
                                    n.N=quote(n.N),
                                    n.P=quote(n.P),
                                    n.total=quote(n.total))),
            p=interp(~(test(count,n.N-count,n.P-count,n.total-n.N-n.P+2*count)[[1]]),
                     .values=list(fisher=quote(fisher),
                                  count=as.name(count),
                                  n.N=quote(n.N),
                                  n.P=quote(n.P),
                                  n.total=quote(n.total))),
            OR=interp(~(test(count,n.N-count,n.P-count,n.total-n.N-n.P+2*count)[[2]]),
                      .values=list(fisher=quote(fisher),
                                   count=as.name(count),
                                   n.N=quote(n.N),
                                   n.P=quote(n.P),
                                   n.total=quote(n.total)))) %>% 
    ungroup() %>%
    mutate_(p_adj=interp(~p.adjust(p, method="BH"),
                         .values=list(p.adjust=quote(p.adjust),
                                      p=quote(p))))

}

Gives:

    N  P  c n.total n.N n.P     score            p         OR       p_adj
1  n2 p1  9      89  23  27 1.2898551 1.856249e-01  2.0197105 0.309374904
2  n1 p2  3      89  21  16 0.7946429 1.000000e+00  0.7458441 1.000000000
3  n4 p3  5      89  20  30 0.7416667 5.917559e-01  0.6561651 0.724442095
4  n3 p1  9      89  25  27 1.1866667 3.053538e-01  1.7087545 0.469775140
5  n2 p3  3      89  23  30 0.3869565 2.237379e-02  0.2365142 0.074579284
6  n3 p4  3      89  25  16 0.6675000 5.428536e-01  0.5696359 0.723804744
7  n2 p1  5      89  23  27 0.7165862 4.412042e-01  0.6216888 0.630291707
8  n4 p3  2      89  20  30 0.2966667 1.503170e-02  0.1733288 0.060126805
9  n4 p3 10      89  20  30 1.4833333 5.406588e-02  2.9136831 0.108131750
10 n3 p4  1      89  25  16 0.2225000 3.524192e-02  0.1410289 0.091433058
11 n2 p1  1      89  23  27 0.1433172 1.312078e-03  0.0731707 0.008747184
12 n1 p3  1      89  21  30 0.1412698 1.168232e-03  0.0704372 0.008747184
13 n2 p4  1      89  23  16 0.2418478 6.108872e-02  0.1598541 0.111070394
14 n3 p1  3      89  25  27 0.3955556 3.793658e-02  0.2475844 0.091433058
15 n1 p2 10      89  21  16 2.6488095 8.710747e-05 10.5125558 0.001742149
16 n4 p2  3      89  20  16 0.8343750 1.000000e+00  0.8027796 1.000000000
17 n1 p4  7      89  21  16 1.8541667 4.114488e-02  3.6049777 0.091433058
18 n2 p4  4      89  23  16 0.9673913 1.000000e+00  1.0173534 1.000000000
19 n2 p2  0      89  23  16 0.0000000 9.115366e-03  0.0000000 0.045576831
20 n3 p3  9      89  25  30 1.0680000 6.157758e-01  1.3880504 0.724442095

Am I not using lazyeval appropriately, or maybe building the function in a stupid way ? Some input is really appreciated here.

回答1:

You have to use lazy evaluation (with the package lazyeval), for example like this:

library(lazyeval)
func2 <- function(df, varname){
     df %>%
       mutate_(v3=interp(~sum(x), x = as.name(varname)))
}
func2(data, "v1")
#  v1 v2 v3
#1  1  3  3
#2  2  4  3

回答2:

With the devel version of dplyr (0.5.0) or in the new version (0.6.0 - awaiting release in April 2017), this can be done using slightly different syntax

library(dplyr)
funcN <- function(dat, varname){
 expr <- enquo(varname)
 dat %>%
     mutate(v3 = sum(!!expr))
     #or
     #mutate(v3 = sum(UQ(expr)))

} 

funcN(data, v1)
#  v1 v2 v3
#1  1  3  3
#2  2  4  3

Here, enquo takes the arguments and returns the value as a quosure (similar to substitute in base R) by evaluating the function arguments lazily and inside the summarise, we ask it to unquote (!! or UQ) so that it gets evaluated.