opposite of mutate_at dplyr

Question

We know mutate_at function from dplyr allows us to mutate selected multiple columns and apply a function to each of them. I need opposite of it. I mean to say, apply multiple functions to same column or apply same function multiple times to the same column. Take the following reproducible example.

    > main <- structure(list(PolygonId = c(0L, 1L, 1612L, 3L, 2L, 1698L), Area = c(3.018892, 
1.995702, 0.582808, 1.176975, 2.277057, 0.014854), Perimeter = c(10.6415, 
8.6314, 4.8478, 6.1484, 9.2226, 0.6503), h0 = c(1000,500,700,1000,200,1200)), .Names = c("PolygonId", 
"Area", "Perimeter", "h0"), row.names = c(NA, 6L), class = "data.frame")

> main
  PolygonId     Area Perimeter     h0
1         0 3.018892   10.6415   1000
2         1 1.995702    8.6314    500
3      1612 0.582808    4.8478    700
4         3 1.176975    6.1484   1000
5         2 2.277057    9.2226    200
6      1698 0.014854    0.6503   1200

I am only concerned about h0 column in the df main.

Expected outcome:
The h10 field is h0 + 10% of h0 and h_10 is h0 - 10% of h0

  PolygonId     Area Perimeter      h0     h10      h20    h_10   h_20
1         0 3.018892   10.6415    1000    1100     1200    900     800
2         1 1.995702    8.6314     500     550      600    450     400
3      1612 0.582808    4.8478     700     770      840    630     560
4         3 1.176975    6.1484    1000    1100     1200    900     800
5         2 2.277057    9.2226     200     220      240    180     160
6      1698 0.014854    0.6503    1200    1320     1440   1080     960

I'd usually do this::

calcH <- function(h, pc){
h + pc / 100 * h
}

new_main <- mutate ( main,
                     h10 = calcH(h0, 10),
                     h20 = calcH(h0, 20),
                     h_10 = calcH(h0, -10),
                     h_20 = calcH(h0, -20)
                    )

But this is going to be hectic and long code since I have to do this calculation for 1%, 2.5%, 5%, 7.5%, 10%, 12.5%, 15%... 30% in both positive and negative ways.

What is `df`? Please provide reproducible example and expected output — Sotos, Apr 30 '18 at 06:57
@Sotos okay, gimme a few minutes to edit the question. I thought it didnot need reproducible example, since what I was asking was just a concept. — anup, Apr 30 '18 at 06:59
Checkout this SO post, it may be helpful: https://stackoverflow.com/questions/4391592/run-a-function-with-multiple-values-for-more-than-one-arguments-that-are-not-the and this one: https://stackoverflow.com/questions/6827299/r-apply-function-with-multiple-parameters — Amar, Apr 30 '18 at 07:08

score 1 · Answer 1 · answered Apr 30 '18 at 07:40

I like to solve these kind of problems using long data representation:

library(dplyr)
library(tidyr)

# create data frame with join helper and multiplier-values:
bla <- data.frame(mult = seq(-.1, .1, .01), 
                  join = TRUE)

# join, calculate values, create names, transform to wide:
main %>% 
  mutate(join = TRUE) %>% 
  left_join(bla) %>% 
  mutate(h0 = h0*(1+mult),
         mult = sub(x = paste0("h", mult*100), pattern = "-", replacement = "_")) %>% 
  select(-join) %>% 
  spread(mult, h0)

this is the sexiest code I've even seen. It works just fine and exactly the output as I expected. thanks ! — anup, Apr 30 '18 at 08:43

score 1 · Answer 2 · answered Apr 30 '18 at 07:55

1

This is easy in base R. The idea is to create a vector with the required percentages, loop over that vector and calculate your metric, i.e.

v1 <- c(1, seq(2.5, 30, by = 2.5), seq(-30, -2.5, by = 2.5), -1)
sapply(v1, function(i) calcH(main$h0, i))

answered Apr 30 '18 at 07:55

Sotos

51,121
6
32
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it gives the output slightly off-bound as expected. The output obtained in my original df is `num [1:2796, 1:26]`. Now it would be really troublesome to keep track of columns. I did apply colnames and rownames to the output vector but it would be great if I could get the output in the original df itself as new columns. It would be easier to perform further operations for me. – anup Apr 30 '18 at 08:05

andyyy · Accepted Answer · 2018-04-30T08:24:48.493

1

mutate_at can use multiple functions, but they need to exist in the environment as named functions (can't be anonymous functions) So something like

pcts<-rep(c(1,2.5*1:12),2)*c(-1,1)
for(i in pcts){
    assign(gsub("-","_",paste0("h",i)),eval(parse(text=sprintf("function(x) x*(100+%f)/100",i))))    }

main %>% mutate_at(vars(h0),gsub("-","_",paste0("h",pcts)))

would work

edited Apr 30 '18 at 08:24

answered Apr 30 '18 at 08:17

andyyy

986
8
8

this works. but I'm not able to understand the code, the one inside for loop. I understood the gsub() portion but not the `eval` thing. – anup Apr 30 '18 at 08:49
eval(parse(text="some R code")) will evaluate (i.e. run) the R code inside the quotes. assign stores the result of that code in a variable. So assign("h10",eval(parse(text="function(x) x*(100+10)/100" ))) defines a function called h10 that takes x and returns x*(100+10)/100 – andyyy Apr 30 '18 at 15:40
so it is a programmatic way of writing h10<-function(x) x*(100+10)/100 etc for all the different functions that we want to define – andyyy Apr 30 '18 at 15:41
ahh! okay got it. We are creating multiple functions named h1, h2.5, h5 ... etc and called them from the last line with mutate_at. But why is there `sprintf()` and `%f` instead of `i`. Is it the C/C++ way (using printf and sprintf and %d, %f, etc) ? – anup Apr 30 '18 at 16:31
Yes, it is the same as the C++ way. By the way, using assign and eval is generally not viewed as good R programming practice. There is a new package called zeallot which may be of use to create the named functions, but I have not tried it yet. For this problem, I would follow the long format method that Tino uses, but form more general problems, where the functions could be more varied, using mutate_at with a vector of function names is probably best. – andyyy May 01 '18 at 07:29
oh okay. Tino's code responded slower than this, btw. It is excellent though. – anup May 01 '18 at 07:43

score 0 · Answer 4 · answered Aug 13 '18 at 21:31

Here's another approach similar to @andyyy's, but uses rlang instead:

library(dplyr)
library(rlang)

percent <- c(1, 2.5*1:12)

calc_expr <- function(percent_vec){
  parse_exprs(paste(paste0("h0+(",percent_vec,"/100*h0)"), collapse = ";"))
}

main %>%
  mutate(!!!calc_expr (percent), !!!calc_expr (percent*-1)) %>%
  setNames(c(colnames(main), paste0("h", percent), paste0("h_", percent)))

Result:

  PolygonId     Area Perimeter   h0   h1   h2.5   h5   h7.5  h10  h12.5  h15  h17.5  h20  h22.5  h25  h27.5
1         0 3.018892   10.6415 1000 1010 1025.0 1050 1075.0 1100 1125.0 1150 1175.0 1200 1225.0 1250 1275.0
2         1 1.995702    8.6314  500  505  512.5  525  537.5  550  562.5  575  587.5  600  612.5  625  637.5
3      1612 0.582808    4.8478  700  707  717.5  735  752.5  770  787.5  805  822.5  840  857.5  875  892.5
4         3 1.176975    6.1484 1000 1010 1025.0 1050 1075.0 1100 1125.0 1150 1175.0 1200 1225.0 1250 1275.0
5         2 2.277057    9.2226  200  202  205.0  210  215.0  220  225.0  230  235.0  240  245.0  250  255.0
6      1698 0.014854    0.6503 1200 1212 1230.0 1260 1290.0 1320 1350.0 1380 1410.0 1440 1470.0 1500 1530.0
   h30  h_1  h_2.5  h_5  h_7.5 h_10 h_12.5 h_15 h_17.5 h_20 h_22.5 h_25 h_27.5 h_30
1 1300  990  975.0  950  925.0  900  875.0  850  825.0  800  775.0  750  725.0  700
2  650  495  487.5  475  462.5  450  437.5  425  412.5  400  387.5  375  362.5  350
3  910  693  682.5  665  647.5  630  612.5  595  577.5  560  542.5  525  507.5  490
4 1300  990  975.0  950  925.0  900  875.0  850  825.0  800  775.0  750  725.0  700
5  260  198  195.0  190  185.0  180  175.0  170  165.0  160  155.0  150  145.0  140
6 1560 1188 1170.0 1140 1110.0 1080 1050.0 1020  990.0  960  930.0  900  870.0  840

Notes:

Using the vector of percentages, I construct multiple expressions using paste0 and parse_exprs then unquote and splice them in mutate using !!!. Finally, rename the columns using setNames.

opposite of mutate_at dplyr

4 Answers4