Making a pick verb

[Kyle Andrews]

I'd like to a create a new verb for dplyr flollowing this StackOverflow post and am seeking help on getting started.

While my "production code" uses the pick function shown in the question inside of mutate (hard coded to only choose from 4 columns), I'd rather be able to choose from any set of columns. Thinking about the problem some more, I came up with the pipeline below (at the bottom of this post). However, instead of typing that block of code every time, I'd like to turn it into a function. How would I go about it? I'm especially confused the non-standard evaluation of columns.

What I want:

df %>% consolidate(cols = y1:y4, pick = x)

What I have:

df %>%
  mutate(row = row_number()) %>%
  gather(n, y, y1:y4) %>%
  mutate(n = as.integer(str_extract(n, "[0-9]+"))) %>%
  filter(x == n) %>%
  arrange(row) %>%

  select(-c(row, n))

[Hadley Wickham]

The link didn't come through. Can you just explain inline what you're
trying to do?
Hadley

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[Kyle Andrews]

Sorry about that. The link is below:

http://stackoverflow.com/questions/28593265/is-there-a-function-like-switch-which-works-inside-of-dplyrmutate

Basically, I am using a hacked up window function called pick inside of mutate to select meaningful columns of data into another column. It works great, but limits me to four choices and the only way I see to extend it to work for more choices is to expand the ifelse spaghetti.

pick <- function(x, v1, v2, v3, v4) {
    ifelse(x == 1, v1,
           ifelse(x == 2, v2,
                  ifelse(x == 3, v3,
                         ifelse(x == 4, v4, NA))))
}

Later, I came up with a tidyr based workflow which accomplishes largely the same thing, but requires a lot more manipulation.

df %>%
  mutate(row = row_number()) %>%
  gather(n, y, y1:y4) %>%
  mutate(n = as.integer(str_extract(n, "[0-9]+"))) %>%
  filter(x == n) %>%
  arrange(row) %>%
  select(-c(row, n))

Let me start with my original, immediately practical question was: is there a better implementation of pick that works for n columns?

[Hadley Wickham]

Have a look at the discussion at https://github.com/hadley/dplyr/issues/631

Hadley