spread with a data.frame of only two columns key and value about tidyr

[joshualande]

To be clear, the output is:

> df <- data.frame( x = c(1, 2), y = c(3, 4))
> spread(df, x, y) 
   1  2
1  3 NA
2 NA  4

I would have expected the output to have one row:

   1  2
1  3  4

[alexbbrown]

Me too: in the absence of an identifying column it should all be assumed to have the same identifier. This is important in nested (group) processing. Right now it's necessary to ungroup before spread to bring the identifiers into scope.

[kleinschmidt]

I've also just run into this (tidyr 0.3.1). @hadley, can you weigh in on whether this is a bug or expected behavior?

FWIW I was trying to use spread on the output of broom::tidy to get the estimate values into a single row with columns named for term, as in:

library(tidyr)
library(broom)
library(dplyr)

d <- data_frame(x = runif(100, 0, 10),
                y = 5 - x + rnorm(100))

d %>%
  lm(y ~ x, data=.) %>%
  tidy() %>%
  select(term, estimate) %>%
  spread(term, estimate) %>%
  mutate(x_intercept = -`(Intercept)` / x)
#>   (Intercept)         x x_intercept
#> 1    5.148648        NA          NA
#> 2          NA -1.010463          NA

d %>%
  lm(y ~ x, data=.) %>%
  tidy() %>%
  select(term, estimate) %>%
  mutate(dummy='') %>%
  spread(term, estimate) %>%
  mutate(x_intercept = -`(Intercept)` / x)
#>   dummy (Intercept)         x x_intercept
#> 1          5.148648 -1.010463    5.095337

Seems like a perfectly good use case for tidyr + broom so I was surprised.

[hadley]

More illuminating example:

data_frame(
  key = c("a", "b", "c"),
  value = c(1, 2, 3)
) %>%
  spread(key, value)

This is definitely a bug