Spread should work with drop=FALSE and fill=NA and create columns (not rows) from empty factor levels about tidyr HOT 5 CLOSED

Interestingly, if ONE level is missing, but they others are present, it works as expected:

data.frame(U=c("foo","bar")[c(1,1,2,2)],K=factor(letters[c(1:2,1:2)],levels=letters[1:3]),V=c(1:2,1:2)) %>% spread(K,V,drop=FALSE)
    U a b  c
1 bar 1 2 NA
2 foo 1 2 NA

But fails if no data is present:

data.frame(U=c("foo","bar")[c()],K=factor(c(),levels=letters[1:3]),V=numeric()) %>% spread(K,V,drop=FALSE)

Error in `colnames<-`(`*tmp*`, value = c("a", "b", "c")) : 
  'names' attribute [3] must be the same length as the vector [0]

from tidyr.

This the provided example:

data.frame(U=c("foo","bar")[c()],K=factor(c(),levels=letters[1:3]),V=numeric()) %>% spread(K,V,drop=FALSE)

in spread_.data_frame we get a failure to operate because

col_id <- dplyr::id(col, drop = drop)
integer(0)
attr(,"n")
[1] 0

but

col_labels <- split_labels(col, col_id, drop = drop)
  K
1 a
2 b
3 c

Hence in

> dim(ordered) <- c(attr(row_id, "n"), attr(col_id, "n"))
[1] 0,0
> ordered <- as.data.frame.matrix(ordered, stringsAsFactors = FALSE)
> colnames(ordered) <- as.character(col_labels[[1]])
# FAIL!

colnames fails because it's expecting col_labels column count, but in fact has a column count of 0.

This failure to behave in a similar manner in empty cases (a seasonal problem in R code) starts:

in tidyr::id

.variables <- .variables[lengths != 0]

the id function which nominally handles a data frame (for which lengths would be a repeated number), is actually capable of handling a list (for which other situations might occur). It's hard for me to see how that code - filtering out empty columns - would make sense in a data frame.

It continue in tidy::id_var with

    if (length(x) == 0) 
    return(structure(integer(), n = 0L))

which precedes the factor handling code:

if (is.factor(x) && !drop) {
    id <- as.integer(addNA(x, ifany = TRUE))
    n <- length(levels(x))
}

I suggest that the order of these clauses in id_var be fixed to check factor first.

from tidyr.

Gah, wrong issue.

from tidyr.

Here's my understanding of the issue, in code:

df_c <- data_frame(
  x = c("a", "a", "b", "b"),
  y = c("y", "z", "y", "z"),
  z = 1:4
)
df_f <- df_c %>% mutate(x = factor(x, levels = c("b", "a")), y = factor(y))

# Correct: only differ in order
df_c %>% spread(y, z) %>% str()
df_f %>% spread(y, z) %>% str()

# Correct: only see y
df_c[1,] %>% spread(y, z) %>% str()
df_f[1,] %>% spread(y, z) %>% str()

# Correct: expands out both y and z
df_f[1,] %>% spread(y, z, drop = FALSE) %>% str()

# Correct: don't see any values
df_c[0,] %>% spread(y, z) %>% str()

# Incorrect: from the levels of the factor, should have columns a and b
df_f[0,] %>% spread(y, z, drop = FALSE) %>% str()

from tidyr.

I'm pretty sure I correctly identified the underlying problem. Please let me know if I missed anything.

from tidyr.