How to use the spread function properly in tidyr

Question

How do I change the following table from:

Type    Name    Answer     n
TypeA   Apple   Yes        5
TypeA   Apple   No        10
TypeA   Apple   DK         8         


        

Answer 1

Following on the comment from ayk, I'm providing an example. It looks to me like when you have a data_frame with a column of either a factor or character class that has values of NA, this cannot be spread without either removing them or re-classifying the data. This is specific to a data_frame (note the dplyr class with the underscore in the name), as this works in my example when you have values of NA in a data.frame. For example, a slightly modified version of the example above:

Here is the dataframe

library(dplyr)
library(tidyr)
df_1 <- data_frame(Type = c("TypeA", "TypeA", "TypeB", "TypeB"),
                   Answer = c("Yes", "No", NA, "No"),
                   n = 1:4)
df_1

Which gives a data_frame that looks like this

Source: local data frame [4 x 3]

   Type Answer     n
  (chr)  (chr) (int)
1 TypeA    Yes     1
2 TypeA     No     2
3 TypeB     NA     3
4 TypeB     No     4

Then, when we try to tidy it, we get an error message:

df_1 %>% spread(key=Answer, value=n)
Error: All columns must be named

But if we remove the NA's then it 'works':

df_1 %>%
    filter(!is.na(Answer)) %>%
    spread(key=Answer, value=n)
Source: local data frame [2 x 3]

   Type    No   Yes
  (chr) (int) (int)
1 TypeA     2     1
2 TypeB     4    NA

However, removing the NAs may not give you the desired result: i.e. you might want those to be included in your tidied table. You could modify the data directly to change the NAs to a more descriptive value. Alternatively, you could change your data to a data.frame and then it spreads just fine:

as.data.frame(df_1) %>% spread(key=Answer, value=n)
   Type No Yes NA
1 TypeA  2   1 NA
2 TypeB  4  NA  3

Answer 2

I think only tidyr is needed to get from df_1 to df_2.

library(magrittr)
df_1 <- read.csv(text="Type,Name,Answer,n\nTypeA,Apple,Yes,5\nTypeA,Apple,No,10\nTypeA,Apple,DK,8\nTypeA,Apple,NA,20\nTypeA,Orange,Yes,6\nTypeA,Orange,No,11\nTypeA,Orange,DK,8\nTypeA,Orange,NA,23", stringsAsFactors=F)

df_2 <- df_1 %>% 
  tidyr::spread(key=Answer, value=n)

Output:

   Type   Name DK No Yes NA
1 TypeA  Apple  8 10   5 20
2 TypeA Orange  8 11   6 23