replace NA in a dplyr chain - r

Question

Question has been edited from the original.
After reading this interesting discussion I was wondering how to replace NAs in a column using dplyr in, for example, the Lahman batting data:
Source: local data frame [96,600 x 3]
Groups: teamID
yearID teamID G_batting
1 2004 SFN 11
2 2006 CHN 43
3 2007 CHA 2
4 2008 BOS 5
5 2009 SEA 3
6 2010 SEA 4
7 2012 NYA NA
The following does not work as I expected
library(dplyr)
library(Lahman)
df <- Batting[ c("yearID", "teamID", "G_batting") ]
df <- group_by(df, teamID )
df$G_batting[is.na(df$G_batting)] <- mean(df$G_batting, na.rm = TRUE)
Source: local data frame [20 x 3]
Groups: yearID, teamID
yearID teamID G_batting
1 2004 SFN 11.00000
2 2006 CHN 43.00000
3 2007 CHA 2.00000
4 2008 BOS 5.00000
5 2009 SEA 3.00000
6 2010 SEA 4.00000
7 2012 NYA **49.07894**
> mean(Batting$G_battin, na.rm = TRUE)
[1] **49.07894**
In fact it imputed the overall mean and not the group mean. How would you do this in a dplyr chain? Using transform from base R also does not work as it imputed the overall mean and not the group mean. Also this approach converts the data to a regular dat. a frame. Is there a better way to do this?
df %.%
group_by( yearID ) %.%
transform(G_batting = ifelse(is.na(G_batting),
mean(G_batting, na.rm = TRUE),
G_batting)
)
Edit: Replacing transform with mutate gives the following error
Error in mutate_impl(.data, named_dots(...), environment()) :
INTEGER() can only be applied to a 'integer', not a 'double'
Edit: Adding as.integer seems to resolve the error and does produce the expected result. See also #eddi's answer.
df %.%
group_by( teamID ) %.%
mutate(G_batting = ifelse(is.na(G_batting), as.integer(mean(G_batting, na.rm = TRUE)), G_batting))
Source: local data frame [96,600 x 3]
Groups: teamID
yearID teamID G_batting
1 2004 SFN 11
2 2006 CHN 43
3 2007 CHA 2
4 2008 BOS 5
5 2009 SEA 3
6 2010 SEA 4
7 2012 NYA 47
> mean_NYA <- mean(filter(df, teamID == "NYA")$G_batting, na.rm = TRUE)
> as.integer(mean_NYA)
[1] 47
Edit: Following up on #Romain's comment I installed dplyr from github:
> head(df,10)
yearID teamID G_batting
1 2004 SFN 11
2 2006 CHN 43
3 2007 CHA 2
4 2008 BOS 5
5 2009 SEA 3
6 2010 SEA 4
7 2012 NYA NA
8 1954 ML1 122
9 1955 ML1 153
10 1956 ML1 153
> df %.%
+ group_by(teamID) %.%
+ mutate(G_batting = ifelse(is.na(G_batting), mean(G_batting, na.rm = TRUE), G_batting))
Source: local data frame [96,600 x 3]
Groups: teamID
yearID teamID G_batting
1 2004 SFN 0
2 2006 CHN 0
3 2007 CHA 0
4 2008 BOS 0
5 2009 SEA 0
6 2010 SEA 1074266112
7 2012 NYA 90693125
8 1954 ML1 122
9 1955 ML1 153
10 1956 ML1 153
.. ... ... ...
So I didn't get the error (good) but I got a (seemingly) strange result.

Answer 1

The main issue you're having is that mean returns a double while the G_batting column is an integer. So wrapping the mean in as.integer would work, or you'd need to convert the entire column to numeric I guess.
That said, here are a couple of data.table alternatives - I didn't check which one is faster.
library(data.table)
# using ifelse
dt = data.table(a = 1:2, b = c(1,2,NA,NA,3,4,5,6,7,8))
dt[, b := ifelse(is.na(b), mean(b, na.rm = T), b), by = a]
# using a temporary column
dt = data.table(a = 1:2, b = c(1,2,NA,NA,3,4,5,6,7,8))
dt[, b.mean := mean(b, na.rm = T), by = a][is.na(b), b := b.mean][, b.mean := NULL]
And this is what I'd want to do ideally (there is an FR about this):
# again, atm this is pure fantasy and will not work
dt[, b[is.na(b)] := mean(b, na.rm = T), by = a]
The dplyr version of the ifelse is (as in OP):
dt %>% group_by(a) %>% mutate(b = ifelse(is.na(b), mean(b, na.rm = T), b))
I'm not sure how to implement the second data.table idea in a single line in dplyr. I'm also not sure how you can stop dplyr from scrambling/ordering the data (aside from creating an index column).