On Oct 28, 2009, at 6:22 PM, Mark W. Miller wrote:
I am guessing that your <NA> is not really an NA_character, but
rather a factor with a level of "<NA>".
See if str(s1) confirms my suspicions.
df1 <- read.table(textConnection(" Firstname Lastname Age
+ 1 Bob Smith 20
+ 2 John Clark NA
+ 3 Andy <NA> 40"), header=T)
levels(df1$Lastname)
[1] "<NA>" "Clark" "Smith"
levels(df1$Lastname)[1] <- "qqqq"
levels(df1$Lastname)
[1] "qqqq" "Clark" "Smith"
df1
Firstname Lastname Age
1 Bob Smith 20
2 John Clark NA
3 Andy qqqq 40
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
Dr. Winsemius, I just tried your suggestion of typing str(s1).
Below is the
code and the comments that R returned:
channel <- odbcConnectExcel('u:/test.xls')
sqlTables(channel)
TABLE_CAT TABLE_SCHEM TABLE_NAME TABLE_TYPE REMARKS
1 u:\\test <NA> Sheet1$ SYSTEM TABLE <NA>
2 u:\\test <NA> Sheet2$ SYSTEM TABLE <NA>
3 u:\\test <NA> Sheet3$ SYSTEM TABLE <NA>
s1 <- sqlFetch(channel, "Sheet1")
odbcClose(channel)
s1
Firstname Lastname Age
1 Bob Smith 20
2 John Clark NA
3 Andy <NA> 40
str(s1)
'data.frame': 3 obs. of 3 variables:
$ Firstname: Factor w/ 3 levels "Andy","Bob","John": 2 3 1
$ Lastname : Factor w/ 2 levels "Clark","Smith": 2 1 NA
My guess was wrong.
$ Age : num 20 NA 40
levels(s1$Lastname)
[1] "Clark" "Smith"
When I used the code:
levels(s1$Lastname)[1] <- "qqqq"
R replaced Clark with qqqq and kept <NA> I guess because <NA> is not
considered a level here:
levels(s1$Lastname)[1] <- "qqqq"
When I did that on the dataset I already had examined "df1". I knew
that <NA> was a level. In your case is was not a named level. So you
should not have expected the same outcome as with my efforts.
I am unable to explain why your methods did not work with that
structure:
This seems to work:
> df1$Lastname <- ifelse(is.na(df1$Lastname), "qqqq", df1$Lastname)
> df1
Firstname Lastname Age
1 Bob 3 20
2 John 2 NA
3 Andy qqqq 40
s1
Firstname Lastname Age
1 Bob Smith 20
2 John qqqq NA
3 Andy <NA> 40
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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