Pivot Table "like" structure
On Oct 13, 2012, at 5:38 PM, Bhupendrasinh Thakre wrote:
HI Team,
I am currently working on problem and stumped on "for" loop.
Data:
structure(list(Coutry = structure(c(3L, 3L, 3L, 3L, 2L, 2L, 1L,
1L), .Label = c("J", "M", "U"), class = "factor"), State =
structure(c(1L,
1L, 4L, 2L, 5L, 5L, 3L, 6L), .Label = c("A", "C", "K", "O", "S",
"T"), class = "factor"), City = structure(c(1L, 8L, 7L, 2L, 3L,
6L, 5L, 4L), .Label = c("BEN", "HRD", "JKL", "KK", "KL", "KMM",
"OKC", "TYU"), class = "factor"), Char1 = structure(c(1L, 2L,
1L, 3L, 4L, 2L, 3L, 5L), .Label = c("A", "B", "C", "D", "M"), class
= "factor"),
Char2 = structure(c(1L, 2L, 1L, 2L, 3L, 4L, 4L, 2L), .Label =
c("ABCD",
"EFGH", "FGHJ", "GGGG"), class = "factor"), Char3 = structure(c(1L,
1L, 2L, 3L, 1L, 1L, 2L, 3L), .Label = c("ASDFG", "DDDDD",
"EEEEEE"), class = "factor")), .Names = c("Coutry", "State",
"City", "Char1", "Char2", "Char3"), row.names = c(NA, -8L), class =
"data.frame")
Question:
I am trying to create a pivot table which will count the occurrences
of Char1 : Char4
from the columns Coutry, State, City. I am not sure to use all the
four columns and get something like
structure(list(Group.1 = structure(1:4, .Label = c("ABCD", "EFGH",
"FGHJ", "GGGG"), class = "factor"), x = c(2L, 3L, 1L, 2L)), .Names =
c("Group.1",
"x"), row.names = c(NA, -4L), class = "data.frame")
Code which I tried to use with not best results:
aggregate(State, list(Char2), FUN="count")
You are apparently using attach() on your dataframe. That will often
create confusion. Better to use with():
?with
For your problem without `attach` these are available:
> table(dat$Char2)
ABCD EFGH FGHJ GGGG
2 3 1 2
> aggregate(dat$Char2, dat['Char2'], length)
Char2 x
1 ABCD 2
2 EFGH 3
3 FGHJ 1
4 GGGG 2
With attach() in effect for your dataframe this would have worked as
well:
> aggregate(Char2, list(Char2), length)
Group.1 x
1 ABCD 2
2 EFGH 3
3 FGHJ 1
4 GGGG 2
'count' is not a base R function, although it may be available in some
packages. If you have other packages you are loading, you should name
them.
If you want to get tabulations of all the columns that have "Char" in
their names
> sapply(dat[ grep("Char", names(dat)) ], table)
$Char1
A B C D M
2 2 2 1 1
$Char2
ABCD EFGH FGHJ GGGG
2 3 1 2
$Char3
ASDFG DDDDD EEEEEE
4 2 2
David Winsemius, MD Alameda, CA, USA