On Mon, 25 Sep 2006, Bruce LaZerte wrote:
# R version 2.3.1 (2006-06-01) Debian Linux testing
# Is the following behaviour a bug, feature or just a lack of
# understanding on my part? I see that this was discussed here
# last March with no apparent resolution.
Reference? It is the third
# R version 2.3.1 (2006-06-01) Debian Linux testing
# Is the following behaviour a bug, feature or just a lack of
# understanding on my part? I see that this was discussed here
# last March with no apparent resolution.
d - as.factor(c(1970-04-04,1970-08-11,1970-10-18))
x - c(9,10,11)
ch -
If you want it to act like a date store it as a Date:
dx - as.Date(c(1970-04-04,1970-08-11,1970-10-18)) ###
x - c(9,10,11)
ch - data.frame(Date=dx,X=x)
dy - as.Date(c(1970-06-04,1970-08-11,1970-08-18)) ###
y - c(109,110,111)
sp - data.frame(Date=dy,Y=y)
merge(ch, sp, all = TRUE)
By the way you
Hello!
I am merging two datasets and I have encountered a problem with sort.
Can someone please point me to my error. Here is the example.
## I have dataframes, first one with factor and second one with factor
## and integer
tmp1 - data.frame(col1 = factor(c(A, A, C, C, 0, 0)))
tmp2 -
If you make the levels the same does that give what you want:
levs - c(LETTERS[1:6], 0)
tmp1 - data.frame(col1 = factor(c(A, A, C, C, 0, 0), levs))
tmp2 - data.frame(col1 = factor(c(C, D, E, F), levs), col2 = 1:4)
merge(tmp2, tmp1, all = TRUE, sort = FALSE)
merge(tmp1, tmp2, all = TRUE, sort =
Gabor Grothendieck wrote:
If you make the levels the same does that give what you want:
levs - c(LETTERS[1:6], 0)
tmp1 - data.frame(col1 = factor(c(A, A, C, C, 0, 0), levs))
tmp2 - data.frame(col1 = factor(c(C, D, E, F), levs), col2 = 1:4)
merge(tmp2, tmp1, all = TRUE, sort = FALSE)
If believe that merge is trying to put first whatever cells that are
nonempty. For example if you instead did
tmp2 - data.frame(col1 = factor(c(C, D, E, F,A), levs),
col2 = 1:5)
tmp2
col1 col2
1C1
2D2
3E3
4F4
5A5
merge(tmp2, tmp1, all.y = TRUE,
I think you will need to reorder it:
out - merge( cbind(tmp1, seq = 1:nrow(tmp1)), tmp2, all.x = TRUE, sort = FALSE)
out[out$seq, -2]
On 3/6/06, Gregor Gorjanc [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
If you make the levels the same does that give what you want:
levs -
Actually we don't need sort = FALSE if we are reordering it anyways:
out - merge( cbind(tmp1, seq = 1:nrow(tmp1)), tmp2, all.x = TRUE)
out[out$seq, -2]
On 3/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
I think you will need to reorder it:
out - merge( cbind(tmp1, seq = 1:nrow(tmp1)),
Gabor and Jean thank you for your time and answers. Gabors approach does
not do what I want (with or without sort). Gabor note that when we merge
data.frame, this new data.frame gets new row.names and we can not be
consistent with sort.
out - merge(cbind(tmp1, seq = 1:nrow(tmp1)), tmp2, all.x =
On 3/6/06, Gregor Gorjanc [EMAIL PROTECTED] wrote:
But I want to get out
A NA
A NA
C 1
C 1
0 NA
0 NA
That's what I get except for the rownames. Be sure to
make the factor levels consistent. I have renamed the data frames
tmp1a and tmp2a to distinguish them from the ones in your
post
Sorry, I mixed up out and outa in the last post. Here it is correctly.
levs - c(LETTERS[1:6], 0)
tmp1a - data.frame(col1 = factor(c(A, A, C, C, 0, 0), levs))
tmp2a - data.frame(col1 = factor(c(C, D, E, F), levs), col2 = 1:4)
out - merge( cbind(tmp1a, seq = 1:nrow(tmp1a)), tmp2a, all.x =
One other idea; one could use match instead of merge:
# tmp1a and tmp2a from below
cbind(tmp1a, tmp2a[match(tmp1a$col1, tmp2a$col1), -1, drop = FALSE])
col1 col2
1A NA
2A NA
3C1
4C1
50 NA
60 NA
This avoids having to muck with reordering of rows and
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