Re: [R] stacking consecutive columns
On Wed, Nov 17, 2010 at 10:37 AM, Graves, Gregory wrote:
> Follows is the exact solution to this:
>
> v <- NULL
>
> #note that decreasing is FALSE so preceding year preceeds
>
> for(i in 2:46) {
> kk <- melt(yearmonth[, c(1, i, i+1)], id.vars="month", variable_name="year")
> v[[i-1]] <- kk[order(kk$year, decreasing=FALSE),]
> }
>
> x <- do.call(cbind, v)
> write.table(x,"clipboard",sep=" ",col.names=NA) #export to Excell via Ctrl-V
>
>
I get an error when I run that code so I am not completely sure
precisely what the order of columns etc. should be but here is a
solution that can be tweaked that uses zoo's ability to use vector
lags which seems reasonably straight forward.
First we create a ts series, s and then cbind it with year and month
giving s3. Then we use zoo's lag function together with the vector of
lags 0, 12, 24, ... Finally use write table to create a csv file and
run the csv file (which under the default set up in Windows launches
Excel and reads it in). If yearmonth is a matrix rather than a data
frame or if you want reversed rows or columns use the appropriate
commented out lines.
# create a ts series, s, and then cbind it with the year and month giving s3
# s <- ts(c(yearmonth[, -seq(2)]), start = c(2001, 1), freq = 12)
s <- ts(unlist(yearmonth[, -seq(2)]), start = c(2001, 1), freq = 12)
s3 <- cbind(year = floor(time(s)), month = cycle(s), s)
# use zoo's lag to create the desired layout
library(zoo)
numyears <- length(s) / 12
# sl <- na.omit(lag(as.zoo(s3), seq((numyears-2)*12, 0, -12)))
sl <- na.omit(lag(as.zoo(s3), seq(0, (numyears-2)*12, 12)))
# time(sl) <- - time(sl)
write.table(sl, file = "out.csv", row.names = FALSE,
col.names = rep(c("year", "month", "value"), length = ncol(sl)), sep = ",")
shell("out.csv")
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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Re: [R] stacking consecutive columns
On Nov 17, 2010, at 16:37 , Graves, Gregory wrote:
> Follows is the exact solution to this:
>
> v <- NULL
>
> #note that decreasing is FALSE so preceding year preceeds
>
> for(i in 2:46) {
> kk <- melt(yearmonth[, c(1, i, i+1)], id.vars="month", variable_name="year")
> v[[i-1]] <- kk[order(kk$year, decreasing=FALSE),]
> }
>
> x <- do.call(cbind, v)
> write.table(x,"clipboard",sep=" ",col.names=NA) #export to Excell via
> Ctrl-V
>
Just wondering, was anything wrong with
y1 <- cbind(yearmonth[,-1], foo = NA)
names(y1) <- names(yearmonth)
rbind(yearmonth, y1)
? (untested, of course)
--
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacking consecutive columns
On Nov 17, 2010, at 10:37 AM, Graves, Gregory wrote:
Follows is the exact solution to this:
v <- NULL
#note that decreasing is FALSE so preceding year preceeds
for(i in 2:46) {
kk <- melt(yearmonth[, c(1, i, i+1)], id.vars="month",
variable_name="year")
v[[i-1]] <- kk[order(kk$year, decreasing=FALSE),]
}
x <- do.call(cbind, v)
write.table(x,"clipboard",sep=" ",col.names=NA) #export to Excell
via Ctrl-V
The usual method of separating items in columns when exporting to
Excel would be with tabs so people using this as a template might also
want to try:
write.table(x,"clipboard",sep="\t",col.names=NA) #export to Excell
via Ctrl-V
--
David.
Gregory A. Graves, Lead Scientist
-Original Message-
From: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de]
Sent: Wednesday, November 17, 2010 9:49 AM
Hi Gregory,
is this what you want? Ok, not the most elegant way...
# using 'melt' from the 'reshape' package
library(reshape)
Data <- data.frame(month = 1:12,
x2002 = runif(12),
x2003 = runif(12),
x2004 = runif(12),
x2005 = runif(12))
v <- NULL
for(i in 2:4) {
kk <- melt(Data[, c(1, i, i+1)], id.vars="month",
variable_name="year")
v[[i-1]] <- kk[order(kk$year, decreasing=TRUE),]
}
out <- do.call(cbind, v)
HTH
Patrick
Am 17.11.2010 15:03, schrieb Graves, Gregory:
I have a file, each column of which is a separate year, and each
row of each column is mean precipitation for that month. Looks
like this (except it goes back to 1964).
monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007
X2008 X2009
11.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675
1.315 2.920
22.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380
2.480 2.380
31.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420
2.845 2.460
43.775 1.350 2.745 0.170 0.710 2.570 0.255 0.425
4.470 1.250
54.050 1.385 5.650 1.515 12.005 6.895 7.020 4.060
7.725 2.775
68.635 8.900 15.715 12.680 16.270 12.605 7.095 7.025
10.465 7.345
75.475 7.955 7.880 6.670 7.955 7.355 5.475 5.650
7.255 7.985
88.435 5.525 7.120 6.250 7.150 7.610 5.525 6.500
6.275 10.405
95.855 7.830 7.250 7.355 9.715 7.850 6.385 7.960
4.485 7.250
10 7.965 11.915 6.735 8.125 7.855 10.465 4.340 6.165
2.400 3.240
11 1.705 1.525 0.905 1.670 1.840 2.100 0.255 2.830
4.425 1.645
12 2.335 0.840 0.795 1.890 0.145 1.700 0.260 2.160
2.300 2.220
What I want to do is to stack 2008 data underneath 2009 data, 2007
under 2008, 2006 under 2007, etc. I have so far figured out that I
can do this with the following clumsy approach:
a=stack(yearmonth,select=c(X2009,X2008))
b=stack(yearmonth,select=c(X2008,X2007))
x=as.data.frame(c(a,b))
write.table(x,"clipboard",sep=" ",col.names=NA) #then paste this
back into Excel to get this
values ind values.1ind.1
1 0.275 X2009 1.285 X2008
2 0.41X2009 3.85X2008
3 1.915 X2009 3.995 X2008
4 1.25X2009 3.845 X2008
5 8.76X2009 2.095 X2008
6 8.65X2009 8.29X2008
7 7.175 X2009 9.405 X2008
8 7.19X2009 13.44 X2008
9 8.13X2009 7.245 X2008
10 1.46X2009 5.645 X2008
11 2.56X2009 0.535 X2008
12 5.01X2009 1.225 X2008
13 1.285 X2008 0.72X2007
14 3.85X2008 1.89X2007
15 3.995 X2008 1.035 X2007
16 3.845 X2008 2.86X2007
17 2.095 X2008 3.785 X2007
18 8.29X2008 9.62X2007
19 9.405 X2008 9.245 X2007
20 13.44 X2008 5.595 X2007
21 7.245 X2008 8.4 X2007
22 5.645 X2008 6.705 X2007
23 0.535 X2008 1.47X2007
24 1.225 X2008 1.665 X2007
Is there an easier, cleaner way to do this? Thanks.
Gregory A. Graves, Lead Scientist.
David Winsemius, MD
West Hartford, CT
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Re: [R] stacking consecutive columns
Follows is the exact solution to this:
v <- NULL
#note that decreasing is FALSE so preceding year preceeds
for(i in 2:46) {
kk <- melt(yearmonth[, c(1, i, i+1)], id.vars="month", variable_name="year")
v[[i-1]] <- kk[order(kk$year, decreasing=FALSE),]
}
x <- do.call(cbind, v)
write.table(x,"clipboard",sep=" ",col.names=NA) #export to Excell via Ctrl-V
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
-Original Message-
From: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de]
Sent: Wednesday, November 17, 2010 9:49 AM
To: r-help@r-project.org; Graves, Gregory
Subject: Re: [R] stacking consecutive columns
Hi Gregory,
is this what you want? Ok, not the most elegant way...
# using 'melt' from the 'reshape' package
library(reshape)
Data <- data.frame(month = 1:12,
x2002 = runif(12),
x2003 = runif(12),
x2004 = runif(12),
x2005 = runif(12))
v <- NULL
for(i in 2:4) {
kk <- melt(Data[, c(1, i, i+1)], id.vars="month", variable_name="year")
v[[i-1]] <- kk[order(kk$year, decreasing=TRUE),]
}
out <- do.call(cbind, v)
HTH
Patrick
Am 17.11.2010 15:03, schrieb Graves, Gregory:
> I have a file, each column of which is a separate year, and each row of each
> column is mean precipitation for that month. Looks like this (except it goes
> back to 1964).
>
> monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008
> X2009
> 11.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315
> 2.920
> 22.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380 2.480
> 2.380
> 31.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420 2.845
> 2.460
> 43.775 1.350 2.745 0.170 0.710 2.570 0.255 0.425 4.470
> 1.250
> 54.050 1.385 5.650 1.515 12.005 6.895 7.020 4.060 7.725
> 2.775
> 68.635 8.900 15.715 12.680 16.270 12.605 7.095 7.025 10.465
> 7.345
> 75.475 7.955 7.880 6.670 7.955 7.355 5.475 5.650 7.255
> 7.985
> 88.435 5.525 7.120 6.250 7.150 7.610 5.525 6.500 6.275
> 10.405
> 95.855 7.830 7.250 7.355 9.715 7.850 6.385 7.960 4.485
> 7.250
> 10 7.965 11.915 6.735 8.125 7.855 10.465 4.340 6.165 2.400
> 3.240
> 11 1.705 1.525 0.905 1.670 1.840 2.100 0.255 2.830 4.425
> 1.645
> 12 2.335 0.840 0.795 1.890 0.145 1.700 0.260 2.160 2.300
> 2.220
>
> What I want to do is to stack 2008 data underneath 2009 data, 2007 under
> 2008, 2006 under 2007, etc. I have so far figured out that I can do this
> with the following clumsy approach:
>
> a=stack(yearmonth,select=c(X2009,X2008))
> b=stack(yearmonth,select=c(X2008,X2007))
> x=as.data.frame(c(a,b))
> write.table(x,"clipboard",sep=" ",col.names=NA) #then paste this back
> into Excel to get this
>
>
> values ind values.1ind.1
> 1 0.275 X2009 1.285 X2008
> 2 0.41X2009 3.85X2008
> 3 1.915 X2009 3.995 X2008
> 4 1.25X2009 3.845 X2008
> 5 8.76X2009 2.095 X2008
> 6 8.65X2009 8.29X2008
> 7 7.175 X2009 9.405 X2008
> 8 7.19X2009 13.44 X2008
> 9 8.13X2009 7.245 X2008
> 101.46X2009 5.645 X2008
> 112.56X2009 0.535 X2008
> 125.01X2009 1.225 X2008
> 131.285 X2008 0.72X2007
> 143.85X2008 1.89X2007
> 153.995 X2008 1.035 X2007
> 163.845 X2008 2.86X2007
> 172.095 X2008 3.785 X2007
> 188.29X2008 9.62X2007
> 199.405 X2008 9.245 X2007
> 2013.44 X2008 5.595 X2007
> 217.245 X2008 8.4 X2007
> 225.645 X2008 6.705 X2007
> 230.535 X2008 1.47X2007
> 241.225 X2008 1.665 X2007
>
>
> Is there an easier, cleaner way to do this? Thanks.
>
> Gregory A. Graves, Lead Scientist
> Everglades REstoration COoordination and VERification (RECOVER)
> Restoration Sciences Department
> South Florida Water Management District
> Phones: DESK: 561 / 682 - 2429
> CELL: 561 / 719 - 8157
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacking consecutive columns
Elegant or not ... This approach worked as intended, the objective being to
create individual 'years' each of which was 24 months long (not 12) such that
each 'year' reflected the antecedent rainfall condition. THANKS.
I needed to do this to help us identify years that were similar to one another
with respect to resultant water-level stage pattern in the Everglades. Thus
the need to examine each year in respect to its preceeding year.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
-Original Message-
From: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de]
Sent: Wednesday, November 17, 2010 9:49 AM
To: r-help@r-project.org; Graves, Gregory
Subject: Re: [R] stacking consecutive columns
Hi Gregory,
is this what you want? Ok, not the most elegant way...
# using 'melt' from the 'reshape' package
library(reshape)
Data <- data.frame(month = 1:12,
x2002 = runif(12),
x2003 = runif(12),
x2004 = runif(12),
x2005 = runif(12))
v <- NULL
for(i in 2:4) {
kk <- melt(Data[, c(1, i, i+1)], id.vars="month", variable_name="year")
v[[i-1]] <- kk[order(kk$year, decreasing=TRUE),]
}
out <- do.call(cbind, v)
HTH
Patrick
Am 17.11.2010 15:03, schrieb Graves, Gregory:
> I have a file, each column of which is a separate year, and each row of each
> column is mean precipitation for that month. Looks like this (except it goes
> back to 1964).
>
> monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008
> X2009
> 11.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315
> 2.920
> 22.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380 2.480
> 2.380
> 31.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420 2.845
> 2.460
> 43.775 1.350 2.745 0.170 0.710 2.570 0.255 0.425 4.470
> 1.250
> 54.050 1.385 5.650 1.515 12.005 6.895 7.020 4.060 7.725
> 2.775
> 68.635 8.900 15.715 12.680 16.270 12.605 7.095 7.025 10.465
> 7.345
> 75.475 7.955 7.880 6.670 7.955 7.355 5.475 5.650 7.255
> 7.985
> 88.435 5.525 7.120 6.250 7.150 7.610 5.525 6.500 6.275
> 10.405
> 95.855 7.830 7.250 7.355 9.715 7.850 6.385 7.960 4.485
> 7.250
> 10 7.965 11.915 6.735 8.125 7.855 10.465 4.340 6.165 2.400
> 3.240
> 11 1.705 1.525 0.905 1.670 1.840 2.100 0.255 2.830 4.425
> 1.645
> 12 2.335 0.840 0.795 1.890 0.145 1.700 0.260 2.160 2.300
> 2.220
>
> What I want to do is to stack 2008 data underneath 2009 data, 2007 under
> 2008, 2006 under 2007, etc. I have so far figured out that I can do this
> with the following clumsy approach:
>
> a=stack(yearmonth,select=c(X2009,X2008))
> b=stack(yearmonth,select=c(X2008,X2007))
> x=as.data.frame(c(a,b))
> write.table(x,"clipboard",sep=" ",col.names=NA) #then paste this back
> into Excel to get this
>
>
> values ind values.1ind.1
> 1 0.275 X2009 1.285 X2008
> 2 0.41X2009 3.85X2008
> 3 1.915 X2009 3.995 X2008
> 4 1.25X2009 3.845 X2008
> 5 8.76X2009 2.095 X2008
> 6 8.65X2009 8.29X2008
> 7 7.175 X2009 9.405 X2008
> 8 7.19X2009 13.44 X2008
> 9 8.13X2009 7.245 X2008
> 101.46X2009 5.645 X2008
> 112.56X2009 0.535 X2008
> 125.01X2009 1.225 X2008
> 131.285 X2008 0.72X2007
> 143.85X2008 1.89X2007
> 153.995 X2008 1.035 X2007
> 163.845 X2008 2.86X2007
> 172.095 X2008 3.785 X2007
> 188.29X2008 9.62X2007
> 199.405 X2008 9.245 X2007
> 2013.44 X2008 5.595 X2007
> 217.245 X2008 8.4 X2007
> 225.645 X2008 6.705 X2007
> 230.535 X2008 1.47X2007
> 241.225 X2008 1.665 X2007
>
>
> Is there an easier, cleaner way to do this? Thanks.
>
> Gregory A. Graves, Lead Scientist
> Everglades REstoration COoordination and VERification (RECOVER)
> Restoration Sciences Department
> South Florida Water Management District
> Phones: DESK: 561 / 682 - 2429
> CELL: 561 / 719 - 8157
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.o
Re: [R] stacking consecutive columns
Hi Gregory,
is this what you want? Ok, not the most elegant way...
# using 'melt' from the 'reshape' package
library(reshape)
Data <- data.frame(month = 1:12,
x2002 = runif(12),
x2003 = runif(12),
x2004 = runif(12),
x2005 = runif(12))
v <- NULL
for(i in 2:4) {
kk <- melt(Data[, c(1, i, i+1)], id.vars="month", variable_name="year")
v[[i-1]] <- kk[order(kk$year, decreasing=TRUE),]
}
out <- do.call(cbind, v)
HTH
Patrick
Am 17.11.2010 15:03, schrieb Graves, Gregory:
I have a file, each column of which is a separate year, and each row of each
column is mean precipitation for that month. Looks like this (except it goes
back to 1964).
monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009
11.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920
22.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380 2.480 2.380
31.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420 2.845 2.460
43.775 1.350 2.745 0.170 0.710 2.570 0.255 0.425 4.470 1.250
54.050 1.385 5.650 1.515 12.005 6.895 7.020 4.060 7.725 2.775
68.635 8.900 15.715 12.680 16.270 12.605 7.095 7.025 10.465 7.345
75.475 7.955 7.880 6.670 7.955 7.355 5.475 5.650 7.255 7.985
88.435 5.525 7.120 6.250 7.150 7.610 5.525 6.500 6.275 10.405
95.855 7.830 7.250 7.355 9.715 7.850 6.385 7.960 4.485 7.250
10 7.965 11.915 6.735 8.125 7.855 10.465 4.340 6.165 2.400 3.240
11 1.705 1.525 0.905 1.670 1.840 2.100 0.255 2.830 4.425 1.645
12 2.335 0.840 0.795 1.890 0.145 1.700 0.260 2.160 2.300 2.220
What I want to do is to stack 2008 data underneath 2009 data, 2007 under 2008,
2006 under 2007, etc. I have so far figured out that I can do this with the
following clumsy approach:
a=stack(yearmonth,select=c(X2009,X2008))
b=stack(yearmonth,select=c(X2008,X2007))
x=as.data.frame(c(a,b))
write.table(x,"clipboard",sep=" ",col.names=NA) #then paste this back into
Excel to get this
values ind values.1ind.1
1 0.275 X2009 1.285 X2008
2 0.41X2009 3.85X2008
3 1.915 X2009 3.995 X2008
4 1.25X2009 3.845 X2008
5 8.76X2009 2.095 X2008
6 8.65X2009 8.29X2008
7 7.175 X2009 9.405 X2008
8 7.19X2009 13.44 X2008
9 8.13X2009 7.245 X2008
10 1.46X2009 5.645 X2008
11 2.56X2009 0.535 X2008
12 5.01X2009 1.225 X2008
13 1.285 X2008 0.72X2007
14 3.85X2008 1.89X2007
15 3.995 X2008 1.035 X2007
16 3.845 X2008 2.86X2007
17 2.095 X2008 3.785 X2007
18 8.29X2008 9.62X2007
19 9.405 X2008 9.245 X2007
20 13.44 X2008 5.595 X2007
21 7.245 X2008 8.4 X2007
22 5.645 X2008 6.705 X2007
23 0.535 X2008 1.47X2007
24 1.225 X2008 1.665 X2007
Is there an easier, cleaner way to do this? Thanks.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacking consecutive columns
Dear Gregory,
Is there an easier, cleaner way to do this? Thanks.
There are of course several ways...
(assuming yearmonth to be a data.frame)
--- 1 ---
year <- colnames (yearmonth) [-1]
year <- gsub ("^[^[:digit:]]*([[:digit:]]*[^[:digit:]]*$)", "\\1", year)
year <- as.numeric (year)
month <- yearmonth$month
precip <- as.matrix (yearmonth [, -1])
long.df <- data.frame (month = rep (month, length (year)),
year = rep (year, each = nrow (yearmonth)),
precipitation = as.numeric (precip))
If you're about to do this more often:
--- 2 ---
package hyperSpec (outdated on CRAN, if you want to install it use the version
on rforge)
has a function array2df which helps with this transformation:
long.df <- array2df (precip, label.x = "precipitation",
levels = list (month = month, year = year)
--- 3 ---
depending on your file (are the column names numbers without the Xs?)
you may be able to abuse a hyperSpec object to read your data easily:
x <- read.txt.wide ("filename", ...more options...)
then
as.long.df (x)
is about what you want.
(You'd probably want to rename the columns)
HTH Claudia
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Claudia Beleites
Dipartimento dei Materiali e delle Risorse Naturali
Università degli Studi di Trieste
Via Alfonso Valerio 6/a
I-34127 Trieste
phone: +39 0 40 5 58-37 68
email: cbelei...@units.it
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Re: [R] stacking consecutive columns
one approach is the following: # say 'Data' is your data frame DataNew <- reshape(Data, direction = "long", varying = list(2:length(Data))) DataNew$year <- rep(2000:2009, each = 12) DataNew I hope it helps. Best, Dimitris On 11/17/2010 3:03 PM, Graves, Gregory wrote: I have a file, each column of which is a separate year, and each row of each column is mean precipitation for that month. Looks like this (except it goes back to 1964). monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009 11.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920 22.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380 2.480 2.380 31.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420 2.845 2.460 43.775 1.350 2.745 0.170 0.710 2.570 0.255 0.425 4.470 1.250 54.050 1.385 5.650 1.515 12.005 6.895 7.020 4.060 7.725 2.775 68.635 8.900 15.715 12.680 16.270 12.605 7.095 7.025 10.465 7.345 75.475 7.955 7.880 6.670 7.955 7.355 5.475 5.650 7.255 7.985 88.435 5.525 7.120 6.250 7.150 7.610 5.525 6.500 6.275 10.405 95.855 7.830 7.250 7.355 9.715 7.850 6.385 7.960 4.485 7.250 10 7.965 11.915 6.735 8.125 7.855 10.465 4.340 6.165 2.400 3.240 11 1.705 1.525 0.905 1.670 1.840 2.100 0.255 2.830 4.425 1.645 12 2.335 0.840 0.795 1.890 0.145 1.700 0.260 2.160 2.300 2.220 What I want to do is to stack 2008 data underneath 2009 data, 2007 under 2008, 2006 under 2007, etc. I have so far figured out that I can do this with the following clumsy approach: a=stack(yearmonth,select=c(X2009,X2008)) b=stack(yearmonth,select=c(X2008,X2007)) x=as.data.frame(c(a,b)) write.table(x,"clipboard",sep=" ",col.names=NA) #then paste this back into Excel to get this values ind values.1ind.1 1 0.275 X2009 1.285 X2008 2 0.41X2009 3.85X2008 3 1.915 X2009 3.995 X2008 4 1.25X2009 3.845 X2008 5 8.76X2009 2.095 X2008 6 8.65X2009 8.29X2008 7 7.175 X2009 9.405 X2008 8 7.19X2009 13.44 X2008 9 8.13X2009 7.245 X2008 10 1.46X2009 5.645 X2008 11 2.56X2009 0.535 X2008 12 5.01X2009 1.225 X2008 13 1.285 X2008 0.72X2007 14 3.85X2008 1.89X2007 15 3.995 X2008 1.035 X2007 16 3.845 X2008 2.86X2007 17 2.095 X2008 3.785 X2007 18 8.29X2008 9.62X2007 19 9.405 X2008 9.245 X2007 20 13.44 X2008 5.595 X2007 21 7.245 X2008 8.4 X2007 22 5.645 X2008 6.705 X2007 23 0.535 X2008 1.47X2007 24 1.225 X2008 1.665 X2007 Is there an easier, cleaner way to do this? Thanks. Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 Web: http://www.erasmusmc.nl/biostatistiek/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
