Re: [R] how to subtotal by rows
Hello Shyam, This is one way to do it jd1 - read.table(text= fid year rice wheat maize 11995 5NA NA 11995 NA3 NA 11995 NA NA 2 11996 4NA NA 11996 NA2 NA 11996 NANA 6 21995 3NA NA 21995 NA8 NA 21995 NANA 4 21996 7NA NA 21996 NA6 NA 21996 NANA 7 , sep=, header=T) jd1 library(plyr) ddply(jd1,.(fid,year),summarise, rice=sum(rice,na.rm=T),wheat=sum(wheat,na.rm=T),maize=sum(maize,na.rm=T)) Good luck Janesh On Fri, Apr 19, 2013 at 10:59 AM, shyam basnet shyamabc2...@yahoo.comwrote: Dear R-users, I have a dataset as like below, and I want to subtotal the values of rice,wheat and maize by year for each fid. fid year rice wheat maize 11995 5NA NA 11995 NA3 NA 11995 NA NA 2 11996 4NA NA 11996 NA2 NA 11996 NANA 6 21995 3NA NA 21995 NA8 NA 21995 NANA 4 21996 7NA NA 21996 NA6 NA 21996 NANA 7--- And, my output should look like below: fid year rice wheat maize 11995 53 2 11996 42 6 21995 38 4 21996 76 7I am looking for some ideas or r-codes on resolving my problem. I appreciate your kind help, Thanks a lot, Sincerely yours, Shyam Nepal [[alternative HTML version deleted]] __ 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. [[alternative HTML version deleted]] __ 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] how to subtotal by rows
You can also use this short command. library(plyr) ddply(jd1,.(fid,year),colSums,na.rm=T) Janesh On Fri, Apr 19, 2013 at 2:30 PM, Janesh Devkota janesh.devk...@gmail.comwrote: Hello Shyam, This is one way to do it jd1 - read.table(text= fid year rice wheat maize 11995 5NA NA 11995 NA3 NA 11995 NA NA 2 11996 4NA NA 11996 NA2 NA 11996 NANA 6 21995 3NA NA 21995 NA8 NA 21995 NANA 4 21996 7NA NA 21996 NA6 NA 21996 NANA 7 , sep=, header=T) jd1 library(plyr) ddply(jd1,.(fid,year),summarise, rice=sum(rice,na.rm=T),wheat=sum(wheat,na.rm=T),maize=sum(maize,na.rm=T)) Good luck Janesh On Fri, Apr 19, 2013 at 10:59 AM, shyam basnet shyamabc2...@yahoo.comwrote: Dear R-users, I have a dataset as like below, and I want to subtotal the values of rice,wheat and maize by year for each fid. fid year rice wheat maize 11995 5NA NA 11995 NA3 NA 11995 NA NA 2 11996 4NA NA 11996 NA2 NA 11996 NANA 6 21995 3NA NA 21995 NA8 NA 21995 NANA 4 21996 7NA NA 21996 NA6 NA 21996 NANA 7--- And, my output should look like below: fid year rice wheat maize 11995 53 2 11996 42 6 21995 38 4 21996 76 7I am looking for some ideas or r-codes on resolving my problem. I appreciate your kind help, Thanks a lot, Sincerely yours, Shyam Nepal [[alternative HTML version deleted]] __ 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. [[alternative HTML version deleted]] __ 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] how to subtotal by rows
Thanks, Arun. I was also trying to come up with that solution. On Fri, Apr 19, 2013 at 2:43 PM, arun smartpink...@yahoo.com wrote: #or ddply(jd1,.(fid,year),numcolwise(sum,na.rm=TRUE)) # fid year rice wheat maize #1 1 19955 3 2 #2 1 19964 2 6 #3 2 19953 8 4 #4 2 19967 6 7 A.K. - Original Message - From: Janesh Devkota janesh.devk...@gmail.com To: shyam basnet shyamabc2...@yahoo.com Cc: r-help@R-project.org r-help@r-project.org Sent: Friday, April 19, 2013 3:30 PM Subject: Re: [R] how to subtotal by rows Hello Shyam, This is one way to do it jd1 - read.table(text= fid year rice wheat maize 11995 5NA NA 11995 NA3 NA 11995 NA NA 2 11996 4NA NA 11996 NA2 NA 11996 NANA 6 21995 3NA NA 21995 NA8 NA 21995 NANA 4 21996 7NA NA 21996 NA6 NA 21996 NANA 7 , sep=, header=T) jd1 library(plyr) ddply(jd1,.(fid,year),summarise, rice=sum(rice,na.rm=T),wheat=sum(wheat,na.rm=T),maize=sum(maize,na.rm=T)) Good luck Janesh On Fri, Apr 19, 2013 at 10:59 AM, shyam basnet shyamabc2...@yahoo.com wrote: Dear R-users, I have a dataset as like below, and I want to subtotal the values of rice,wheat and maize by year for each fid. fid year rice wheat maize 11995 5NA NA 11995 NA3 NA 11995 NA NA 2 11996 4NA NA 11996 NA2 NA 11996 NANA 6 21995 3NA NA 21995 NA8 NA 21995 NANA 4 21996 7NA NA 21996 NA6 NA 21996 NANA 7--- And, my output should look like below: fid year rice wheat maize 11995 53 2 11996 42 6 21995 38 4 21996 76 7I am looking for some ideas or r-codes on resolving my problem. I appreciate your kind help, Thanks a lot, Sincerely yours, Shyam Nepal [[alternative HTML version deleted]] __ 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. [[alternative HTML version deleted]] __ 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. [[alternative HTML version deleted]] __ 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] Spider Plot
Xing, I cannot open the attachment. Can you attach the png file or pdf or any other format ? -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of XINLI LI Sent: Friday, April 19, 2013 3:30 PM To: r-help Subject: [R] Spider Plot Does any one have a sample code for a Spider Plot as attached? Thanks, Xing __ 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] Subsetting a large number into smaller numbers and find the largest product
Hello, I have a big number lets say of around hundred digits. I want to subset that big number into consecutive number of 5 digits and find the product of those 5 digits. For example my first 5 digit number would be 73167. I need to check the product of the individual numbers in 73167 and so on. The sample number is as follows: 73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698747158523863050715693290963295227443043557 I have a problem subsetting the small numbers out of the big number. Any help is highly appreciated. Best Regards, Janesh Devkota [[alternative HTML version deleted]] __ 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] Merge
Hi, I have a quick question here. Lets say he has three data frames and he needs to combine those three data frame using merge. Can we simply use merge to join three data frames ? I remember I had some problem using merge for more than two dataframes. Thanks. On Wed, Apr 17, 2013 at 1:05 AM, Farnoosh farnoosh...@yahoo.com wrote: Thanks a lot:) Sent from my iPad On Apr 16, 2013, at 10:15 PM, arun smartpink...@yahoo.com wrote: Hi Farnoosh, YOu can use either ?merge() or ?join() DataA- read.table(text= ID v1 1 10 2 1 3 22 4 15 5 3 6 6 7 8 ,sep=,header=TRUE) DataB- read.table(text= ID v2 2 yes 5 no 7 yes ,sep=,header=TRUE,stringsAsFactors=FALSE) merge(DataA,DataB,by=ID,all.x=TRUE) # ID v1 v2 #1 1 10 NA #2 2 1 yes #3 3 22 NA #4 4 15 NA #5 5 3 no #6 6 6 NA #7 7 8 yes library(plyr) join(DataA,DataB,by=ID,type=left) # ID v1 v2 #1 1 10 NA #2 2 1 yes #3 3 22 NA #4 4 15 NA #5 5 3 no #6 6 6 NA #7 7 8 yes A.K. From: farnoosh sheikhi farnoosh...@yahoo.com To: smartpink...@yahoo.com smartpink...@yahoo.com Sent: Wednesday, April 17, 2013 12:52 AM Subject: Merge Hi Arun, I want to merge a data set with another data frame with 2 columns and keep the sample size of the DataA. DataA DataB DataCombine ID v1 ID V2 ID v1 v2 1 10 2 yes 1 10 NA 2 1 5 no 2 1 yes 3 22 7 yes 3 22 NA 4 15 4 15 NA 5 3 5 3 no 6 6 6 6 NA 7 8 7 8 yes Thanks a lot for your help and time. __ 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. [[alternative HTML version deleted]] __ 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] Merge
Hello Arun, Thank you so much for the prompt reply. I have one simple question here. DOes three dots (...) in the reduce function means we are applying for three dataframes here ? So, if we were to combine four would that dots be four dots ? Thanks. On Wed, Apr 17, 2013 at 12:16 PM, arun smartpink...@yahoo.com wrote: HI Janesh, YOu can use: library(plyr) ?join_all() #From the help page: dfs - list( a = data.frame(x = 1:10, a = runif(10)), b = data.frame(x = 1:10, b = runif(10)), c = data.frame(x = 1:10, c = runif(10)) ) join_all(dfs) join_all(dfs, x) join_all(dfs, x) #x a b c #1 1 0.7113766 0.1348978 0.1153703 #2 2 0.2520057 0.7249154 0.2362936 #3 3 0.5670157 0.8166805 0.3049683 #4 4 0.7441726 0.4929165 0.6779029 #5 5 0.5616914 0.5272339 0.6202915 #6 6 0.2858429 0.1203205 0.8399356 #7 7 0.9910520 0.1251815 0.4729418 #8 8 0.7079778 0.5465055 0.8951371 #9 9 0.0564100 0.1837211 0.6451289 #10 10 0.7169663 0.1328287 0.2467554 Reduce(function(...) merge(...,by=x),dfs) #x a b c #1 1 0.7113766 0.1348978 0.1153703 #2 2 0.2520057 0.7249154 0.2362936 #3 3 0.5670157 0.8166805 0.3049683 #4 4 0.7441726 0.4929165 0.6779029 #5 5 0.5616914 0.5272339 0.6202915 #6 6 0.2858429 0.1203205 0.8399356 #7 7 0.9910520 0.1251815 0.4729418 #8 8 0.7079778 0.5465055 0.8951371 #9 9 0.0564100 0.1837211 0.6451289 #10 10 0.7169663 0.1328287 0.2467554 A.K. From: Janesh Devkota janesh.devk...@gmail.com To: Farnoosh farnoosh...@yahoo.com Cc: arun smartpink...@yahoo.com; R help r-help@r-project.org Sent: Wednesday, April 17, 2013 1:05 PM Subject: Re: [R] Merge Hi, I have a quick question here. Lets say he has three data frames and he needs to combine those three data frame using merge. Can we simply use merge to join three data frames ? I remember I had some problem using merge for more than two dataframes. Thanks. On Wed, Apr 17, 2013 at 1:05 AM, Farnoosh farnoosh...@yahoo.com wrote: Thanks a lot:) Sent from my iPad On Apr 16, 2013, at 10:15 PM, arun smartpink...@yahoo.com wrote: Hi Farnoosh, YOu can use either ?merge() or ?join() DataA- read.table(text= ID v1 1 10 2 1 3 22 4 15 5 3 6 6 7 8 ,sep=,header=TRUE) DataB- read.table(text= ID v2 2 yes 5 no 7 yes ,sep=,header=TRUE,stringsAsFactors=FALSE) merge(DataA,DataB,by=ID,all.x=TRUE) # ID v1 v2 #1 1 10 NA #2 2 1 yes #3 3 22 NA #4 4 15 NA #5 5 3 no #6 6 6 NA #7 7 8 yes library(plyr) join(DataA,DataB,by=ID,type=left) # ID v1 v2 #1 1 10 NA #2 2 1 yes #3 3 22 NA #4 4 15 NA #5 5 3 no #6 6 6 NA #7 7 8 yes A.K. From: farnoosh sheikhi farnoosh...@yahoo.com To: smartpink...@yahoo.com smartpink...@yahoo.com Sent: Wednesday, April 17, 2013 12:52 AM Subject: Merge Hi Arun, I want to merge a data set with another data frame with 2 columns and keep the sample size of the DataA. DataA DataB DataCombine ID v1 ID V2 ID v1 v2 1 10 2 yes 1 10 NA 2 1 5 no 2 1 yes 3 22 7 yes 3 22 NA 4 15 4 15 NA 5 3 5 3 no 6 6 6 6 NA 7 8 7 8 yes Thanks a lot for your help and time. __ 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. [[alternative HTML version deleted]] __ 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] Change the default resolution for plotting figures?
I have been using the following so far without having any problems: dev.copy(png,sample.png,width=8, height=10, units=in,res=500) dev.off() On Tue, Apr 16, 2013 at 6:32 PM, jt...@mappi.helsinki.fi wrote: Hi, I want to save a plot in the windows device as png and the default resolution is 72dpi. Is it possible to increase the default resolution to for example 300 dpi? I have thought of using function png(..., res=300), but the problem is that the figure produced this way looks different than the one shown in the windows device. One notable difference is the missing of some ticks in the x axis. Therefore I would rather to produce the figure in a window device and then save it as a png. Unfortunately in the device window there is no such an option to change the resolution. Little information can be found so far. Any ideas are appreciated! Best, Jing __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ 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] Read the data from a text file and reshape the data
Hi Arun,
Thank you so much for your answer. It surely does help. Having to know
different approaches for the same problem has given me more insights on R
language. I appreciate your time and effort.
Best,
Janesh Devkota
On Thu, Apr 11, 2013 at 10:00 PM, arun smartpink...@yahoo.com wrote:
Hi,
May be this helps:
lines1- readLines(WAT_DEP.DAT.part)
indx- which(grepl([*],lines1))
indx2-indx[seq(from=indx[2],length(indx),by=2)]+1
lines2-str_trim(lines1[indx2],side=left)
dat1-read.table(text=lines2,sep=,header=FALSE)
library(stringr)
lst1- lapply(split(indx,((seq_along(indx)-1)%/%2)+1), function(x)
{x1-seq(max(x)+2,max(x)+2+49,by=1); x2-
str_trim(lines1[x1][!is.na(lines1[x1])],side=left);
x3-as.vector(as.matrix(read.table(text=x2[x2!=],header=FALSE,sep=)));
x4- if(length(x3) 500) c(x3,rep(NA,500-length(x3))) else x3;x4 })
dat2- as.data.frame(do.call(cbind,lst1))
colnames(dat2)-paste(t,colnames(dat2),_,dat1[,2],sep=)
dim(dat2)
#[1] 500 622
dat2[1:3,1:3]
# t1_0.00208 t2_0.00417 t3_0.00625
#1 3.224 4.124 4.502
#2 3.205 4.118 4.500
#3 3.189 4.114 4.498
A.K.
- Original Message -
From: Janesh Devkota janesh.devk...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Thursday, April 11, 2013 5:55 PM
Subject: [R] Read the data from a text file and reshape the data
I have a data set for different time intervals. The data has three comment
lines before data for each time interval. For each time interval there are
500 data points. I want to change the dataset such that I have the
following
format:
t1t2t3
0.00208 0.00417 0.00625 .
a1 a2 a3 ...
b1 b2 b3 ...
c1c2c3 .
...
The link to the file is as follows:
https://www.dropbox.com/s/hc8n3qcai1mlxca/WAT_DEP.DAT
As you will see on the file, time for each interval is the second data of
the third line before the data starts. For the first time, t= 0.00208. I
need to change the data in several rows into one column. At last I need to
create a dataframe with the format shown above. In the sample above, a1,
b1,
c1 are the data for time t1, and so on.
The sample data is as follows:
** N:SNAPSHOTTIME DELT[S]
** WATER DEPTH [M]: (HP(L),L=2,LA)
18000.00208 0.1
3.224 3.221 3.220 3.217 3.216 3.214 3.212
3.210 3.209 3.207
3.205 3.203 3.202 3.200 3.199 3.197 3.196
3.193 3.192 3.190
3.189 3.187 3.186 3.184 3.184 3.182 3.181
3.179 3.178 3.176
3.175 3.174 3.173 3.171 3.170 3.169 3.168
3.167 3.166 3.164
3.164 3.162 3.162 3.160 3.160 3.158 3.158
3.156 3.156 3.155
3.154 3.153 3.152 3.151 3.150 3.150 3.149
3.149 3.147 3.147
3.146 3.146 3.145 3.145 3.144 3.144 3.143
3.143 3.142 3.142
3.141 3.142 3.141 3.141 3.140 3.141 3.140
3.140 3.139 3.140
3.139 3.140 3.139 3.140 3.139 3.140 3.139
3.140 3.139 3.140
3.139 3.140 3.140 3.140 3.140 3.141 3.141
3.142 3.141 3.142
3.142 3.142 3.143 3.143 3.144 3.144 3.145
3.145 3.146 3.146
3.147 3.148 3.149 3.149 3.150 3.150 3.152
3.152 3.153 3.154
3.155 3.156 3.157 3.158 3.159 3.160 3.161
3.162 3.163 3.164
3.165 3.166 3.168 3.169 3.170 3.171 3.173
3.174 3.176 3.176
3.178 3.179 3.181 3.182 3.184 3.185 3.187
3.188 3.190 3.191
3.194 3.195 3.196 3.198 3.199 3.202 3.203
3.205 3.207 3.209
3.210 3.213 3.214 3.217 3.218 3.221 3.222
3.225 3.226 3.229
3.231 3.233 3.235 3.238 3.239 3.242 3.244
3.247 3.248 3.251
3.253 3.256 3.258 3.261 3.263 3.266 3.268
3.271 3.273 3.276
3.278 3.281 3.283 3.286 3.289 3.292 3.294
3.297 3.299 3.303
3.305 3.307 3.311 3.313 3.317 3.319 3.322
3.325 3.328 3.331
3.334 3.337 3.340 3.343 3.347 3.349 3.353
3.356 3.359 3.362
3.366 3.369 3.372 3.375 3.379 3.382 3.386
3.388 3.392 3.395
3.399 3.402 3.406 3.409 3.413
Re: [R] Solving an integral in R gives the error The integral is probably divergent
Hi Peter,
It is supposed to be t^{3/2} instead of t^{0.5}. The code had a typo. Yes, I
figured out that when x =0, the solution is divergent because it is like
dividing something by zero.
Thanks.
Best,
Janesh
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: Friday, April 12, 2013 1:32 PM
To: Thomas Lumley
Cc: Janesh Devkota; groep R-help
Subject: Re: [R] Solving an integral in R gives the error The integral is
probably divergent
But is it supposed to be t^{-3/2} or t^{-0.5}?? The formula has the former
and the code the latter, and the integral is clearly divergent with the
former.
-pd
On Apr 12, 2013, at 04:51 , Thomas Lumley wrote:
I don't get an error message (after I correct the missing line break
after the comment
b- sapply(a, Cfun, upper=1)
b
[1] 1.583458e-54 7.768026e-50 2.317562e-45 4.206260e-41
4.645737e-37
3.123801e-33 1.279358e-29 3.193257e-26 4.860876e-23 [10]
4.516582e-20 2.564400e-17 8.908932e-15 1.896996e-12 2.481084e-10
1.998561e-08 9.946570e-07 3.067751e-05 5.862075e-04 [19]
6.818952e-03 4.297061e-02 0.00e+00 3.175122e-01 3.723022e-01
2.364930e-01 9.144836e-02 2.190878e-02 3.252754e-03 [28]
2.983763e-04 1.685692e-05 5.849602e-07 1.244158e-08 1.619155e-10
1.287603e-12 6.250149e-15 1.850281e-17 3.338241e-20 [37]
3.668412e-23 2.454192e-26 9.991546e-30 2.474577e-33 3.727226e-37
3.413319e-41 1.900112e-45 6.428505e-50 1.321588e-54 [46]
1.650722e-59 1.252524e-64 5.772750e-70 1.615916e-75 2.746972e-81
2.835655e-87 1.777399e-93 6.764271e-100 1.562923e-106 [55]
2.192373e-113 1.866955e-120 9.651205e-128 3.028623e-135 5.769185e-143
6.670835e-151 4.682023e-159 1.994643e-167 5.157808e-176 [64]
8.095084e-185 7.711162e-194 4.458042e-203 1.564139e-212 3.330362e-222
4.302974e-232 3.373500e-242 1.604721e-252 4.631224e-263 [73]
8.108474e-274 8.611898e-285 5.547745e-296 0.00e+00 0.00e+00
0.00e+00 0.00e+00 0.00e+00 0.00e+00 [82]
0.00e+00 0.00e+00 0.00e+00 0.00e+00 0.00e+00
0.00e+00 0.00e+00 0.00e+00 0.00e+00 [91]
0.00e+00 0.00e+00 0.00e+00 0.00e+00 0.00e+00
0.00e+00 0.00e+00 0.00e+00 0.00e+00 [100]
0.00e+00
-thomas
On Tue, Apr 9, 2013 at 3:14 PM, Janesh Devkota
janesh.devk...@gmail.comwrote:
I am trying to solve an integral in R. However, I am getting an error
when I am trying to solve for that integral.
The equation that I am trying to solve is as follows:
$$ C_m = \frac{{abs{x}}e^{2x}}{\pi^{1/2}}\int_0^t
t^{-3/2}e^{-x^2/t-t}dt $$
[image: enter image description here]
The code that I am using is as follows:
a - seq(from=-10, by=0.5,length=100) ## Create a function to compute
integrationCfun - function(XX, upper){ integrand -
function(x)x^(-0.5)*exp((-XX^2/x)-x)
integrated - integrate(integrand, lower=0, upper=upper)$value
(final - abs(XX)*pi^(-0.5)*exp(2*XX)*integrated) }
b- sapply(a, Cfun, upper=1)
The error that I am getting is as follows:
Error in integrate(integrand, lower = 0, upper = upper) :
the integral is probably divergent
Does this mean I cannot solve the integral ?
Any possible ways to fix this problem will be highly appreciated.The
question can be found on
http://stackoverflow.com/questions/15892586/solving-an-integral-in-r-
gives-error-the-integral-is-probably-divergent
also.
Thanks.
[[alternative HTML version deleted]]
__
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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.
--
Thomas Lumley
Professor of Biostatistics
University of Auckland
[[alternative HTML version deleted]]
__
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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.
--
Peter Dalgaard, Professor,
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.
[R] Read the data from a text file and reshape the data
I have a data set for different time intervals. The data has three comment lines before data for each time interval. For each time interval there are 500 data points. I want to change the dataset such that I have the following format: t1t2t3 0.00208 0.00417 0.00625 . a1 a2 a3 ... b1 b2 b3 ... c1c2c3 . ... The link to the file is as follows: https://www.dropbox.com/s/hc8n3qcai1mlxca/WAT_DEP.DAT As you will see on the file, time for each interval is the second data of the third line before the data starts. For the first time, t= 0.00208. I need to change the data in several rows into one column. At last I need to create a dataframe with the format shown above. In the sample above, a1, b1, c1 are the data for time t1, and so on. The sample data is as follows: ** N:SNAPSHOTTIME DELT[S] ** WATER DEPTH [M]: (HP(L),L=2,LA) 18000.00208 0.1 3.224 3.221 3.220 3.217 3.216 3.214 3.212 3.210 3.209 3.207 3.205 3.203 3.202 3.200 3.199 3.197 3.196 3.193 3.192 3.190 3.189 3.187 3.186 3.184 3.184 3.182 3.181 3.179 3.178 3.176 3.175 3.174 3.173 3.171 3.170 3.169 3.168 3.167 3.166 3.164 3.164 3.162 3.162 3.160 3.160 3.158 3.158 3.156 3.156 3.155 3.154 3.153 3.152 3.151 3.150 3.150 3.149 3.149 3.147 3.147 3.146 3.146 3.145 3.145 3.144 3.144 3.143 3.143 3.142 3.142 3.141 3.142 3.141 3.141 3.140 3.141 3.140 3.140 3.139 3.140 3.139 3.140 3.139 3.140 3.139 3.140 3.139 3.140 3.139 3.140 3.139 3.140 3.140 3.140 3.140 3.141 3.141 3.142 3.141 3.142 3.142 3.142 3.143 3.143 3.144 3.144 3.145 3.145 3.146 3.146 3.147 3.148 3.149 3.149 3.150 3.150 3.152 3.152 3.153 3.154 3.155 3.156 3.157 3.158 3.159 3.160 3.161 3.162 3.163 3.164 3.165 3.166 3.168 3.169 3.170 3.171 3.173 3.174 3.176 3.176 3.178 3.179 3.181 3.182 3.184 3.185 3.187 3.188 3.190 3.191 3.194 3.195 3.196 3.198 3.199 3.202 3.203 3.205 3.207 3.209 3.210 3.213 3.214 3.217 3.218 3.221 3.222 3.225 3.226 3.229 3.231 3.233 3.235 3.238 3.239 3.242 3.244 3.247 3.248 3.251 3.253 3.256 3.258 3.261 3.263 3.266 3.268 3.271 3.273 3.276 3.278 3.281 3.283 3.286 3.289 3.292 3.294 3.297 3.299 3.303 3.305 3.307 3.311 3.313 3.317 3.319 3.322 3.325 3.328 3.331 3.334 3.337 3.340 3.343 3.347 3.349 3.353 3.356 3.359 3.362 3.366 3.369 3.372 3.375 3.379 3.382 3.386 3.388 3.392 3.395 3.399 3.402 3.406 3.409 3.413 3.416 3.420 3.423 3.427 3.430 3.435 3.438 3.442 3.445 3.449 3.453 3.457 3.460 3.464 3.468 3.472 3.475 3.479 3.483 3.486 3.491 3.494 3.498 3.502 3.506 3.510 3.514 3.518 3.522 3.526 3.531 3.534 3.539 3.542 3.547 3.551 3.555 3.559 3.564 3.567 3.572 3.576 3.581 3.584 3.589 3.593 3.598 3.602 3.606 3.610 3.615 3.619 3.624 3.628 3.633 3.637 3.642 3.646 3.651 3.655 3.660 3.664 3.669 3.673 3.678 3.682 3.686 3.691 3.695 3.700 3.704 3.710 3.714 3.719 3.723 3.728 3.733 3.738 3.742 3.747 3.752 3.757 3.761 3.766 3.771 3.776 3.780 3.786 3.790 3.795 3.800 3.805 3.810 3.815 3.819 3.825 3.829 3.835 3.839 3.845 3.849 3.855 3.859 3.865 3.869 3.875 3.879 3.885 3.889 3.895 3.900 3.905 3.910 3.915 3.920 3.926 3.930 3.935 3.941 3.945 3.951 3.956 3.961 3.966 3.972 3.976 3.982 3.987 3.993 3.997 4.003 4.008 4.014 4.018 4.024 4.029 4.035 4.039 4.045
Re: [R] Read the data from a text file and reshape the data
Hi Jim,
Yes, this is exactly what I was looking for. The code works perfect. Thanks
a lot for your time and effort.
Best,
Janesh Devkota
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Thursday, April 11, 2013 7:32 PM
To: Janesh Devkota
Cc: R mailing list
Subject: Re: [R] Read the data from a text file and reshape the data
Is this what you are looking for:
input - readLines(C:\\Users\\Owner\\Downloads\\WAT_DEP.DAT)
start - grep(N:SNAPSHOT, input) # find start of the data
# add index of what would have been the last block
start - c(start, tail(start, 1) + 53L)
# now read in the data using 'text' parameter of 'scan'
columns - lapply(seq(length(start) - 1L)
+ , function(indx){
+ scan(text = input[(start[indx] + 3L):(start[indx + 1L]
- 1L)]
+ , what = 0 # read in numeric
+ , quiet = TRUE
+ )
+ }
+ )
# create the columns
columns - do.call(cbind, columns)
# add column names based on the time
cNames - vapply(seq(length(start) - 1L)
+ , function(indx){
+ scan(text = input[start[indx] + 2L]
+ , what = 0
+ , quiet = TRUE
+ )[2L] # only the second value
+ }
+ , 1.0 # numeric
+ )
colnames(columns) - cNames
# sample data
columns[1:10, 1:10]
0.00208 0.00417 0.00625 0.00833 0.01042 0.0125 0.01458 0.01667 0.01875
0.02083
[1,] 3.224 4.124 4.502 4.649 4.705 4.726 4.734 4.737 4.738
4.738
[2,] 3.221 4.123 4.501 4.649 4.704 4.725 4.733 4.736 4.737
4.738
[3,] 3.220 4.123 4.502 4.649 4.705 4.726 4.734 4.737 4.738
4.738
[4,] 3.217 4.122 4.501 4.648 4.704 4.725 4.733 4.736 4.737
4.738
[5,] 3.216 4.122 4.501 4.649 4.705 4.726 4.734 4.737 4.738
4.738
[6,] 3.214 4.121 4.500 4.648 4.704 4.725 4.733 4.736 4.737
4.738
[7,] 3.212 4.121 4.501 4.649 4.705 4.726 4.734 4.737 4.738
4.738
[8,] 3.210 4.120 4.500 4.648 4.704 4.725 4.733 4.736 4.737
4.738
[9,] 3.209 4.120 4.500 4.649 4.705 4.726 4.734 4.737 4.738
4.738
[10,] 3.207 4.119 4.500 4.648 4.704 4.725 4.733 4.736 4.737
4.738
On Thu, Apr 11, 2013 at 5:55 PM, Janesh Devkota janesh.devk...@gmail.com
wrote:
I have a data set for different time intervals. The data has three comment
lines before data for each time interval. For each time interval there are
500 data points. I want to change the dataset such that I have the following
format:
t1t2t3
0.00208 0.00417 0.00625 .
a1 a2 a3 ...
b1 b2 b3 ...
c1c2c3 .
...
The link to the file is as follows:
https://www.dropbox.com/s/hc8n3qcai1mlxca/WAT_DEP.DAT
As you will see on the file, time for each interval is the second data of
the third line before the data starts. For the first time, t= 0.00208. I
need to change the data in several rows into one column. At last I need to
create a dataframe with the format shown above. In the sample above, a1, b1,
c1 are the data for time t1, and so on.
The sample data is as follows:
** N:SNAPSHOTTIME DELT[S]
** WATER DEPTH [M]: (HP(L),L=2,LA)
18000.00208 0.1
3.224 3.221 3.220 3.217 3.216 3.214 3.212
3.210 3.209 3.207
3.205 3.203 3.202 3.200 3.199 3.197 3.196
3.193 3.192 3.190
3.189 3.187 3.186 3.184 3.184 3.182 3.181
3.179 3.178 3.176
3.175 3.174 3.173 3.171 3.170 3.169 3.168
3.167 3.166 3.164
3.164 3.162 3.162 3.160 3.160 3.158 3.158
3.156 3.156 3.155
3.154 3.153 3.152 3.151 3.150 3.150 3.149
3.149 3.147 3.147
3.146 3.146 3.145 3.145 3.144 3.144 3.143
3.143 3.142 3.142
3.141 3.142 3.141 3.141 3.140 3.141 3.140
3.140 3.139 3.140
3.139 3.140 3.139 3.140 3.139 3.140 3.139
3.140 3.139 3.140
3.139 3.140 3.140 3.140 3.140 3.141 3.141
3.142 3.141 3.142
3.142 3.142 3.143 3.143 3.144 3.144 3.145
3.145 3.146 3.146
3.147 3.148 3.149 3.149 3.150 3.150 3.152
3.152 3.153 3.154
3.155 3.156 3.157 3.158
Re: [R] Plot two separate curves in R Graphics and R Lattice package
Hi, This should be fairly easy by using base R graphics. Lets suppose your first data is represented by (x1,y1) and second data is represented by (x2,y2) You can use the following command. plot(x1,y1,type=l) points(x2,y2) Hope it helps. On Tue, Apr 9, 2013 at 8:24 PM, jpm miao miao...@gmail.com wrote: Hi, How can I plot two curves with distinct x and y vectors? I would like to join one of them by regular lines and plot the other just by points (no lines). Can it be done in regular R graphic tools, say plot function? Can it be done in Lattice package, say xyplot function? Thanks, Miao My data look like this: two curves with different vector size x y 3973730 0.00322 2391576 0.003487 2840944 0.005145 2040943 0.006359 1982715 0.006253 1618162 0.00544 820082.3 0.004213 1658597 0.004883 1762794 0.006216 93439.5 0.004255 218481.3 0.006924 2332477 0.004862 725835.5 0.00089 811575.3 0.012962 292223 0.002614 153862.3 0.007524 1272367 0.006899 734199 0.00988 421404.5 0.005048 189047.5 0.004821 529102 0.009637 56833 0.006171 125856.3 0.00839 598893.8 0.006622 258240 0.00613 159086.3 0.008819 122863 0.010093 404699.5 0.008148 453514.5 0.008407 545028 0.006096 1233366 0.006111 1192758 0.008162 147563.3 0.00838 247293 0.010283 1074838 0.007413 459227.5 0.00862 202332 0.009061 377401.3 0.006923 1876753 0.010226 and x y 50118.72 0.012117 51286.14 0.012054 52480.75 0.011991 53703.18 0.011928 54954.09 0.011865 56234.13 0.011803 57543.99 0.011742 58884.37 0.01168 60255.96 0.011619 61659.5 0.011559 63095.73 0.011498 64565.42 0.011438 66069.34 0.011379 67608.3 0.011319 69183.1 0.01126 [[alternative HTML version deleted]] __ 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. [[alternative HTML version deleted]] __ 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] Solving tridiagonal matrix in R
Dear R Users, I am trying to solve a tridiagonal matrix in R. I am wondering if there is an inbuilt R function or package to solve that. I tried looking on google but couldn't find something that would help directly. Any help is highly appreciated. Thanks. Janesh [[alternative HTML version deleted]] __ 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] Solving an integral in R gives the error “The integral is probably divergent”
I am trying to solve an integral in R. However, I am getting an error when
I am trying to solve for that integral.
The equation that I am trying to solve is as follows:
$$ C_m = \frac{{abs{x}}e^{2x}}{\pi^{1/2}}\int_0^t t^{-3/2}e^{-x^2/t-t}dt $$
[image: enter image description here]
The code that I am using is as follows:
a - seq(from=-10, by=0.5,length=100)
## Create a function to compute integrationCfun - function(XX, upper){
integrand - function(x)x^(-0.5)*exp((-XX^2/x)-x)
integrated - integrate(integrand, lower=0, upper=upper)$value
(final - abs(XX)*pi^(-0.5)*exp(2*XX)*integrated) }
b- sapply(a, Cfun, upper=1)
The error that I am getting is as follows:
Error in integrate(integrand, lower = 0, upper = upper) :
the integral is probably divergent
Does this mean I cannot solve the integral ?
Any possible ways to fix this problem will be highly appreciated.The
question can be found on
http://stackoverflow.com/questions/15892586/solving-an-integral-in-r-gives-error-the-integral-is-probably-divergent
also.
Thanks.
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Read ADCP data in R
I have ADCP measured data for a river and I am wondering if it is possible to read the ADCP file in R. I found a package called oce but I couldn't read the ADCP file. The function I found in oce package is as follows: read.oce read.adp I have uploaded the sample file here https://www.dropbox.com/sh/owian354auah6h3/379D5spA2X. If anyone could help me how to read this kind of ADCP, I would highly appreciate. Thanks. Janesh Devkota [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Convert to date and time of the year
Dear R Users,
I have data for more than 3 years. For each year I want to find the day
corresponding to Jaunary 1 of that year. For example:
x - c('5/5/2007','12/31/2007','1/2/2008')
#Convert to day of year (julian date) -
strptime(x,%m/%d/%Y)$yday+1
[1] 125 365 2
I want to know how to do the same thing but with time added. But I still get
the day not time. Can anyone suggest what is the better way to find the
julian date with date and time ?
x1 - c('5/5/2007 02:00','12/31/2007 05:58','1/2/2008 16:25')
#Convert to day of year (julian date) -
strptime(x1,%m/%d/%Y %H:%M)$yday+1
[1] 125 365 2
Thank you so much.
Janesh
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[R] Interpret R-squared and cor in R
Hi I am trying to find the relationship between two variables. First I fitted a linear model between two variables and I found the following results: Residual standard error: 0.03253 on 2498 degrees of freedom Multiple R-squared: 0.5551, Adjusted R-squared: 0.5549 F-statistic: 3116 on 1 and 2498 DF, p-value: 2.2e-16 Then I used the cor function to see the correlation between two variable I get the following result -0.7450344 How can we interpret the result based on R-squared and correlation ? From the p-value we can see that there is very strong relationship between variables as it is way less that 0.001 Can anyone kindly explain the difference between Multiple R squared, adjusted R-squared and correlation and how to report these values while writing a report ? Thank you so much. [[alternative HTML version deleted]] __ 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] Clip a contour with shapefile while using contourplot
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Re: [R] low pass filter analysis in R
Dear Franklin Bretschneider, Thank you so much for your reply and explanation about the filter using the stats and signal package. I decided to opt the filter method in signal package. I have a simple question about the cut off frequency here. I have 30 minute collected tidal data and I want to use the 48 hour low pass filter to my data to remove the fluctuations and then get only the residuals. What should be the cutoff frequency in my case ? I have tried to figure out cut off frequency with the following rationale : . The parameters for butter filter are n, Wn and type. In the help, W is defined as critical frequencies of the filter. W must be a scalar for low-pass and high-pass filters, and W must be a two-element vector c(low, high) specifying the lower and upper bands. For digital filters, W must be between 0 and 1 where 1 is the Nyquist frequency. A value of 1 corresponds to half the sampling frequency. In my case the sampling frequency is 2 hr^-1. Hence a value of 0.01 corresponds to a frequency cutoff of .01*1 = .01 hr^-1 or 100 hrs time. Using unitary method, if 100 hours cut off frequency is 0.01 then 48 hours cut off frequency is 0.01/100*48 = 0.0048 hr^-1 . Is that correct ? Thank you so much Janesh On Thu, Feb 7, 2013 at 6:07 AM, Bretschneider SIG-R brets...@xs4all.nlwrote: Dear Janesh Devkota, Sorry, I forgot an edit. The last command should read: yfiltered = signal:::filter(myfilter, y) # apply filter Best wishes, Franklin Bretschneider -- Dept Biologie Kruytgebouw W711 Padualaan 8 3584 CH Utrecht The Netherlands [[alternative HTML version deleted]] __ 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] low pass filter analysis in R
Hello R users, I am trying to use R to do the low pass filter analysis for the tidal data. I am a novice in R and so far been doing only simple stuffs on R. I found a package called signal but couldn't find the proper tutorial for the low pass filter. Could anyone point me to the proper tutorial or starting point on how to do low pass filter analysis in R ? Thank you so much. Janesh [[alternative HTML version deleted]] __ 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] Summary of data for each year
Hello All, I have a data with two columns. In one column it is date and in another column it is flow data. I was able to read the data as date and flow data. I used the following code: creek - read.csv(creek.csv) library(ggplot2) creek[1:10,] colnames(creek) - c(date,flow) creek$date - as.Date(creek$date, %m/%d/%Y) The link to my data is https://www.dropbox.com/s/eqpena3nk82x67e/creek.csv Now, I want to find the summary of each year. I want to especially know mean, median, maximum etc. Thanks. Janesh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to assign time series to a vector with one leap year
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Re: [R] How to read a file with two data sets in text format
I was able to read the data using the following code:
jd1 - read.table('Practicedata.dat',header=T,sep=\t,nrow=6240)
jd2 - read.table('Practicedata.dat',header=T,sep=\t,skip=6241)
colnames(jd1) - c(Date,Mod)
colnames(jd2) - c(Date, Obs)
p - ggplot(jd1,aes(x=Date,y=Mod))+geom_line()
p - p + geom_line(data=jd2,aes(x=Date,y=Obs),color=red)
p
Now, I want to make a scatter plot between jd1$Mod and jd2$Obs. But I
cannot create one since both of them have different number of rows. Since I
have less number of rows for Mod I am planning to use the date of Mod and
then find the corresponding values of Obs for those time periods. How can I
find the corresponding values of Obs for the give date in jd1 ?
Or is there any way to create a scatter plot and put the regression
equation and correlation coefficient.
Thank you so much.
Best Regards,
Janesh
On Mon, Jan 21, 2013 at 1:19 AM, Jd Devkota janesh.devk...@gmail.comwrote:
Hello All,
I have a data file in a text format and there are two data sets. The data
set are continuous.
For each data set there is a header which has the number of data rows and
the name of data series.
For example first data set has 6240 Terry Cove-Model. Then the data for
that series follows upto 6240 rows. Then another data would start and it
will have the header such as 5200 Terry-Observed
The sample data would look like:
6240 Terry Cove-Model
300 .300110459327698
300.041656494141 .289277672767639
300.083343505859 .276237487792969
300.125 .258902788162231
300.166656494141 .236579895019531
300.208343505859 .221315026283264
300.25 .214318037033081
300.291656494141 .190926909446716
300.43505859 .158144593238831
300.375 .113302707672119
300.416656494141 .103684902191162
300.458343505859 9.72903966903687E-02
300.5 8.76948833465576E-02
300.541656494141 8.42459201812744E-02
300.583343505859 .078397274017334
300.625 8.44632387161255E-02
300.56494141 9.32939052581787E-02
300.708343505859 .113663911819458
300.75 .123064398765564
300.791656494141 .157548069953918
300.833343505859 .148393034934998
300.875 .135645747184753
300.916656494141 .137590646743774
300.958343505859 .133154153823853
301 .131152510643005
301.041656494141 .114152908325195
301.083343505859 8.04083347320557E-02
301.125 5.53587675094604E-02
301.166656494141 3.17397117614746E-02
301.208343505859 4.07266616821289E-03
301.25 -2.15455293655396E-02
301.291656494141 -4.07489538192749E-02
301.43505859 -5.85414171218872E-02
301.375 -7.53517150878906E-02
301.416656494141 -8.49723815917969E-02
301.458343505859 -7.91778564453125E-02
301.5 -7.02846050262451E-02
301.541656494141 -7.24701881408691E-02
301.583343505859 -7.76907205581665E-02
301.625 -6.82642459869385E-02
62401 Terry Cove-Data
300 .216407993
300.0042 .204216005
300.0083 .210311999
300.0125 .195071996
300.0167 .192023999
300.0208 .179831992
300.025 .188976001
300.0292 .185928004
300.0333 .195071996
300.0375 .219456009
300.0417 .210311999
300.0458 .204216005
300.05 .195071996
300.0542 .188976001
300.0583 .195071996
300.0625 .195071996
300.0667 .185928004
300.0708 .173735998
300.075 .170688001
300.0792 .167640004
300.0833 .167640004
300.0875 .167640004
300.0917 .167640004
300.0958 .161543991
300.1 .1524
300.1042 .158495994
300.1083 .149352003
300.1125 .158495994
300.1167 .1524
300.1208 .1524
300.125 .149352003
300.1292 .143256
300.1333 .146303997
300.1375 .149352003
300.1417 .146303997
300.1458 .137159996
300.15 .131064002
300.1542 .124967999
300.1583 .128015996
300.1625 .124967999
300.1667 .131064002
300.1708 .124967999
300.175 .124967999
300.1792 .134111999
300.1833 .118871996
300.1875 .128015996
300.1917 .131064002
300.1958 .128015996
300.2 .131064002
300.2042 .128015996
300.2083 .121920002
300.2125 .115823999
300.2167 .112776001
300.2208 .103632001
300.225 .097535998
300.2292 .103632001
300.2333 .094488001
300.2375 .082296003
300.2417 .0762
300.2458 .079247997
300.25 .067056
300.2542 .064007998
300.2583 .045720002
300.2625 .033528
300.2667 .036575999
300.2708 .036575999
300.275 .036575999
300.2792 .027432001
300.2833 .027432001
300.2875 .021336
300.2917 .012192
300.2958 .009144
300.3 .009144
300.3042 .003048
300.3083 0
300.3125 -.003048
300.3167 -.006096
300.3208 0
300.325 .006096
300.3292 -.003048
300. .006096
The full data set can be downloaded from
https://www.dropbox.com/s/chhw3vz6ru1godk/Practicedata.Dat
I want to make a comparison graph between modeled and observed. Once I am
able to read two data sets as two sets of data or combined in one I would
be able to create the time series graph.
Another thing I need to do is create another sub data set where both the
series have common data. One data might have more intervals than another.
After I find two data sets of same interval then I want to plot a
correlation graph.
I hope I made it clear what I want to do.
Thank you so much.
Best Regards,
Janesh
[R] Very slow in processing the equation in the scatter plot ggplot
Hello All,
I have plotted a scatter plot in ggplot2 and added a regression line and a
regression equation. But the processing is very very slow. One reason might
be because I have so many data pairs. Is there any way to speed up this
code ? I need to create a multiple layout as well.
The code I have used is as follows:
setwd(C:/Users/jzd0009/Documents/R software)
mydata - read.table(dataset.csv,header=TRUE,sep=,)
library(ggplot2)
p -
ggplot(mydata,aes(date))+geom_line(aes(y=modeled,colour=modeled))+geom_line(aes(y=observed,colour=observed))
p
p1 - ggplot(mydata, aes(modeled,observed))+geom_point(aes(y=observed))
#p1 - p1+stat_smooth()
lm_eqn = function(mydata){
m = lm(modeled ~ observed, mydata);
eq - substitute(italic(y) == a + b %.%
italic(x)*,~~italic(r)^2~=~r2,
list(a = format(coef(m)[1], digits = 2),
b = format(coef(m)[2], digits = 2),
r2 = format(summary(m)$r.squared, digits = 3)))
as.character(as.expression(eq));
}
p1 - p1 + geom_text(aes(x = -0.1, y = 0.5, label = lm_eqn(mydata)), parse
= TRUE)
p1 - p1+geom_smooth(method=lm,se=FALSE,color=green,formula=y~x,lwd=2)
p1
#For multiple layout
library(grid)
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,2)))
vplayout - function(x,y)
viewport(layout.pos.row=x,layout.pos.col=y)
print(p,vp=vplayout(1,1))
print(p1,vp=vplayout(1,2))
print(p,vp=vplayout(2,1))
print(p1,vp=vplayout(2,2))
The data for the above code can be found on
https://www.dropbox.com/s/1xrgvnge0prf0a6/dataset.csv
Thank you so much.
Best Regards,
Janesh Devkota
[[alternative HTML version deleted]]
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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] Very slow in processing the equation in the scatter plot ggplot
Dear Ista,
Thank you so much for the prompt reply. Yes, using the annotate instead of
the geom_text definitely speed up my graph and look much better.
Thanks.
Best Regards,
Janesh Devkota
On Mon, Jan 21, 2013 at 3:46 PM, Ista Zahn istaz...@gmail.com wrote:
Hi,
One problem is that you are printing the regression equation multiple
times. Compare:
system.time({
+ p2 - p1 + annotate(x = -0.1, y = 0.5, geom=text, label =
lm_eqn(mydata), parse = TRUE)
+ print(p2)
+ })
user system elapsed
0.930 0.006 0.976
system.time({
+ p3 - p1 + geom_text(aes(x = -0.1, y = 0.5, label =
lm_eqn(mydata)), parse = TRUE)
+ print(p3)
+ })
user system elapsed
38.667 0.046 38.956
So, use annotate instead of geom_text (as a bonus the equation will
look better). Further speedups are possible, but this is probably the
biggest problem.
Best,
Ista
On Mon, Jan 21, 2013 at 4:21 PM, Janesh Devkota
janesh.devk...@gmail.com wrote:
Hello All,
I have plotted a scatter plot in ggplot2 and added a regression line and
a
regression equation. But the processing is very very slow. One reason
might
be because I have so many data pairs. Is there any way to speed up this
code ? I need to create a multiple layout as well.
The code I have used is as follows:
setwd(C:/Users/jzd0009/Documents/R software)
mydata - read.table(dataset.csv,header=TRUE,sep=,)
library(ggplot2)
p -
ggplot(mydata,aes(date))+geom_line(aes(y=modeled,colour=modeled))+geom_line(aes(y=observed,colour=observed))
p
p1 - ggplot(mydata, aes(modeled,observed))+geom_point(aes(y=observed))
#p1 - p1+stat_smooth()
lm_eqn = function(mydata){
m = lm(modeled ~ observed, mydata);
eq - substitute(italic(y) == a + b %.%
italic(x)*,~~italic(r)^2~=~r2,
list(a = format(coef(m)[1], digits = 2),
b = format(coef(m)[2], digits = 2),
r2 = format(summary(m)$r.squared, digits = 3)))
as.character(as.expression(eq));
}
p1 - p1 + geom_text(aes(x = -0.1, y = 0.5, label = lm_eqn(mydata)),
parse
= TRUE)
p1 -
p1+geom_smooth(method=lm,se=FALSE,color=green,formula=y~x,lwd=2)
p1
#For multiple layout
library(grid)
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,2)))
vplayout - function(x,y)
viewport(layout.pos.row=x,layout.pos.col=y)
print(p,vp=vplayout(1,1))
print(p1,vp=vplayout(1,2))
print(p,vp=vplayout(2,1))
print(p1,vp=vplayout(2,2))
The data for the above code can be found on
https://www.dropbox.com/s/1xrgvnge0prf0a6/dataset.csv
Thank you so much.
Best Regards,
Janesh Devkota
[[alternative HTML version deleted]]
__
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.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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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.
[R] Faceting in ggplot with 3 series in each plot
I have created a graph with basic R plot with 2 rows and 3 columns. With the basic plot feature of R I had to write several lines of code to come up with the graph https://www.dropbox.com/s/f7d6ei6krtcrtti/hello.png I was wondering whether we can plot multiple series in each plot using ggplot as created in the above link. If anyone could give me any suggestion then I could try on ggplot. I know how to create basic plot with 3 series in ggplot but don't know how to use facet and then use 3 series in each sub plot. As I am a new user in R and ggplot and on a learning curve please excuse me if this question is very naive. Best Regards, Janesh Devkota [[alternative HTML version deleted]] __ 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.
