Re: [R] uneven vector length issue with read.zoo?
For simplicity sake though, yes I understand the issue and solution, and the solution using read.table, na.omit, and read.zoo is sound. Thanks Gabor! :) -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4613983.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
Yeah, I was unclear about what I mean by "uneven vector lengths". I should say "uneven valid vectors" instead where "valid" refers to (1) a field containing a value that is not NA, for this specific case, and (2) a value that is compatible with the vector class assigned through colClasses etc., and therefore avoids the read.zoo error. I understand and agree that the error is clear. I have no issue with that. My issue is with the need to use read.table and then read.zoo shortly after (this seems inefficient). I was simply pushing toward the idea of where this type of situation could be avoided for future users in where if there are uneven valid vectors that there would be a logical argument saying that it's okay to truncate to the shortest valid vector (in this case columns 1 and 2). My raw data consisted of a lot of uneven valid vectors. My expected thought of nulling out columns 3:5 would be that there would have no need for read.zoo to try to read in the bad data entry rows in columns 1:2 containing NA's that's already outside of the valid vector length. Anyway, this is probably trivial now considering that this problem is already solved haha, and also I don't mean to offend and criticize. I simply see an efficiency opportunity and an opportunity to create more robust source code. Why use read.table with read.zoo if you can just do it all with read.zoo? Do you not agree? -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4613975.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
Right, but it seems to me that the error being the NA's in the index field are caused by the longer vector lengths of columns 4 and 5. I would think that the EOF in the scanf() (assuming C is used for the source code) would be called where the NA's begin in columns 1 and 2 since columns 3:5 are nulled out. Does this sound like a possible case? So, if the read in data only contained columns 1 and 2, it wouldn't even look at columns 3:5 and thus, rows 14 and so on wouldn't even be looked at and that would be EOF already - resulting in no error. -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4613384.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
"However, I do wonder why it still complains of the vector length even though I nulled out the other columns. It's an interesting error to run into. Probably looks at FUN before nulling out the other columns was my theory. " Referring to just a straight up read.zoo in this case ^ -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4609509.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
Hey Gabor, just trying to understand this here..sorry for the noob question: DF1 <- read.table(URL, skip = 1, header = TRUE, sep = ",", fill = TRUE, as.is = TRUE) I'm not to familiar with as.is, however I quickly read the R documentation on that. From my understanding it converts character to factor in terms of atomic vector class/mode...sorta like what colClasses would do. Why is it needed here for this specific case? -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4609382.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
Thank you for the suggestion Gabor. It's definitely more elegant than what I had above. Instead of going from character representation to POSIXct to chron, it looks at the character representation and goes straight to chron. It's good. However, I do wonder why it still complains of the vector length even though I nulled out the other columns. It's an interesting error to run into. Probably looks at FUN before nulling out the other columns was my theory. -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4609468.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
So with case (6) here's the general structure of what I have:
chw = read.table("crac.csv", skip = 1, header = TRUE,
colClasses = rep(c("NULL", NA, "numeric", "NULL"),
c(3, 1, 1, 24)),
sep = ",")
chw$Time.1 = as.POSIXct(chw$Time.1, format = fmt, tz = TZ)
chw = na.omit(chw)
chw = read.zoo(chw, header = TRUE,
colClasses = rep(c(NA, "numeric"), c(1, 1)),
FUN = chr, aggregate = tail1)
You don't have to try this, but the main point is that
read.table -> POSIXct -> na.omit -> read.zoo and chron
I guess this alternative solution is adequate along with using readLines.
Initially I was hoping just a simple read.zoo would do the trick. The catch
is that I need the index/timestamp column to be in chron format for an easy
na.approx function to deal with things. Thank you for the readLines
suggestion Rui. Much appreciated.
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Re: [R] uneven vector length issue with read.zoo?
case (6) - regress back to read.table apparently -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4604537.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
blank.lines.skip is not working either... -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4604360.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
Actually case (4) didn't work. The issue is also with the index.."fill" only seems to work with the dimensions/columns that contain the data associated to the index. Dang.yeah, I need help here. -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4604349.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
Case (4) - use the fill argument in ?read.tablethis looks useful...guess I answered my own question...going to delete this thread now... -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4604332.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
So far I see two options: (1) nrows argument to specify max number of rows to read in or (2) go into excel, and put a bunch of NA's . Both which are inefficient in that they're not so "automated". For case (1), I have to wait till an error pops up each time and deal with each one individually taking into account the skip and header args, and for case (2), now I'm just not even using R to do the dirty work...anyway, I'm going to continue to go through this R documentation to see if I find anything else for ?read.table and ?read.zoo. -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4604323.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
Make that 3 options actually. In case (3) I would have to take each category on the spreadsheet and isolate each to its own csv file using excel. Fun stuff... -- View this message in context: http://r.789695.n4.nabble.com/uneven-vector-length-issue-with-read-zoo-tp4604287p4604329.html Sent from the R help mailing list archive at Nabble.com. __ 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] uneven vector length issue with read.zoo?
I truncated and simplified my code and the read in data that I'm working with
to isolate the issue. Here is the read in data and R script respectively:
http://r.789695.n4.nabble.com/file/n4604287/test.csv test.csv
http://pastebin.com/rCdaDqPm
Here is the terminal/R shell output that I hope the above replicates on your
screen:
> source("elecLoad.r", echo = TRUE)
> #Load packages
> library(zoo)
> library(chron)
> #Initial assignments for format (fmt), timezone (TZ), and user
> #defined chron function (chr)
> fmt = "%m/%d/%y %I:%M %p"
> TZ = "PDT"
> chr = function(x) as.chron(x, fmt)
> #Read in data as zoo object using relevant arguments in read.zoo()
> #for details of arguments, see Kevin Navero or see ?read.zoo
> #and ?read.table [TRUNCATED]
Error in read.zoo("http://dl.dropbox.com/u/41922443/test.csv";, skip = 1, :
index has bad entries at data rows: 14 15 16 17 18 19 20 21 22 23 24 25 26
27 28
I was hoping that the "NULL" in colClasses() would've taken care of this
uneven vector length issue, however, that was not the case. Any ideas?
Thanks in advance. Sorry if my post didn't follow the forum rules exactly. I
tried to make small scale reproducible code and what not. I'm still a bit of
a noob here and there.
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Re: [R] simple read in with zoo using POSIXlt
Thanks Gabor. Much appreciated. -- View this message in context: http://r.789695.n4.nabble.com/simple-read-in-with-zoo-using-POSIXlt-tp4557138p4559762.html Sent from the R help mailing list archive at Nabble.com. __ 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] simple read in with zoo using POSIXlt
Achim Zeileis-4 wrote
>
> I don't know why you make this so complicated. Either use
>
> read.zoo("test.txt", header = FALSE, sep = "\t",
>format = "%d/%m/%Y %H:%M", tz = "")
>
> which yields a POSIXct time index. Alternatively, you can produce POSIXlt
> via strptime:
>
> read.zoo("test.txt", header = FALSE, sep = "\t",
>FUN = function(x) strptime(x, "%d/%m/%Y %H:%M"))
>
> The former is recommended for use in zoo.
>
Sorry, it's not that I'm trying to make it complicated, but rather specific.
As Gabor said in the earlier post, it seems POSIXlt is not a suitable
argument for read.zoo, and therefore explains the problem that I have been
having. Like I said, I was able to produce the solution that I wanted;
however, it was not as efficient and elegant as I was hoping it to be. When
going through 2 year span data sets, there is a noticeable difference in
speed when you use POSIXlt separately outside the read.zoo function. Anyway,
it's fine. I appreciate the feedback and suggestion. It definitely helps
bounce ideas around and possibly offers ways to expand R source code in the
future maybe. Thanks!
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Re: [R] simple read in with zoo using POSIXlt
Achim Zeileis-4 wrote
>
> You just need to declare that the index is in two columns (1 and 2) and
> then provide a function that extracts a suitable object from it:
>
> read.zoo("test.txt", header = FALSE, index = 1:2,
>FUN = function(x, y) strptime(paste(x, y), "%d/%m/%Y %H:%M"))
>
> Use an additional as.POSIXct(...) around the strptime() call if you want
> to use POSIXct instead of POSIXlt which is typically recommended.
>
> See vignette("zoo-read", package = "zoo") for more examples.
> Z
>
Unfortunately, it's not working as I hoped for. Let me elaborate,
new code:
http://pastebin.com/axpPB6M8
So for this, I understand that the read in works very well with POSIXct, but
I want to utilize the vectors contained with the POSIXlt class (wday, yday,
mon, etc.). Here's how the POSIXct read.zoo looks like in the shell when
copy pasted:
> test = read.zoo("http://dl.dropbox.com/u/41922443/test.txt";,
+header = FALSE, sep = "\t",
+FUN = function(idx) as.POSIXct(strptime(idx,
+ format = fmt, tz = "PDT"), format = fmt, tz = "PDT"),
+colClasses = rep(c(NA, "numeric", "NULL"), c(1, 1, 0)),
+aggregate = tail1)
> test
2010-01-07 00:15:00 2010-01-07 00:30:00 2010-01-07 00:45:00 2010-01-07
01:00:00
1333.6201333.3881335.343
1334.251
2010-01-07 01:15:00 2010-01-07 01:30:00 2010-01-07 01:45:00 2010-01-07
02:00:00
1331.5891328.6951329.151
1329.077
2010-01-07 02:15:00 2010-01-07 02:30:00
1327.6491326.789
This is good when you just eyeball it, HOWEVER, when the date/time is looked
at by the machine, it doesn't see vectors that can be accessed, but the lame
numerical/double that is the UTC time from 1960 or whatever in seconds.
Proof of this is the following:
> unclass(index(test))
[1] 1262823300 1262824200 1262825100 1262826000 1262826900 1262827800
[7] 1262828700 1262829600 1262830500 1262831400
attr(,"tzone")
[1] "PDT"
Now with this code:
http://pastebin.com/pr2X78sX
This just pisses me offlet me elaborate...or well, eff it. I'll just
copy paste and you'll get my point:
> test = read.zoo("http://dl.dropbox.com/u/41922443/test.txt";,
+header = FALSE, sep = "\t",
+FUN = function(idx) as.POSIXlt(strptime(idx,
+ format = fmt, tz = "PDT"), format = fmt, tz = "PDT"),
+colClasses = rep(c(NA, "numeric", "NULL"), c(1, 1, 0)),
+aggregate = tail1)
> test
0
1326.789
Basically, what the hell is 0 and 1326.789 doing there?.right?
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Re: [R] simple read in with zoo using POSIXlt
Achim Zeileis-4 wrote
>
> You just need to declare that the index is in two columns (1 and 2) and
> then provide a function that extracts a suitable object from it:
>
> read.zoo("test.txt", header = FALSE, index = 1:2,
>FUN = function(x, y) strptime(paste(x, y), "%d/%m/%Y %H:%M"))
>
> Use an additional as.POSIXct(...) around the strptime() call if you want
> to use POSIXct instead of POSIXlt which is typically recommended.
>
Ah I see. Yeah I'm familiar with the index = 1:2 argument of read.zoo.
However, I'm not sure if it's necessary here since I'm specifying one of the
arguments to be sep = "\t", and thus it sees one timestamp column as
"%m/%d/%Y %H:%M" format and the second column, being the data column,
separated via tab. So I'll try using strptime() and POSIXlt() on that column
as FUN = function(idx) as.POSIXlt(strptime(idx, format = "%m/%d/%Y %H:%M",
tz = "PDT")).
The reason I plan on using POSIXlt here as opposed to POSIXct is for
functionality reasons such that the class itself contains a list of vectors
that can be accessed i.e. sec, min, hour, ...wday, yday, etc. I'll post the
results in a few mins after testing it out on the shell here. Thanks for the
suggestion!!!
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[R] simple read in with zoo using POSIXlt
Easy question a bit. So here's my code: http://pastebin.com/F4iQPVy5 I am trying to read in a series of timestamps. However with POSIXlt as FUN in read.zoo, the output is merely two numbers and is not the output that I'm hoping for. The code above should reproduce the error. Here is code that shows what I want it to do: http://pastebin.com/GEPZ5R9B The problem though, is that it's not that elegant since POSIXlt is outside of read.zoo. Wondering if there's a way to put POSIXlt inside read.zoo and compress things a bit to make the code run faster. Any ideas? Thanks. Also, if there's any issues reading the file i uploaded to dropbox directly off of the net with the code above, just use this: http://r.789695.n4.nabble.com/file/n4557138/test.txt test.txt Hope I covered everything in a compact reproducible manner. Sorry if I made a noob mistake. Still kind of new. -- View this message in context: http://r.789695.n4.nabble.com/simple-read-in-with-zoo-using-POSIXlt-tp4557138p4557138.html Sent from the R help mailing list archive at Nabble.com. __ 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] Trying to merge new data set to bottom of old data set. Both are zoo objects.
Okay, will do. Thanks for all the handy advice Gabor. Ugh, it's such a stupid bug once I actually know what is going on. I need to go over my Unix date/time format specifiers, and I'll probably use the rep function to simplify and reducing the amount of code. A lot of that is definitely new to me. As for shortening the read in data, I do it find it tricky sometimes since you have to incrementally test it in the sense that you want to shorten it to the point that it still reproduces the problem. Anyway, I'll try to make the data significantly shorter in my next post if possible. Thanks again. -- View this message in context: http://r.789695.n4.nabble.com/Trying-to-merge-new-data-set-to-bottom-of-old-data-set-Both-are-zoo-objects-tp4530888p4532484.html Sent from the R help mailing list archive at Nabble.com. __ 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] Trying to merge new data set to bottom of old data set. Both are zoo objects.
Here's a case where it doesn't work. Again, the problem is that when I use the rbind or concatenate functions, the 2012 data set seems to go ahead of the 2010 and 2011 portions of the data set. The problem seems dependent on the text files I read in: http://r.789695.n4.nabble.com/file/n4531011/old.txt old.txt http://r.789695.n4.nabble.com/file/n4531011/new.txt new.txt using this code: http://pastebin.com/8W6KaaPQ In a case where it works, and the data seemed to be in the right order, I read in a different old.txt named old1.txt and somehow it seemed to work. The contents and format were similar to that of new.txt where there was 18 columns with the same headers. Here are the files to use: http://r.789695.n4.nabble.com/file/n4531011/old1.txt old1.txt http://r.789695.n4.nabble.com/file/n4531011/new.txt new.txt using this code: http://pastebin.com/6iNF5bPd That should clarify the issue I'm having. Let me know if a dput is necessary here. However all the vectors and vector modes seem to check out okay. -- View this message in context: http://r.789695.n4.nabble.com/Trying-to-merge-new-data-set-to-bottom-of-old-data-set-Both-are-zoo-objects-tp4530888p4531011.html Sent from the R help mailing list archive at Nabble.com. __ 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] Trying to merge new data set to bottom of old data set. Both are zoo objects.
Here is the data I'm working with: http://r.789695.n4.nabble.com/file/n4530888/new.txt new.txt http://r.789695.n4.nabble.com/file/n4530888/old.txt old.txt My code is here: http://pastebin.com/9jjs6Ahr I'm looking for away to simply attach the new.txt to the bottom of old.txt through R, else I'll just throw it in Excel to do some preprocessing. I've looked into using merge, cbind, concatenate, and rbind. However, I'm running into problems where the 2012 data keeps ending up on top before the 2010 and 2011 data or the function just adds more extra columns to the right side. Is there a simple method of doing this? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Trying-to-merge-new-data-set-to-bottom-of-old-data-set-Both-are-zoo-objects-tp4530888p4530888.html Sent from the R help mailing list archive at Nabble.com. __ 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.zoo - combining two columns with date and time respectively into one index column?
Ah yes, sorry about that Gabor. Found a pretty good example in ?read.zoo where it has index = 1:3 and takes in times objects. I must have overlooked it out of impatience. Thanks for the quick and simple feedback however. -- View this message in context: http://r.789695.n4.nabble.com/read-zoo-combining-two-columns-with-date-and-time-respectively-into-one-index-column-tp4495326p4496743.html Sent from the R help mailing list archive at Nabble.com. __ 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.zoo - combining two columns with date and time respectively into one index column?
I actually just found an alternative solution which involves setting Excel's or Libre Office Calc's fixed width option then saving it again as a .txt or .csv. If there's a cleaner way of doing this within the R shell, I'm open to ideas. -- View this message in context: http://r.789695.n4.nabble.com/read-zoo-combining-two-columns-with-date-and-time-respectively-into-one-index-column-tp4495326p4495345.html Sent from the R help mailing list archive at Nabble.com. __ 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] read.zoo - combining two columns with date and time respectively into one index column?
I have three columns in my raw data: date, time, and dry bulb temperature: http://r.789695.n4.nabble.com/file/n4495326/weathDataSDCoron.txt weathDataSDCoron.txt The date format is %Y%m%d and the time format is %H:%M. Any ideas on how to read it in such that it looks at the first two columns and then merges it into one column combining both the date and time? -- View this message in context: http://r.789695.n4.nabble.com/read-zoo-combining-two-columns-with-date-and-time-respectively-into-one-index-column-tp4495326p4495326.html Sent from the R help mailing list archive at Nabble.com. __ 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] Wrong output due to what I think might be a data type issue (zoo read in problem)
Ah I see. Thank you very much Gabor and Joshua. Yes that makes sense since in C, alpha characters are represented in single quotes as to represent the ASCII value hence 'M'. I would've never imagined the raw data would be so lame like that though. Thanks again! -- View this message in context: http://r.789695.n4.nabble.com/Wrong-output-due-to-what-I-think-might-be-a-data-type-issue-zoo-read-in-problem-tp4487682p4490172.html Sent from the R help mailing list archive at Nabble.com. __ 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] Wrong output due to what I think might be a data type issue (zoo read in problem)
update temporary fix: http://pastebin.com/dzj0W89H -- View this message in context: http://r.789695.n4.nabble.com/Wrong-output-due-to-what-I-think-might-be-a-data-type-issue-zoo-read-in-problem-tp4487682p4488179.html Sent from the R help mailing list archive at Nabble.com. __ 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] Wrong output due to what I think might be a data type issue (zoo read in problem)
found a temporary fix (I'm sure it's redundant and not as elegant, but here
it is):
require(zoo)
require(chron)
setwd("/home/knavero/Desktop/")
fmt = "%m/%d/%Y %H:%M"
tail1 = function(x) tail(x, 1)
rawData = read.zoo("weatherData.txt", header = T, FUN = as.chron,
format = fmt, sep = "\t", aggregate = tail1)
#colClasses = c(NA, "matrix"))
rawData = zoo(cbind(temp = as.vector(rawData)), time(rawData))
oneMin = seq(start(rawData), end(rawData), by = times("01:00:00"))
intData = na.approx(rawData, xout = oneMin)
par(mfrow = c(3, 1), oma = c(0, 0, 2, 0), mar = c(2, 4, 1, 1))
plot(rawData, type = "p", ylim = c(0, 100))
grid(col = "darkgrey")
plot(intData, type = "p", ylim = c(0, 100))
grid(col = "darkgrey")
Silly coding huh? It works thoughthe plots were just to double check
btw...nothing significant obviously
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[R] Wrong output due to what I think might be a data type issue (zoo read in problem)
Here's the small scale version of the R script: http://pastebin.com/sEYKv2Vv Here's the file that I'm reading in: http://r.789695.n4.nabble.com/file/n4487682/weatherData.txt weatherData.txt I apologize for the length of the data. I tried to cut it down to 12 lines, however, it wasn't reproducing the bad output that I wanted to show. The problem is that my whole data set shifts down. For example, I have this when the raw data is read in or scanned in as a zoo object: "> rawData (12/01/10 00:53:00) (12/01/10 01:53:00) (12/01/10 02:53:00) (12/01/10 03:53:00) 41 40 39 38 (12/01/10 04:53:00) (12/01/10 05:53:00) (12/01/10 06:53:00) (12/01/10 07:53:00) 38 37 36 39 (12/01/10 08:53:00) (12/01/10 09:53:00) (12/01/10 10:53:00) (12/01/10 11:53:00) 43 47 50 " Then when I run it through my code, which should feed out the exact same thing (the values at least), the output is this: "> intData (12/01/10 00:53:00) (12/01/10 01:53:00) (12/01/10 02:53:00) (12/01/10 03:53:00) 11.010.0 9.0 8.0 (12/01/10 04:53:00) (12/01/10 05:53:00) (12/01/10 06:53:00) (12/01/10 07:53:00) 8.0 7.0 6.0 9.0 (12/01/10 08:53:00) (12/01/10 09:53:00) (12/01/10 10:53:00) (12/01/10 11:53:00) 13.017.020.0 24.0 " Finally, my dput(rawData) and dput(intData): "> dput(rawData) structure(c(11L, 10L, 9L, 8L, 8L, 7L, 6L, 9L, 13L, 17L, 20L, 24L, 27L, 27L, 27L, 26L, 23L, 21L, 20L, 21L, 18L, 16L, 14L, 14L, 12L, 10L, 12L, 11L, 10L, 10L, 11L, 14L, 16L, 20L, 23L, 27L, 25L, 26L, 29L, 28L, 27L, 26L, 24L, 24L, 25L, 24L, 23L, 23L, 21L, 20L, 18L, 19L, 18L, 18L, 16L, 18L, 21L, 24L, 25L, 27L, 27L, 29L, 29L,..." "> dput(intData) structure(c(11, 10, 9, 8, 8, 7, 6, 9, 13, 17, 20, 24, 27, 27, 27, 26, 23, 21, 20, 21, 18, 16, 14, 14, 12, 10, 12, 11, 10, 10, 11, 14, 16, 20, 23, 27, 25, 26, 29, 28, 27, 26, 24, 24, 25, 24, 23, 23, 21, 20, 18, 19, 18, 18, 16, 18, 21, 24, 25, 27, 27, 29, 29, 28, 26, 25, 24, 22, 22, 22, 21, 21, 21, 20, 21, 21, 20, 21,..." I am not sure how to interpret this, however I have tried researching on what the "L" following the number is, and it seems they are "list" values? Also, I have read ?colClasses in the R manual, and have tried colClasses. >From experience using C, there seems to be a related error message saying: "scan() expected 'a real', got 'M'" What is "M"? Is that matrix? Any clarification of the issue and solution is appreciated. I apologize in advance for any noob mistake related to asking questions correctly according to forum specifications. Thanks for any help! I will keep messing around with colClassesI feel like I am close to a solution..however, am very far from understanding the problem. -- View this message in context: http://r.789695.n4.nabble.com/Wrong-output-due-to-what-I-think-might-be-a-data-type-issue-zoo-read-in-problem-tp4487682p4487682.html Sent from the R help mailing list archive at Nabble.com. __ 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] logical test not functioning correctly on zoo series...what the hell?
Thanks Gabor. colClasses did the trick. Also, I'll be sure to minimize the reproducible example the next time around using a text quote. Sorry about the noob mistake. -- View this message in context: http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4473370.html Sent from the R help mailing list archive at Nabble.com. __ 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] logical test not functioning correctly on zoo series...what the hell?
Heh, a one liner fix huh? This was going to be my alternative solution (got
this coded as soon as the character strings theory arrived, and yes I'll use
dput next time, sorry about that again and for all the miscommunication):
require(zoo); require(chron)
setwd("/home/knavero/Desktop")
rawData = read.table("dataout_2471_843.csv", header = T, sep = ",")
rawData$Meter.ID = NULL
pwr = as.vector(rawData$KW.ch..1..set.0.)
idx = as.vector(rawData$Date...Time)
idx = as.chron(idx, format = "%m/%d/%Y %H:%M")
newData = zoo(cbind(pwr = pwr), idx)
newData = newData[newData$pwr > 0]
Guess I won't be needing that ^ anymore. And here I thought i redefined
logic where 0 > 0 ha...I knew it was too good to be true. Thanks for all the
help. Much appreciated.
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Re: [R] logical test not functioning correctly on zoo series...what the hell?
"If you showed the output of dput(yourData) or even str(yourData)
others could see what types the columns are. Ordinary printed output can
make numeric, character, and factor data look the same. Comparisons
involving strings do not always return the same value as comparisons
involving the equivalent numbers. E.g.,
> y <- c("0.00", "0.01", "10.1", "20.2", "0.00")
> z <- zoo(y, as.POSIXct("2012-03-14")+(1:5)*24*60*60)
> z # looks like numeric data, but it is not
2012-03-15 2012-03-16 2012-03-17 2012-03-18 2012-03-19
0.00 0.01 10.1 20.2 0.00
> str(z)
'zoo' series from 2012-03-15 to 2012-03-19
Data: chr [1:5] "0.00" "0.01" "10.1" "20.2" "0.00"
Index: POSIXct[1:5], format: "2012-03-15" "2012-03-16" "2012-03-17"
"2012-03-18" "2012-03-19"
> z[z>0]
2012-03-15 2012-03-16 2012-03-17 2012-03-18 2012-03-19
0.00 0.01 10.1 20.2 0.00
> z[z>"0.00"]
2012-03-16 2012-03-17 2012-03-18
0.01 10.1 20.2
> "0.0" > 0
[1] TRUE "
Ah I see. So it's seeing the ASCII values I'm assuming. Yeah, I think this
was the case I was running into. It seems it's just dependent on the raw
data because recently when I was reading in raw data as zoo, this didn't
happen. However, my dput results show quoted values, technically strings.
Thank you for pointing this out. I'll remember to use dput next time. Noob
mistake. Won't happen again.
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Re: [R] logical test not functioning correctly on zoo series...what the hell?
Additionally, the dput(rawData) output:
dput(rawData)
structure(list(Date...Time = structure(1:1040, .Label = c("2/11/12 12:45
",
"2/11/12 13:00 ", "2/11/12 13:15 ", "2/11/12 13:30
",
"2/11/12 13:45 ", "2/11/12 14:00 ", "2/11/12 14:15
",
"2/11/12 14:30 ", "2/11/12 14:45 ", "2/11/12 15:00
",
"2/11/12 15:15 ", "2/11/12 15:30 ", "2/11/12 15:45
",
"2/11/12 16:00 ", "2/11/12 16:15 ", "2/11/12 16:30
",
"2/11/12 16:45 ", "2/11/12 17:00 ", "2/11/12 17:15
",
"2/11/12 17:30 ", "2/11/12 17:45 ", "2/11/12 18:00
",
"2/11/12 18:15 ", "2/11/12 18:30 ", "2/11/12 18:45
",
"2/11/12 19:00 ", "2/11/12 19:15 ", "2/11/12 19:30
",
"2/11/12 19:45 ", "2/11/12 20:00 ", "2/11/12 20:15
",
"2/11/12 20:30 ", "2/11/12 20:45 ", "2/11/12 21:00
",
"2/11/12 21:15 ", "2/11/12 21:30 ", "2/11/12 21:45
",
"2/11/12 22:00 ", "2/11/12 22:15 ", "2/11/12 22:30
",
"2/11/12 22:45 ", "2/11/12 23:00 ", "2/11/12 23:15
",
"2/11/12 23:30 ", "2/11/12 23:45 ", "2/12/12 00:00
",
"2/12/12 00:15 ", "2/12/12 00:30 ", "2/12/12 00:45
",
"2/12/12 01:00 ", "2/12/12 01:15 ", "2/12/12 01:30
",
"2/12/12 01:45 ", "2/12/12 02:00 ", "2/12/12 02:15
",
"2/12/12 02:30 ", "2/12/12 02:45 ", "2/12/12 03:00
",
"2/12/12 03:15 ", "2/12/12 03:30 ", "2/12/12 03:45
",
"2/12/12 04:00 ", "2/12/12 04:15 ", "2/12/12 04:30
",
"2/12/12 04:45 ", "2/12/12 05:00 ", "2/12/12 05:15
",
"2/12/12 05:30 ", "2/12/12 05:45 ", "2/12/12 06:00
",
"2/12/12 06:15 ", "2/12/12 06:30 ", "2/12/12 06:45
",
"2/12/12 07:00 ", "2/12/12 07:15 ", "2/12/12 07:30
",
"2/12/12 07:45 ", "2/12/12 08:00 ", "2/12/12 08:15
",
"2/12/12 08:30 ", "2/12/12 08:45 ", "2/12/12 09:00
",
"2/12/12 09:15 ", "2/12/12 09:30 ", "2/12/12 09:45
",
"2/12/12 10:00 ", "2/12/12 10:15 ", "2/12/12 10:30
",
"2/12/12 10:45 ", "2/12/12 11:00 ", "2/12/12 11:15
",
"2/12/12 11:30 ", "2/12/12 11:45 ", "2/12/12 12:00
",
"2/12/12 12:15 ", "2/12/12 12:30 ", "2/12/12 12:45
",
"2/12/12 13:00 ", "2/12/12 13:15 ", "2/12/12 13:30
",
"2/12/12 13:45 ", "2/12/12 14:00 ", "2/12/12 14:15
",
"2/12/12 14:30 ", "2/12/12 14:45 ", "2/12/12 15:00
",
"2/12/12 15:15 ", "2/12/12 15:30 ", "2/12/12 15:45
",
"2/12/12 16:00 ", "2/12/12 16:15 ", "2/12/12 16:30
",
"2/12/12 16:45 ", "2/12/12 17:00 ", "2/12/12 17:15
",
"2/12/12 17:30 ", "2/12/12 17:45 ", "2/12/12 18:00
",
"2/12/12 18:15 ", "2/12/12 18:30 ", "2/12/12 18:45
",
"2/12/12 19:00 ", "2/12/12 19:15 ", "2/12/12 19:30
",
"2/12/12 19:45 ", "2/12/12 20:00 ", "2/12/12 20:15
",
"2/12/12 20:30 ", "2/12/12 20:45 ", "2/12/12 21:00
",
"2/12/12 21:15 ", "2/12/12 21:30 ", "2/12/12 21:45
",
"2/12/12 22:00 ", "2/12/12 22:15 ", "2/12/12 22:30
",
"2/12/12 22:45 ", "2/12/12 23:00 ", "2/12/12 23:15
",
"2/12/12 23:30 ", "2/12/12 23:45 ", "2/13/12 00:00
",
"2/13/12 00:15 ", "2/13/12 00:30 ", "2/13/12 00:45
",
"2/13/12 01:00 ", "2/13/12 01:15 ", "2/13/12 01:30
",
"2/13/12 01:45 ", "2/13/12 02:00 ", "2/13/12 02:15
",
"2/13/12 02:30 ", "2/13/12 02:45 ", "2/13/12 03:00
",
"2/13/12 03:15 ", "2/13/12 03:30 ", "2/13/12 03:45
",
"2/13/12 04:00 ", "2/13/12 04:15 ", "2/13/12 04:30
",
"2/13/12 04:45 ", "2/13/12 05:00 ", "2/13/12 05:15
",
"2/13/12 05:30 ", "2/13/12 05:45 ", "2/13/12 06:00
",
"2/13/12 06:15 ", "2/13/12 06:30 ", "2/13/12 06:45
",
"2/13/12 07:00 ", "2/13/12 07:15 ", "2/13/12 07:30
",
"2/13/12 07:45 ", "2/13/12 08:00 ", "2/13/12 08:15
",
"2/13/12 08:30 ", "2/13/12 08:45 ", "2/13/12 09:00
",
"2/13/12 09:15 ", "2/13/12 09:30 ", "2/13/12 09:45
",
"2/13/12 10:00 ", "2/13/12 10:15 ", "2/13/12 10:30
",
"2/13/12 10:45 ", "2/13/12 11:00
Re: [R] logical test not functioning correctly on zoo series...what the hell?
"I'm getting a 404 on the dropbox linklooking at your pastebin link, my guess is that those numbers are in fact greater than zero, but are very small so they appear as zeros in the print method. If you want to send your data, it's easier to simply give us dput(rawData) and copy and paste the result into the email thread. I sincerely doubt that the world's most powerful and commonly used statistical software, after 20 years of development, can't test for something being greater than a literal." Right, sorry about the miscommunication. Did you try the attached file I uploaded 2 hours before the post above? I just noticed the more button here can attach files. Here it is: http://r.789695.n4.nabble.com/file/n4473011/dataout_2471_843.csv dataout_2471_843.csv -- View this message in context: http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4473011.html Sent from the R help mailing list archive at Nabble.com. __ 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] logical test not functioning correctly on zoo series...what the hell?
attached http://r.789695.n4.nabble.com/file/n4472408/dataout_2471_843.csv dataout_2471_843.csv -- View this message in context: http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4472408.html Sent from the R help mailing list archive at Nabble.com. __ 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] logical test not functioning correctly on zoo series...what the hell?
Here's the exact error I'm receiving: http://pastebin.com/mNsPauwk Tracked each output along the way. Starting to think there's a bug in the source code. -- View this message in context: http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4472337.html Sent from the R help mailing list archive at Nabble.com. __ 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] logical test not functioning correctly on zoo series...what the hell?
"You're missing a comma between 0 and "]" in the last line if your goal is to retrieve the rows that satisfy that condition (and if the condition makes any sense). Haven't tested the rest of your code, though. ?Extract > The outputs that I'm getting, however, are > printing out 0's down the columns. I've tried various methods assuming > various theories, read the R manual via "?" for different possible > solutions, Googled stuff, tried the ifelse function which produces > the same > error, tried creating logical vectors to get a better visual of what > the > process is, tried doing all of the above on a small sample data set > (which > actually works correctly for some reason, but doesn't work on the > actual raw > data), and basically have done a lot of research and trial and error > before > posting on here. Help is much appreciated. Thank you. > > P.S. I hope I asked this correctly on the forum (putting > reproducible code > that reproduces the error, etc.). ... [show rest of quote] If there is an error then you should post the complete error message." About to post the error because it's still showing up. I understand the "," is supposed to act as an (x, y) or (i, j) separator such that [,1] would be column 1 similar to how rawData$KW.ch..1..set.0. looks at the same column. Let me know if you're able to retrieve the csv file from the dropbox link. Again, it should prompt you to download it. I would've put a smaller sample size, but the logical condition seemed to work on a small scale version unfortunately, thus being unable to reproduce the problem. -- View this message in context: http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4472301.html Sent from the R help mailing list archive at Nabble.com. __ 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] logical test not functioning correctly on zoo series...what the hell?
"> Here's the raw data I'm working with (will be available temporarily): > > http://dl.dropbox.com/u/41922443/dataout_2471_843.csv Nothing appears. " ^ Clicking on the link should prompt you to download a csv file and save it somewhere in your HDD. -- View this message in context: http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4472215.html Sent from the R help mailing list archive at Nabble.com. __ 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] logical test not functioning correctly on zoo series...what the hell?
"> The outputs that I'm getting, however, are > printing out 0's down the columns. I've tried various methods assuming > various theories, read the R manual via "?" for different possible > solutions, Googled stuff, tried the ifelse function which produces > the same > error, tried creating logical vectors to get a better visual of what > the > process is, tried doing all of the above on a small sample data set > (which > actually works correctly for some reason, but doesn't work on the > actual raw > data), and basically have done a lot of research and trial and error > before > posting on here. Help is much appreciated. Thank you. > > P.S. I hope I asked this correctly on the forum (putting > reproducible code > that reproduces the error, etc.). ... [show rest of quote] If there is an error then you should post the complete error message." Hm, technically there is no "error message"? It just prints zeros (the wrong output) as a result of explicitly making the logical condition " > 0". But anyway, I'll be sure to copy paste the error output next time. Thanks for the advice. -- View this message in context: http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4472236.html Sent from the R help mailing list archive at Nabble.com. __ 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] logical test not functioning correctly on zoo series...what the hell?
Here's the raw data I'm working with (will be available temporarily): http://dl.dropbox.com/u/41922443/dataout_2471_843.csv Next, here's the code I want to impose (error that I'm seeing SHOULD reproduce on your shell when script is executed...btw using Ubuntu Linux 11.10 if that makes a difference): http://pastebin.com/LDNw8UT0 The issue specifically is the last line of code in which I want to print out any value greater than 0. The outputs that I'm getting, however, are printing out 0's down the columns. I've tried various methods assuming various theories, read the R manual via "?" for different possible solutions, Googled stuff, tried the ifelse function which produces the same error, tried creating logical vectors to get a better visual of what the process is, tried doing all of the above on a small sample data set (which actually works correctly for some reason, but doesn't work on the actual raw data), and basically have done a lot of research and trial and error before posting on here. Help is much appreciated. Thank you. P.S. I hope I asked this correctly on the forum (putting reproducible code that reproduces the error, etc.). If I made some noob mistake, I apologize in advance so please don't be mad at me. I will clarify if necessary. -- View this message in context: http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4471654.html Sent from the R help mailing list archive at Nabble.com. __ 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] Filter/Ceiling for unwanted data - zoo
thinking of using an "ifelse" statement, curious if there's a more elegant way of going about this -- View this message in context: http://r.789695.n4.nabble.com/Filter-Ceiling-for-unwanted-data-zoo-tp4447595p4447791.html Sent from the R help mailing list archive at Nabble.com. __ 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] Filter/Ceiling for unwanted data - zoo
Here's my script: http://pastebin.com/zx3TCtXg I want to draw attention to the code block where the read in of the raw data is located. Is there a function that filters out unwanted data with respect to a ceiling limit. For example, I want to remove any value over 500 kW in the rawCool variable. Any ideas where to go with that? I figure it would be an argument within read.zoo or an external function that manipulates the zoo vector that was read in prior. I plan on looking into the zoo FAQ and the R manual '?'. Any help pushing me towards the right direction is much appreciated. -- View this message in context: http://r.789695.n4.nabble.com/Filter-Ceiling-for-unwanted-data-zoo-tp4447595p4447595.html Sent from the R help mailing list archive at Nabble.com. __ 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] Noob question - Identity argument within aggregate function?
Sorry, in regards to the previous post where I said aggregate(z, identity, tail, 1), replace it with aggregate(z, identity, mean) -- View this message in context: http://r.789695.n4.nabble.com/Noob-question-Identity-argument-within-aggregate-function-tp4439806p4440424.html Sent from the R help mailing list archive at Nabble.com. __ 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] Noob question - Identity argument within aggregate function?
I've also searched "?identity" in the R shell and it doesn't seem to be the definition I'm looking for for this particular usage of "identity" as an argument in the aggregate function. I simply would appreciate a conceptual explanation of what it does here and how it relates to the error. -- View this message in context: http://r.789695.n4.nabble.com/Noob-question-Identity-argument-within-aggregate-function-tp4439806p4440420.html Sent from the R help mailing list archive at Nabble.com. __ 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] Noob question - Identity argument within aggregate function?
z is a zoo object as a result from reading in the following series z = suppressWarnings(zoo(1:8), c(1, 2, 2, 2, 3, 4, 5, 5)) This is what z is in the aggregate function. So then that brings us to "aggregate(z, identity, tail, 1)". All I was trying to accomplish was trying to reproduce an example shown on the zoo faq. I've read ?aggregate via terminal and used "/identity" to search through the documentation for the specific term "identity". I'm just trying to understand what identity is used for because I do not understand the error statement. -- View this message in context: http://r.789695.n4.nabble.com/Noob-question-Identity-argument-within-aggregate-function-tp4439806p4440413.html Sent from the R help mailing list archive at Nabble.com. __ 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] Noob question - Identity argument within aggregate function?
>aggregate(z, identity, mean) 1 2 3 4 5 1.0 3.0 5.0 6.0 7.5 > aggregate(z, mean) Error: length(time(x)) == length(by[[1]]) is not TRUE Can someone help me understand the error above and why "identity" is necessary to satisfy the error -- View this message in context: http://r.789695.n4.nabble.com/Noob-question-Identity-argument-within-aggregate-function-tp4439806p4439806.html Sent from the R help mailing list archive at Nabble.com. __ 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] aggregating specific parts in zoo index column to perform sliding average
Thanks Gabor. Answering all my noob questions haha. Interesting strategy to adjust the input raw data as opposed to Excel where one offsets the output data instead. I'll remember that. -- View this message in context: http://r.789695.n4.nabble.com/aggregating-specific-parts-in-zoo-index-column-to-perform-sliding-average-tp4426798p4429257.html Sent from the R help mailing list archive at Nabble.com. __ 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] aggregating specific parts in zoo index column to perform sliding average
Here's my code:
http://pastebin.com/0yRxEVtm
The important parts are uncommented and should be easy to find using the
link above. For the following line of code, I plan on looking for a way to
offset it up 7 rows so that the 15 minute timestamp would be considered the
"median" of the subset being averaged to find the mean:
avgCool = aggregate(intCool, trunc(time(intCool), times("00:15:00")), mean)
Currently the issue is that, with the truncate function, it "truncates" but
really rounds down the time series values to the 15 minute time stamp
earlier in the series. For example, let's say we have one minute intervals
0:00, 0:01, 0:02,,0:37. It takes 0:00 - 0:14 and replaces that with
0:00. Then it sees 0:15, and changes values from 0:15 - 0:29 to 0:15. In
effect, aggregating the values and creating subsets.
What I want to do here is change 0:00 - 0:07 to 0:00, change 0:08 - 0:22 to
0:15, and change 0:23 - 0:37 to 0:30 in which 0:15 and 0:30 are the medians
of each subset. Anyway, I hope that makes sense. Any ideas on which function
will make this an easy job? Much thanks in advance.
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[R] question about user written function (newb question)
Quick newb question about R relating to the line of code below:
rawCool = read.zoo("cooling.txt", FUN = as.chron, format = "%m/%d/%Y %H:%M",
sep = "\t", aggregate = function(x) tail(x, 1))
I'm wondering what the specifics are for the argument where it has
"aggregate = function(x) tail(x, 1)". I understand that it removes the last
row of duplicates/aggregates in the zoo series. I'm confused as to why
"tail(x, 1)", a built in function in the utils package, requires the coder
to treat it as a user written function thus defining the assignment, in this
case an argument, with "function(x)". Why can't the coder just write
"tail(x, 1)" instead? Also, with the argument 'x', within tail, I'm assuming
it's looking at all columns simultaneously within the zoo series? Is that
correct to say? Thanks.
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