[R] readBin into a data frame
Hello. readBin is designed to read a batch of data with the same spec, e.g.
read 1 floats into a vector. In practise I read into data frame, not
vector. For each data frame, I need to read a integer and a float.
for (i in 1:1000) {
dataframe$int[i] <- readBin(con, integer(), size=2)
dataframe$float[i] <- readBin(con, numeric(), size=4)
}
And I need to read 100 such data files, ending up with a for loop in a for
loop. Something feels wrong here, as it is being said if you use double-FOR
you are not speaking R.
What is the R way of doing this? I can think of writing the content of the
loop into a function, and vectorize it -- But, the result would be a list of
list, not exactly data-frame, and the list grows incrementally, which is
inefficient, since I know the size of my data frame at the outset. I am a
new learner, not speaking half of R vocabulary, kindly provide some hint
please:)
Best.
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[R] use Vectorized function as range of for statement
I guess this has been discussed before, but I don't know the name of this
problem, thus had to ask again.
Consider this scenario:
fun <- function(x) { print(x)}
for (i in Vectorize(fun, "x")(1:3)) print("OK")
[1] 1
[1] 2
[1] 3
[1] "OK"
[1] "OK"
[1] "OK"
The optimal behaviour is:
fun <- function(x) { print(x)}
for (i in Vectorize(fun, "x")(1:3)) print("OK")
[1] 1
[1] "OK"
[1] 2
[1] "OK"
[1] 3
[1] "OK"
That is, each iteration of vectorized function should yield some result for
the 'for' statement, rather than having all results collected beforehand.
The intention of such a pattern, is to separates the data generation logic
from data processing logic.
The latter mechanism, I think, is more efficient because it doesn't cache
all data before processing -- and the interpreter has the sure knowledge
that caching is not needed, since the vectorized function is not used in
assignment but as a range.
The difference may be trivial, but this pseud code demonstrates otherwise:
readSample <- function(x) {
sampling_time <- readBin(con, integer(), 1, size=4)
sample_count <- readBin(con, integer(), 1, size=2)
samples <- readBin(con, float(), sample_count, size=4)
matrix # return a big matrix representing a sample
}
for (sample in Vectorize(readSample, "x")(1:1)) {
# process sample
}
The data file is a few Gigabytes, and caching them is not effortless. Not
having to cache them would make a difference.
This email asks to 1. validate this need of the langauge; 2. alternative
design pattern to workaround it; 3. Ask the proper place to discuss this.
Thanks and best...
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Re: [R] use Vectorized function as range of for statement
On Thu, 1 Aug 2013, Jeff Newmiller wrote: The Vectorize function is essentially a wrapped up for loop, so you are really executing two successive for loops. Note that the Vectorize function is not itself vectorised, so there is no particular advantage to using it in this way. You might as well call fun as a statement in the for loop. Thanks all who answered me! Now it is answered. __ 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] why Vectorize conjures a list, not a vector?
The manual seems to suggest, with the SIMPLIFY = TRUE default option, Vectorize would conjure a vector if possible. Quote: SIMPLIFY: logical or character string; attempt to reduce the result to a vector, matrix or higher dimensional array; see the ‘simplify’ argument of ‘sapply’. I assume, if each run of the function results a vector of the same type, the result should be a vector as well; there is a need of list only when data are of different type. Or, given vectors of the same type, conjure vectors of the same type. But it doesn't work that way -- see below -- so what's the magic inside? REPRODUCE: First, to make sure each run of the function always return vector of the same time: for (datafile in list.files(full.names=TRUE,"16b")) print(mode(list.files(full.names=TRUE,datafile))) [1] "character" [1] "character" [1] "character" [1] "character" [1] "character" [1] "character" [1] "character" [1] "character" [1] "character" [1] "character" [1] "character" Then, vectorize it: datafiles <- c(Vectorize(list.files, "path")(full.names=TRUE,path = list.files(base_dir,full.names=TRUE))) mode(datafiles) [1] "list" The same happened with sapply, which should generate list only if a vector is impossible -- it generated a list when every result is a vector: mode(sapply(list.files(base_dir,full.names=TRUE), list.files)) [1] "list" __ 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] why Vectorize conjures a list, not a vector?
On Wed, 14 Aug 2013, Hervé Pagès wrote: Hi Zhang, First note that a list is a vector (try is.vector(list())). The documentation for sapply() and Vectorize() should say *atomic* vector instead of vector in the desccription of the 'simplify' and 'SIMPLIFY' arguments. So in order for sapply() to be able to simplify the result, all runs of the function not only need to produce an atomic vector of the same type, but also of the same length. If this common length is 1, then the final result can be simplified to an atomic vector of the same length as the input. Thanks. Thanks to Jeff Newmiller as well, you answered clearly and the answer solves the problem.__ 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 retain dimension when selecting one row from a matrix?
When you select a single row from a matrix, the dimsion is lost: n <- matrix(nrow = 3, ncol = 5) dim(n) [1] 3 5 dim(n[1,]) NULL dim(n[2,]) NULL This doesn't happen if you select more than one row: dim(n[1:2,]) [1] 2 5 This is causing trouble. SCENARIO: when I filter out unqualified sampled data, I was not unaware that only one row of data qualifies, and the resulted selected data thus doesn't have dimension. I would call matrix[,3] and get an error. One way to mend this is to re-assign the dimension to the result set, but that destories column names... selected = n[1,] selected unixtime agiocount ask bid NA NA NA NA NA dim(selected) NULL dim(selected) <- c(1,5) selected [,1] [,2] [,3] [,4] [,5] [1,] NA NA NA NA NA Is there a way to retain dimension when selecting only one row from a matrix? __ 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] to match samples by minute
Perhaps this is simple and common, but it took me quite a while to admit I cannot solve it in a simple way. The data frame `df` has the following columns: unixtime, value, factor Now I need a matrix of: unixtime, value-difference-between-factor1-and-factor2 The naive solution is: df[df$factor == "factor1",] - df[df$factor == "factor2",] It won't work, because factor1 has 1000 valid samples, factor2 has 1400 valid samples. The invalid samples are dropped on-site, i.e. removed before piped into R. To solve it, I got 2 ideas. 1. create a new data.frame with 24*60 records, each record represent a minute in the day, because sampling is done once per minute. Now fit all records into their 'slots' by their nearest minute. 2. pair each record with another that has similar unixtime but different factor. Both ideas require for loop into individual records. It feels to C-like to write a program that way. Is there a professional way to do it in R? If not, I'd even prefer to rewrite the sampler (in C) to not to discard invalid samples on-site, than to mangle R. Thanks. __ 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] on how to make a skip-table
I've got two data frames, as shown below: (NR means Number of Record) record.lenths NR length 1 100 2 130 3 150 4 148 5 100 683 760 valida.records NR factor 1 3 2 4 4 8 7 9 And I intend to obtain the following skip-table: skip.table NR skip factor 1 0 3 2 0 4 4 150 8 7 183 9 The column 'skip' is the space needed to skip invalid records. For example, the 3rd element of skip.table has skip of '150', intended to skip the invalid record No.3 in record.lengths For example, the 4th element of skip.table has skip of '183', intended to skip the invalid record No.5 and No.6, together is 100+83. It's rather apparently intended for reading huge data files, and looks simple math, and I admit I couldn't find an R-ish way doing it. Thanks in advance and also thanks for pointing out if I had been on the right track to start with. __ 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] on how to make a skip-table
It is a nice surprise to wake up receiving three answers, all producing
correct results. Many thanks to all of you.
Jim Holtman solved it with amazing clarity. Gang Peng using a traditioanl
C-like pointer style and Arun with awesome tight code thanks to diff().
I am embrassed to see my mis-spellings inherited in the answers ('lenths'
should be 'lengths' and 'valida' should be 'valid'). This experience is to
behove me to not to code in midnight again.
For anyone wishing to test these methods, I have compiled them all into one
R script file, pasted at the end of this email.
Jim Holtman asked me to elaborate the problem:
It is a common problem in reading sparse variable-lenght record data
file. Records are stored in file one next to another. The length of
each record is known in advance, but a lot of them records are invalid,
and should be skipped to make efficient use of memory.
Ideally the datafile-reading routine should receive a skip-table. Before
reading each wanted/valid record, it seeks forward for the distance
given in the skip-table. The problem is how to obtain such a skip table.
What we have at hand to produce the skip table, is a set of two data
frames: a record.lengths data frame about each record's length, and a
valid.records data frame about which records are significant and should
be read.
--
## input data:
record.lengths <- read.table(text = "NR length
1 100
2 130
3 150
4 148
5 100
683
760", header = TRUE)
valid.records <- read.table(text = " NR factor
1 3
2 4
4 8
7 9", header = TRUE)
### Jim Holtman's method:
x <- merge(record.length, valid.records, by = "NR", all.x = TRUE)
x$seq <- cumsum(!is.na(x$factor))
# need to add 1 to lines with NA to associate with next group
x$seq[is.na(x$factor)] <- x$seq[is.na(x$factor)] + 1
# split by 'seq', output last record and sum of preceeding records
skip.table <- do.call(rbind
, lapply(split(x, x$seq), function(.sk){
if (nrow(.sk) > 1) .sk$skip <- sum(.sk$length[1:(nrow(.sk) - 1L)])
else .sk$skip <- 0
.sk[nrow(.sk), ] # return first value
})
)
print(skip.table)
### Gang Peng's method:
n.record <- length(record.lengths$NR)
index<- record.lengths$NR %in% valid.records$NR
tmp <- 1:n.record
ind <- tmp[index]
st <- 1
skip <- rep(0,length(ind))
for (i in 1:length(ind)) {
if(sthttps://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 get values within a threshold
input:
> values
[1] 0.854400 1.648465 1.829830 1.874704 7.670915 7.673585 7.722619
> thresholds
[1] 1 3 5 7 9
expected output:
[1] 1 4 4 4 7
That is, need a vector of indexes of the maximum value below the threshold.
e.g.
First element is "1", because value[1] is the largest below threshold "1".
Second element is "4", because value[4] is the largest below threshold "3".
The way I do it is:
sapply(1:length(threshold), function(x) { length(values[values <
threshold[x]])})
[1] 1 4 4 4 7
It just seem to me too long and stupid to be like R. Is it already the best way?
Somehow I feel which() was designed for a purpose like this, but I couldn't
figure out a way to apply which here.
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Re: [R] how to get values within a threshold
On Fri, 13 Sep 2013, William Dunlap wrote: findInterval(thresholds, values) [1] 1 4 4 4 7 Thanks a lot! But now I have a new problem, a typical R issue perhaps. First, let's look at a successful case: > thresholds <- c(1,3,5,7,9) > values <- c(0.854, 1.648, 1.829, 1.874, 7.670, 7.673, 7.722) > values[findInterval(thresholds, values)] [1] 0.854 1.874 1.874 1.874 7.722 Then a new batch of values came, notice only the first element of new values differ: > thresholds <- c(1,3,5,7,9) > values <- c(1.254, 1.648, 1.829, 1.874, 7.670, 7.673, 7.722) > findInterval(thresholds, values) [1] 0 4 4 4 7 > values[findInterval(thresholds, values)] [1] 1.874 1.874 1.874 7.722 This is a surprise. The desirable output is: [1] 0 1.874 1.874 1.874 7.722 This is desirable, because so maintains the same number of elements during calculation. (You may suggest leaving out the indices and try to calculate maximum-values-below-threshold directly, but the indices are useful to address other fields in the data frame whence values came.) This problem can be simplified as following: in R, we have: > a <- 1:10 > a[c(1,3)] [1] 1 3 > a[c(0,3)] [1] 3 While I was hoping to get: > a <- 1:10 > a[c(1,3)] [1] 1 3 > a[c(0,3)] [1] 0 3 The straightforward solution, is to shift the whole test values one position, so that the first value is always zero: > values <- c(0, 1.254, 1.648, 1.829, 1.874, 7.670, 7.673, 7.722) This solution, despite begetting a train of changes elsewhere in the code, is semantically wrong, since the first element of values should be the first value, now it is actually the 0-th value. What would you do in the case? __ 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 get values within a threshold
On Fri, 13 Sep 2013, William Dunlap wrote: You may want to append -Inf (or 0 if you know the data cannot be negative) to the start of your 'values' vector so you don't have to write code to catch the cases when a threshold is below the range of the values. > findInterval(thresholds, c(0,values,Inf)) [1] 1 5 5 5 8 > c(0, values, Inf)[.Last.value] [1] 0.000 1.874 1.874 1.874 7.722 Thanks a lot! I'll stick with this method for this project. Thanks a lot to arun as well, for profiling different methods. __ 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] the problem of buying and selling
I own a lot to the folks on r-help list, especially arun who answered every
of my question and was never wrong. I am disinclined to once again ask this
question, since it is more arithmatic than technical. But, having worked 2
days on it, I realized my brain is just not juicy enough
Here is the problem.
Trust not for freedom to the Franks---
They have a king who buys and sells.
- Lord Byron: The Isles of Greece
Suppose the French King commands you to buy and sell, and tells you only to
deal if the profit is higher than 2%. Question: how much quantity will be
dealt, and what is the actual profit? In fact, the King wants to see the
relationship between his minimum-profit requirement and your result, in
order to better his decision.
Let's look at the input data - a dump of which is attached to this mail.
Column 1 is the price of the market where you buy goods from, column 2 is
the quantity of goods that is being sold at that price.
Column 3 is the price of the market where you sell goods to, column 4 is the
quantity the buyers willing to buy at that price.
cbind(t(to_buy_from), t(to_sell_to))
[,1] [,2] [,3] [,4]
[1,] 61.7050 190 63.170 2500
[2,] 61.750029 63.150799
[3,] 61.8050 166 63.110500
[4,] 61.8950 166 63.060 1
[5,] 61.9450 166 63.020 7840
[6,] 61.9805 6150 62.995 2000
[7,] 62. 3069 62.930 2000
[8,] 62.0600 166 62.860 10811
[9,] 62.1100 166 62.780 18054
[10,] 62.1450 166 62.755 9000
[11,] 62.1750 166 62.690 10960
[12,] 62.2250 166 62.635100
[13,] 62.2450 166 62.585 2380
[14,] 62.2720 100 62.550 2119
[15,] 62.2830 4000 62.525 108091
[16,] 62.2875 100 62.505 2000
[17,] 62.2955 100 62.485816
[18,] 62.3250 307 62.435600
[19,] 62.3800 2906 62.400300
[20,] 62.3940 1969 62.375 4611
[21,] 62.4250 166 62.355 5111
[22,] 62.4505 2000 62.335 1969
[23,] 62.4700 259 62.315500
[24,] 62.475550 62.250 5142
[25,] 62.4800 166 62.165660
[26,] 62.4935 305 62.115 2428
[27,] 62.4975 7786 62.085779
[28,] 62.4995 50049 62.050 12811
[29,] 62.5045 914 62.015192
[30,] 62.5150 1110 61.975 1200
[31,] 62.5285 400 61.895 4
[32,] 62.5500 6352 61.835100
[33,] 62.5750 9 61.775133
[34,] 62.6000 394 61.750 7723
For the simpliest case, if the King had commanded that the minimum profit
should be 2.3742%, which is equal to 63.170/61.7050 (look at the first row),
then you can easily project that 190 quantity of goods will be dealt (the
minmum of [1,2] and [1,4]), and that the actual profit is 2.3742%.
If the king, however, has commanded that a deal should only be carried out
if the profit is higher than 2%, the calculation will be more complicated. I
don't know the right method, but I can demonstrate the wrong method and
explain why it is wrong.
The wrong approach is the following:
The idea is to write a function that asks how much volume (total quantity)
you want to deal, and returns the profit. This generates a relationship
between volume and profit, and with interpolation you can get the volumen
for any given minimum-profit requirement.
revenues <- function(open_orders, volumes) {
# calculate revenue using a list of open orders and desirable "volumes" of goods
# expecting volumnes as a vector, to test the revenue (total amont of money)
# for each volume (total amount of goods to deal) in the 'volumes'
volume <- sapply(1:length(open_orders[2,]),
function(x) { sum(open_orders[2,1:x])})
revenue <- sapply(1:length(open_orders[2,]),
function(x) { sum(open_orders[1, 1:x] * open_orders[2,1:x])})
i <- findInterval(volumes, c(0, volume))
c(0, revenue)[i] + c(open_orders[1,], 0)[i]*(
volumes - c(0, volume)[i])
}
data.frame(volume = volumes, profit = revenues(to_sell_to, volumes) /
revenues(to_buy_from, volumes) - 1)
With the above routine, let us test the profit with the following volumes:
volumes = c(10, 100, 500, 1000, 5000, 1, 3, 5, 7, 9)
And the result:
data.frame(volume = volumes, profit = revenues(to_sell_to, volumes) /
+ revenues(to_buy_from, volumes) - 1)
volume profit
1 10 0.023741938
2 100 0.023741938
3 500 0.022424508
41000 0.020974612
55000 0.018972785
6 1 0.018087976
7 3 0.012223652
8 5 0.009288480
9 7 0.007729286
10 9 0.006204251
So, by looking up the table, if the king requires minimum profit of 2%, the
volume (total quantity) of goods being dealt should be a bit more than 1000.
This answer is inexact, but our French King should get by with it. After
all, he remembers nothing more than the number of digits.
Now let's look at why it is wrong. This answer is, actually, correct, but
the method won'
Re: [R] the problem of buying and selling
On Sat, 14 Sep 2013, Zhang Weiwu wrote:
I own a lot to the folks on r-help list, especially arun who answered
every of my question and was never wrong. I am disinclined to once again
ask this question, since it is more arithmatic than technical. But, having
worked 2 days on it, I realized my brain is just not juicy enough
Here is the problem.
Trust not for freedom to the Franks---
They have a king who buys and sells.
- Lord Byron: The Isles of Greece
Suppose the French King commands you to buy and sell, and tells you only to
deal if the profit is higher than 2%. Question: how much quantity will be
dealt, and what is the actual profit? In fact, the King wants to see the
relationship between his minimum-profit requirement and your result, in order
to better his decision.
Let's look at the input data - a dump of which is attached to this mail.
Column 1 is the price of the market where you buy goods from, column 2 is the
quantity of goods that is being sold at that price.
Column 3 is the price of the market where you sell goods to, column 4 is the
quantity the buyers willing to buy at that price.
Forgive my carelessness. I should emphasize that there is only one type
goods to be dealt. The below table should be read in this way: there is 190
quantity of goods being sold at 61.7050 that you can buy from, and 29
quantity of goods beign sold at 61.7500 that you can buy from (row 1 and 2,
column 1 and 2). They are exactly the same type of goods, that you can sell
the total volume of 219 at the price of 63.170.
cbind(t(to_buy_from), t(to_sell_to))
[,1] [,2] [,3] [,4]
[1,] 61.7050 190 63.170 2500
[2,] 61.750029 63.150799
[3,] 61.8050 166 63.110500
[4,] 61.8950 166 63.060 1
[5,] 61.9450 166 63.020 7840
[6,] 61.9805 6150 62.995 2000
[7,] 62. 3069 62.930 2000
[8,] 62.0600 166 62.860 10811
[9,] 62.1100 166 62.780 18054
[10,] 62.1450 166 62.755 9000
[11,] 62.1750 166 62.690 10960
[12,] 62.2250 166 62.635100
[13,] 62.2450 166 62.585 2380
[14,] 62.2720 100 62.550 2119
[15,] 62.2830 4000 62.525 108091
[16,] 62.2875 100 62.505 2000
[17,] 62.2955 100 62.485816
[18,] 62.3250 307 62.435600
[19,] 62.3800 2906 62.400300
[20,] 62.3940 1969 62.375 4611
[21,] 62.4250 166 62.355 5111
[22,] 62.4505 2000 62.335 1969
[23,] 62.4700 259 62.315500
[24,] 62.475550 62.250 5142
[25,] 62.4800 166 62.165660
[26,] 62.4935 305 62.115 2428
[27,] 62.4975 7786 62.085779
[28,] 62.4995 50049 62.050 12811
[29,] 62.5045 914 62.015192
[30,] 62.5150 1110 61.975 1200
[31,] 62.5285 400 61.895 4
[32,] 62.5500 6352 61.835100
[33,] 62.5750 9 61.775133
[34,] 62.6000 394 61.750 7723
For the simpliest case, if the King had commanded that the minimum profit
should be 2.3742%, which is equal to 63.170/61.7050 (look at the first row),
then you can easily project that 190 quantity of goods will be dealt (the
minmum of [1,2] and [1,4]), and that the actual profit is 2.3742%.
If the king, however, has commanded that a deal should only be carried out if
the profit is higher than 2%, the calculation will be more complicated. I
don't know the right method, but I can demonstrate the wrong method and
explain why it is wrong.
The wrong approach is the following:
The idea is to write a function that asks how much volume (total quantity)
you want to deal, and returns the profit. This generates a relationship
between volume and profit, and with interpolation you can get the volumen for
any given minimum-profit requirement.
revenues <- function(open_orders, volumes) {
# calculate revenue using a list of open orders and desirable "volumes" of
goods
# expecting volumnes as a vector, to test the revenue (total amont of money)
# for each volume (total amount of goods to deal) in the 'volumes'
volume <- sapply(1:length(open_orders[2,]),
function(x) { sum(open_orders[2,1:x])})
revenue <- sapply(1:length(open_orders[2,]),
function(x) { sum(open_orders[1, 1:x] * open_orders[2,1:x])})
i <- findInterval(volumes, c(0, volume))
c(0, revenue)[i] + c(open_orders[1,], 0)[i]*(
volumes - c(0, volume)[i])
}
data.frame(volume = volumes, profit = revenues(to_sell_to, volumes) /
revenues(to_buy_from, volumes) - 1)
With the above routine, let us test the profit with the following volumes:
volumes = c(10, 100, 500, 1000, 5000, 1, 3, 5, 7, 9)
And the result:
data.frame(volume = volumes, profit = revenues(to_sell_to, volumes) /
+ revenues(to_buy_from, volumes) - 1)
volume profit
1 10 0.023741938
2 100 0.023741938
3 500 0.022424508
41000 0.020974612
55000 0.018972785
6
Re: [R] Instructions for upgrading R on ubuntu
On Sun, 15 Sep 2013, Andrew Crane-Droesch wrote: The c2d4u PPA is the main search result when googling "upgrade R 3.0.1 ubuntu". And it should be, because it is more likely that a PPA re-distribution works better for Ubuntu than a general distribution, even if it is an exceptional case with this software this time. You must be in bad need of 64-bit memory access or long vectors†, to do a manual upgrade just one month ahead of Ubuntu's own maintenance upgrade (13.10) with contains R-3.0.1. † They are the major features offered by R-3, so I guess most users are unlikely argued into hurrying an upgrade from R-2.x http://www.r-bloggers.com/r-3-0-0-is-released-whats-new-and-how-to-upgrade/__ 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.
