Displaying 14 results from an estimated 14 matches for "10000l".
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1000l
2018 Mar 13
4
Possible Improvement to sapply
...if (USE.NAMES && is.character(X) && is.null(names(answer)))
names(answer) <- X
if (simplify && length(answer))
simplify2array(answer, higher = (simplify == "array"))
else answer
}
> microbenchmark(sapply(myList, length), times = 10000L)
Unit: microseconds
expr min lq mean median uq max neval
sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46 10000
> microbenchmark(mySapply(myList, length), times = 10000L)
Unit: microseconds
expr min lq mean m...
2018 Mar 13
0
Possible Improvement to sapply
...cter(X) && is.null(names(answer)))
> names(answer) <- X
> if (simplify && length(answer))
> simplify2array(answer, higher = (simplify == "array"))
> else answer
> }
>
>
> > microbenchmark(sapply(myList, length), times = 10000L)
> Unit: microseconds
> expr min lq mean median uq max neval
> sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46 10000
> > microbenchmark(mySapply(myList, length), times = 10000L)
> Unit: microseconds
>...
2018 Mar 13
1
Possible Improvement to sapply
...if (USE.NAMES && is.character(X) && is.null(names(answer)))
names(answer) <- X
if (simplify && length(answer))
simplify2array(answer, higher = (simplify == "array"))
else answer
}
> microbenchmark(sapply(myList, length), times = 10000L)
Unit: microseconds
expr min lq mean median uq max neval
sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46 10000
> microbenchmark(mySapply(myList, length), times = 10000L)
Unit: microseconds
expr min lq mean m...
2018 Mar 13
0
Possible Improvement to sapply
...X) && is.null(names(answer)))
> names(answer) <- X
> if (simplify && length(answer))
> simplify2array(answer, higher = (simplify == "array"))
> else answer
> }
>
>
>> microbenchmark(sapply(myList, length), times = 10000L)
> Unit: microseconds
> expr min lq mean median uq max neval
> sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46 10000
>> microbenchmark(mySapply(myList, length), times = 10000L)
> Unit: microseconds
>...
2018 Mar 13
2
Possible Improvement to sapply
...X) && is.null(names(answer)))
> names(answer) <- X
> if (simplify && length(answer))
> simplify2array(answer, higher = (simplify == "array"))
> else answer
> }
>
>
>> microbenchmark(sapply(myList, length), times = 10000L)
> Unit: microseconds
> expr min lq mean median uq max neval
> sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46
> 10000
>> microbenchmark(mySapply(myList, length), times = 10000L)
> Unit: microseconds
>...
2010 May 11
1
merge two data frames
...ER", "ROBUSTA COFFEE (10)", "SILVER ", "SOYBEAN MEAL ",
"SOYBEAN OIL", "SOYBEANS ", "SPCL HIGH GRADE ZINC USD", "STANDARD LEAD USD",
"SUGAR WHITE", "WHEAT"), class = "factor"), VALUE = c(500L, 10000L,
420L, 420L, 100L, 5000L, 100L, 50L, 250L, 25L, 25L, 6L, 25L,
10L, 375L, 10L, 500L, 150L, 1120L, 50L, 50L, 50L, 100L, 600L,
50L, 500L, 400L, 400L, 110L)), .Names = c("SHORTDESCRIPTION",
"VALUE"), class = "data.frame", row.names = c(NA, -29L))
I want to merge these...
2018 Mar 13
1
Possible Improvement to sapply
...> > names(answer) <- X
> > if (simplify && length(answer))
> > simplify2array(answer, higher = (simplify == "array"))
> > else answer
> > }
> >
> >
> >> microbenchmark(sapply(myList, length), times = 10000L)
> > Unit: microseconds
> > expr min lq mean median uq max
> neval
> > sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46
> > 10000
> >> microbenchmark(mySapply(myList, length), times = 10000L)
> > Uni...
2018 Mar 13
0
Possible Improvement to sapply
...r(X) && is.null(names(answer)))
> names(answer) <- X
> if (simplify && length(answer))
> simplify2array(answer, higher = (simplify == "array"))
> else answer
> }
>
>
>> microbenchmark(sapply(myList, length), times = 10000L)
> Unit: microseconds
> expr min lq mean median uq max neval
> sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46
> 10000
>> microbenchmark(mySapply(myList, length), times = 10000L)
> Unit: microseconds
>...
2008 Nov 19
1
more efficient small subsets from moderate vectors?
This creates a named vector of length nx, then repeatedly draws a
single sample from it.
lkup <- function(nx, m=10000L) {
tbl <- seq_len(nx)
names(tbl) <- as.character(tbl)
v <- sample(names(tbl), m, replace=TRUE)
system.time(for(k in v) tbl[k], gcFirst=TRUE)
}
There is an abrupt performance degredation at nx=1000
> lkup(1000)
user system elapsed
0.180 0.000 0.179
> lkup(1...
2014 Mar 05
1
[PATCH] Code coverage support proof of concept
...roject.org
+% Copyright 1995-2010 R Core Team
+% Distributed under GPL 2 or later
+
+\name{Rcov_start}
+\alias{Rcov_start}
+\title{Start Code Coverage analysis of R's Execution}
+\description{
+ Start Code Coverage analysis of the execution of \R expressions.
+}
+\usage{
+Rcov_start(nb_lines = 10000L, growth_rate = 2)
+}
+\arguments{
+ \item{nb_lines}{
+ Initial max number of lines per source file.
+ }
+ \item{growth_rate}{
+ growth factor of the line numbers vectors per filename.
+ If a reached line number L is greater than nb_lines, the vector will
+ be reallocated with prov...
2018 May 18
0
Error message truncation
...170)
error(_("invalid value for '%s'"), CHAR(namei));
R_WarnLength = k;
SET_VECTOR_ELT(value, i, SetOption(tag, argi));
}
Further, it appears there's a physical limit on the length of the error
message itself which is only slightly larger than 8170:
set.seed(1023)
NN = 10000L
str = paste(sample(letters, NN, TRUE), collapse = '')
# should of course be 10000
tryCatch(stop(str), error = function(e) nchar(e$message))
# [1] 8190
My questions are:
- Can we add some information to the help pages indicating valid values
of options('warning.length')?...
2018 Mar 13
2
Possible Improvement to sapply
...swer)))
>> names(answer) <- X
>> if (simplify && length(answer))
>> simplify2array(answer, higher = (simplify == "array"))
>> else answer
>> }
>>
>>
>>> microbenchmark(sapply(myList, length), times = 10000L)
>> Unit: microseconds
>> expr min lq mean median uq max neval
>> sapply(myList, length) 14.156 15.572 16.67603 15.926 16.634 650.46
>> 10000
>>> microbenchmark(mySapply(myList, length), times = 10000L)
>> Unit: microsec...
2012 Feb 21
2
Dataframes in PLS package
I have been working with the pls procedure and have problems getting the
procedure to work with matrix or frame data. I suspect the problem lies in
my understanding of frames, but can't find anything in the documentation
that will help.
Here is what I have done:
I read in an 10000 x 8 table of data, and assign the first four columns to
matrix A and the second four to matrix B
pls <-
2012 Oct 18
4
speeding read.table
R 2.15.1
OS X
Colleagues,
I am reading a 1 GB file into R using read.table. The file consists of 100 tables, each of which is headed by two lines of characters.
The first of these lines is:
TABLE NO. 1
The second is a list of column headers.
For example:
TABLE NO. 1
COL1 COL2 COL3 COL4 COL5 COL6 COL7 COL8 COL9 COL10