Displaying 20 results from an estimated 236 matches for "q4".
2004 Apr 29
1
I'm trying to use package ts (decompose). How do you set up the data/ See attached. thanks
InDATA <-read.table("C:/Data/May 2004/season.txt",header=T)
X <- decompose(InDATA)
print(X)
Period Connections
Q1 67519
Q2 69713
Q3 68920
Q4 69452
Q1 70015
Q2 59273
Q3 57063
Q4 65596
Q1 73527
Q2 58586
Q3 69522
Q4 60091
Q1 51686
Q2 63490
Q3 55702
Q4 53200
Q1 51033
Q2 48175
Q3 52709
Q4 50106
Q1 50855
Q2 43466
Q3 48190
Q4 41702
Q1 48747
Q2 51441
Q3 42537
Q4 49145
Q1 41457
Q2 39306
Q3 43121
Q4 42777
Q1 40631
Q2...
2010 Mar 11
4
Forecast
sample report data that i want to forecast
quarter quarter_index Revenue
2007 Q1 1 $3,856,799
2007 Q2 2 $4,243,328
2007 Q3 3 $4,930,369
2007 Q4 4 $5,443,579
2008 Q1 5 $5,164,830
2008 Q2 6 $5,104,413
2008 Q3 7 $5,713,240
2008 Q4 8 $6,509,331
2009 Q1 9 $6,271,951
2009 Q2 10 $6,657,190...
2012 Feb 03
3
Cannot get "==" operator to return TRUE
...ot;GOOG?", "GOOG?", "GOOG?",
"GOOG?", "GOOG?", "GOOG?", "GOOG?", "GOOG?", "GOOG?", "GOOG?",
"GOOG?", "GOOG?", "GOOG?", "GOOG?", "GOOG?"), PERIOD = c("Q4?2011",
"Q3?2011", "Q2?2011", "Q1?2011", "Q4?2010", "Q3?2010", "Q2?2010",
"Q1?2010", "Q4?2009", "Q3?2009", "Q2?2009", "Q1?2009", "Q4?2008",
"Q3?2008", "Q2?2008&...
2010 Mar 10
3
see the example and help me
sample report data that i want to forecast
quarter quarter_index Revenue
2007 Q1 1 $3,856,799
2007 Q2 2 $4,243,328
2007 Q3 3 $4,930,369
2007 Q4 4 $5,443,579
2008 Q1 5 $5,164,830
2008 Q2 6 $5,104,413
2008 Q3 7 $5,713,240
2008 Q4 8 $6,509,331
2009 Q1 9 $6,271,951
2009 Q2 10 $6,657,190...
2012 Apr 12
4
Definition of "lag" is opposite in ts and xts objects!
Example:
Will ts objects be obsolete or modified?
> a [,1]
1983 Q1 2.747365190
1983 Q2 2.791594762
1983 Q3 -0.009953715
1983 Q4 -0.015059485
1984 Q1 -1.190061246
1984 Q2 -0.553031799
1984 Q3 0.686874720
1984 Q4 0.953911035> lag(a,4) [,1]
1983 Q1 NA
1983 Q2 NA
1983 Q3 NA
1983 Q4 NA
1984 Q1 2.747365190
1984 Q2 2.791594762
1984 Q3 -0.009953715
1984 Q4 -0.015059485&g...
2008 Mar 02
1
question on lag.zoo
...g what I assumed.
> x <- zoo(11:21)
> z <- zoo(1:10, yearqtr(seq(1959.25, 1961.5, by = 0.25)), frequency = 4)
> x
1 2 3 4 5 6 7 8 9 10 11
11 12 13 14 15 16 17 18 19 20 21
> lag(x)
1 2 3 4 5 6 7 8 9 10
12 13 14 15 16 17 18 19 20 21
> z
1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4 1961 Q1 1961 Q2 1961 Q3
1 2 3 4 5 6 7 8 9 10
> lag(z)
1959 Q1 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4 1961 Q1 1961 Q2
1 2 3 4 5 6 7 8...
2005 Nov 12
1
computation on a table
Hello,
I have a table (1) of the form
q1 q3 q4 q8 q9
A 5 2 0 1 3
B 2 0 2 4 4
I have another table (2):
q1 q2 q3 q4 q5 q6 q7 q8 q9
C 10 7 4 2 6 9 3 1 2
I would like to divide the numbers in table (1) by the number of the
appropriate column in table (2):
q1 q3 q4 q8 q9
A 5/10 2/4 0/2...
2009 Feb 19
2
table with 3 variables
I have the initial matrice:
> *data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste("Q",1:4,
sep=""),2), Boolean = rep(c("Y","N"),4))*
Subject Quarter Boolean
1 100 Q1 Y
2 100 Q2 N
3 100 Q3 Y
4 100 Q4 N
5 101 Q1 Y
6 101 Q2 N
7 101 Q3 Y
8 101 Q4 N
...
>
And I would like to group the Subject by Quarter using as a result in the
table the value of the third variable (Boolean). The final result would
give:
Subjet Q1 Q2 Q3 Q4
1 10...
2010 Jan 24
2
Creating directories & folders
Dear R users,
I would like to create the following 3 folders (FUND1, FUND2, MARINE) within
the 'parent.dir' as defined below.
FUND1 <- "FD1 Q4 2009"
FUND2 <- "FD2 Q4 2009"
MARINE <- "MARINE Q4 2009"
parent.dir <- "D:/....................."
folders <- c("FUND1", "FUND2", "MARINE")
for (i in 1:length(folders)) {
dir.create(do.call("paste", c(paren...
2008 Feb 01
2
the "union" of several data frame rows
Hi,
I have a question about how to obtain the union of several data frame
rows. I'm trying to create a common key for several tests composed of
different items. Here is a small scale version of the problem. These
are keys for 4 different tests, not all mutually exclusive:
id q1 q2 q3 q4 q5 q6
1 A C
2 B D
3 A D B
4 C D B D
I would like to create a single key all test versions, the "union" of
the above:
id q1 q2 q3 q4 q5 q6
key A C D B B D
Here is what I have (unsuccessfully) tried so far:
> key <-
+...
2011 Sep 01
0
[PATCH 5/5] resample: Add NEON optimized inner_product_single for floating point
...s when len % 4 == 0 */
+static inline float inner_product_single(const float *a, const float *b, unsigned int len)
+{
+ float ret;
+ uint32_t remainder = len % 16;
+ len = len - remainder;
+
+ asm volatile (" cmp %[len], #0\n"
+ " bne 1f\n"
+ " vld1.32 {q4}, [%[b]]!\n"
+ " vld1.32 {q8}, [%[a]]!\n"
+ " subs %[remainder], %[remainder], #4\n"
+ " vmul.f32 q0, q4, q8\n"
+ " bne 4f\n"
+ " b 5f\n"
+ "1:"
+ " vld1.32 {q4, q5}, [%[b]]!\n"
+ " v...
2009 Feb 19
2
table with 3 varialbes
I have the initial matrice:
> *data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste("Q",1:4,
sep=""),2), Boolean = rep(c("Y","N"),4))*
Subject Quarter Boolean
1 100 Q1 Y
2 100 Q2 N
3 100 Q3 Y
4 100 Q4 N
5 101 Q1 Y
6 101 Q2 N
7 101 Q3 Y
8 101 Q4 N
...
>
And I would like to group the Subject by Quarter using as a result in the
table the value of the third variable (Boolean). The final result would
give:
Subjet Q1 Q2 Q3 Q4
1 10...
2011 Mar 14
2
data.frame transformation
Hi R users,
I have following data frame
df<-data.frame(q1=c(0,0,33.33,"check"),q2=c(0,33.33,"check",9.156),
q3=c("check","check",25,100),q4=c(7.123,35,100,"check"))
and i would like to replace every element that is less then 10 with . (dot)
in order to obtain this:
q1 q2 q3 q4
1 . . check .
2 . 33.33 check 35
3 33.33 check 25 100
4 check . 100 check
I had a lot of difficulties be...
2009 Jul 11
3
Reading data entered within an R program
...ould you present?
Thanks,
Bob
http://RforSASandSPSSusers.com
# R Program to Read Data Within a Program.
# Very similar to SAS datalines or cards statements,
# and SPSS BEGIN DATA / END DATA commands.
# This stores the data as one long text string.
mystring <-
"workshop,gender,q1,q2,q3,q4
01,1,f,1,1,5,1
02,2,f,2,1,4,1
03,1,f,2,2,4,3
04,2, ,3,1, ,3
05,1,m,4,5,2,4
06,2,m,5,4,5,5
07,1,m,5,3,4,4
08,2,m,4,5,5,5"
# The textConnection function allows read.csv to
# read data from the text string just as it would
# from a file.
# The leading zero on first column helps show that
# R i...
2007 Jul 23
0
Problem w/ MySQL update from perl AGI script
...DBI;
$DATETIME = '123188765' ;
$TIMESTAMP = '8675309';
$dsn = "DBI:mysql:DBname;localhost;3306";
$dbh = DBI->connect($dsn,username,password);
$drh = DBI->install_driver("mysql");
my $Q1 = "8";
my $Q2 = "6";
my $Q3 = "7";
my $Q4 = "5";
my $Q5 = "3";
my $query = "INSERT into caller_data (DateTime, ANI, Q1, Q2, Q3, Q4, Q5)
VALUES($TIMESTAMP,$ANI,$Q1,$Q2,$Q3,$Q4,$Q5)";
$sth = $dbh->prepare($query);
$sth->execute;
$sth->finish;
Then, When I try an put it together with the AGI com...
2012 Feb 17
4
How can I tabulate time series data (in RStudio or any other R editor)?
Hello,
I have a question on how to tabulate the time series data. I use
RStudio, but if can be done in any other R editor, it should work in
RStudio as well.
> a1<-11:22
> a1ts<-ts(a1, frequency=4, start=c(1978,1))
> a1ts Qtr1 Qtr2 Qtr3 Qtr4
1978 11 12 13 14
1979 15 16 17 18
1980 19 20 21 22
If I click the variable "a1ts" on the
2011 Jul 06
3
Tables and merge
...join the 21 files into one, to construct
> tables for each question by CODE.
>
> I tried the command (8 files only):
>
> require(foreign)
> q1 = read.epiinfo('Dados/Q1.rec')
> q2 = read.epiinfo('Dados/Q2.rec')
> q3 = read.epiinfo('Dados/Q3.rec')
> q4 = read.epiinfo('Dados/Q4.rec')
> q5 = read.epiinfo('Dados/Q5.rec')
> q6 = read.epiinfo('Dados/Q6.rec')
> q7 = read.epiinfo('Dados/Q7.rec')
> q8 = read.epiinfo('Dados/Q8.rec')
>
> juntos = merge(q1,q2,q3,q4,q5,q6,q7,q8)
>
> But it didn&...
2018 Jan 28
2
Plotting quarterly time series
I have a data set with quarterly time series for several variables. The
time index is recorded in column 1 of the dataframe as a character vector
"Q1 1961", "Q2 1961","Q3 1961", "Q4 1961", "Q1 1962", etc. I want to
produce line plots with ggplot2, but it seems I need to convert the time
index from character to date class. Is that right? If so, how do I make
the conversion?
2020 Sep 14
2
Cross compiling for ARMv7-m
...libm.a are
under $SYSROOT/libc/usr/lib
(and I did try --sysroot=...../libc --sysroot=.../libc/usr and
--sysroot=../libc/usr/lib, none of which worked)
Any idea?
Thanks,
Christophe
> > cmake -G "Unix Makefiles"
> > -DBAREMETAL_ARMV6M_SYSROOT=../../gcc-arm-none-eabi-9-2019-q4-major/arm-none-eabi/
> > -DBAREMETAL_ARMV7M_SYSROOT=../../gcc-arm-none-eabi-9-2019-q4-major/arm-none-eabi/
> > -DBAREMETAL_ARMV7EM_SYSROOT=../../gcc-arm-none-eabi-9-2019-q4-major/arm-none-eabi/
> > -DCMAKE_BUILD_TYPE=Release -C ../clang/cmake/caches/BaremetalARM.cmake
> > .....
2020 Feb 26
2
Cross compiling for ARMv7-m
...n5.nabble.com/llvm-dev-Compiling-for-baremetal-ARMv4-on-Ubuntu-Linux-td124226.html
After going through this I figured my best bet would be using GCC
sysroot. After which my cmake command looks like this.
cmake -G "Unix Makefiles"
-DBAREMETAL_ARMV6M_SYSROOT=../../gcc-arm-none-eabi-9-2019-q4-major/arm-none-eabi/
-DBAREMETAL_ARMV7M_SYSROOT=../../gcc-arm-none-eabi-9-2019-q4-major/arm-none-eabi/
-DBAREMETAL_ARMV7EM_SYSROOT=../../gcc-arm-none-eabi-9-2019-q4-major/arm-none-eabi/
-DCMAKE_BUILD_TYPE=Release -C ../clang/cmake/caches/BaremetalARM.cmake
../llvm
But with this, I am getting this...