, :
[RODBC] ERROR: Could not SQLExecDirect
37000 -3553 [Microsoft][ODBC Excel Driver] Syntax error in field definition.
Am I doing it wrong way or is there a problem with the Excel driver?
Thank you in advance,
Moshe Olshansky
Chimaera Capital Group
Moshe Olshansky
Chimaera Capital Limited
Level 4
: Saturday, 24 March 2007 3:44 AM
To: Moshe Olshansky
Cc: R Help
Subject: Re: [R] Updating a worksheet in Excel file using RODBC
Hi,
2007/3/23, Moshe Olshansky [EMAIL PROTECTED]:
Hello!
I have no problem reading Excel files (each worksheet in the file is
a table which can be read - at least in my
. manually create an Excel file with a
small VBA macro, make many copies of this file (under appropriate names), write
an appropriate data to each file and then the macro will work on the right data
(different for each file).
Thanky you!
Moshe.
Moshe Olshansky
Chimaera Capital Limited
that there
was no such table!
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: Tuesday, 27 March 2007 4:44 PM
To: Moshe Olshansky
Cc: [EMAIL PROTECTED]
Subject: Re: Updating a worksheet in Excel file using RODBC
Yes, sqlDrop does not work correctly for Excel worksheet
Hello!
I have a short question: Is it possible to create a
(non-existing) Access database using R (and if yes,
how)? I need to create a new database and then insert
a few tables into it.
Thank you in advance,
Moshe Olshansky
[EMAIL PROTECTED]
__
R
).
Hope this helps,
Moshe Olshansky.
__
R-help@stat.math.ethz.ch 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
).
Regards,
Moshe Olshansky
[EMAIL PROTECTED]
__
R-help@stat.math.ethz.ch 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
Is there a convenient way to respond to a particular
posting which is a part of the digest?
I mean something that will automatically quote the
original message, subject, etc.
Thank you!
Moshe Olshansky
[EMAIL PROTECTED]
__
R-help@stat.math.ethz.ch
Hi Yuchen,
In R, if you do not put an explicit return statement
in the function, the value the function returns is the
value of the last statement in the function. Unlike
VB, it does not matter whether you assign this value
to aaa (which is identical to your function name) or b
or c or x etc.
So
a few minutes!
Languages like R and Matlab are extreemely convenient
but if performance is a very important issue you shoul
use C, Fortran, C++, etc.
Regards,
Moshe Olshansky.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman
Hi Tolga,
I do not see any problem with:
max {a1*x1 + a2*x2 + ... + a800*x800}
subject to:
x1+x2+ ... + x800 = 100
b1*x1+b2*x2+ ... +b800*x800 = B
c1*x1+c2*x2+ ... +c800*x800 = C
and an additional condition that x1,x2,...,x800 are
binary 0-1.
Regards,
Moshe Olshansky
--- Tolga Uzuner [EMAIL
An ugly solution (but which works) would be something
like this (assuming that d is a data.frame with 4
columns named Tank, Trt, Fish, Size):
x -1:12
for (i in 1:12) x[i] - mean(d$Size[d$Tank == i])
Regards,
Moshe Olshansky
--- Ronaldo Reis Junior [EMAIL PROTECTED] wrote:
Em Quinta 21 Junho
Try ReadWriteXls package from CRAN.
--- dala [EMAIL PROTECTED] wrote:
I have a 2 columns, Date and Number, in Excel.
I copy and paste them into Notepad.
I can use scan() to import the file but how do I
plot this data with Date as
the x-axis?
--
View this message in context:
Also try xlsReadWrite package on CRAN.
--- Erika Frigo [EMAIL PROTECTED] wrote:
Good morning to everybody,
I have a problem : how can I import excel files in
R???
thank you very much
Dr.sa. Erika Frigo
Università degli Studi di Milano
Facoltà di Medicina Veterinaria
Dipartimento
This is not surprizing at all!
The exact result is log(8,2) = 3, but the numerical
procedure which calculates the logarithm may produce a
result which is a few ULPs different from the exact
one, i.e. you can get that log(8,2) = 2.99
and then floor(2.99) = 2.
--- Fausto
All the nontrivial solutions to AX = 0 are the
eigenvectors of A corresponding to eigenvalue 0 (try
eigen function).
The non-negative solution may or may not exist. For
example, if A is a 2x2 matrix Aij = 1 for 1 =i,j =2
then the only non-trivial solution to AX = 0 is X =
a*(1,-1), where a is any
If your original matrix is A then
unique(A$People) and unique(A$Desc)
will produce a vector of different people and a vector
of different descriptions.
--- yoo [EMAIL PROTECTED] wrote:
Hi all,
let's say I have matrix
PeopleDescValue
Mary Height50
Mary
This is partially true since both the function to be
maximized and the constraint are non-linear. One may
substitute 1-x1-x2 for x3 and use (let say) Lagrange
multipliers to get two non-linear equations with 2
unknowns for which there should be a function solving
them. Then you must find the
Try the xlsReadWrite package from CRAN.
--- Mr Natural [EMAIL PROTECTED] wrote:
Dear Natalia:
I have had the same problem. Solved it by keeping my
dates in Excel files
with the rest of my data. When I want to do
something in R, highlight the
date column and format it as general You will
[EMAIL PROTECTED] wrote:
G'day Moshe,
On Tue, 17 Jul 2007 17:32:52 -0700 (PDT)
Moshe Olshansky [EMAIL PROTECTED] wrote:
This is partially true since both the function to
be
maximized and the constraint are non-linear.
I am not sure what your definition of non-linear is,
but in my book
Hi Tobias,
Could you please explain the role of x and y - are
they somehow related to S1,S2,S3,S4? Are they
constant? Are they additional variable?
What was your original problem (without the slack
variables)?
Regards,
Moshe.
--- Tobias Schlottmann [EMAIL PROTECTED]
wrote:
Hi everybody,
Hi Suzanne,
My solution (which I am sure is not the best) would
be:
heat - read.table('temp.txt')
heat
X1905 X1910 X1950 X1992 X2011 X2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat0.40 0.54 0.52 0.60 0.60 0.63
Cat0.16 0.27 0.29
I was right saying that my solution was not the best
possible!
--- Prof Brian Ripley [EMAIL PROTECTED] wrote:
read.table('temp.txt', check.names = FALSE)
would be easier (and more general, since make.names
can do more than
prepend an 'X').
On Wed, 18 Jul 2007, Moshe Olshansky wrote
After multiplication by 10 you get 6*8 = 48 - the
result is an exact machine number so there is no
roundoff, while 0.6*0.8 = 0.48, where neither of the 3
numbers (0.6, 0.8, 0.48) is an exact machine mumber.
However, (-0.6)*0.8 should be equal EXACTLY to
-(0.6*0.8), and in fact you get that
Hi Montserrat,
What exactly would you like to plot?
Your differential equation can be easily integrated so
that you can get an implicit expression for F(x), i.e.
expression like G(c,x,F(x)) = 0 where G is a known
function and c is an arbitrary constant. For every
value of c and each value of x
Hi Sebastian,
Your equations can be easily solved - no programming
is required!
Let's number you equations:
(1) 0.007= 2VZ
(2) 0.03= W(Y+Z)
(3) 0.034= X(y+Z)
(4) 0.013 = (X+W)Y +(X-W)Z
(5) X = W+V
Substitute (5) into (3) and then divide (2) by (3) to
get
(W+V)/W = 0.034/0.03
so that
(6) V =
Hi Ryan,
I think that the only function you need is the one
which generates random variable. The simplest one is
to generate a uniform (0,1) variable U and then do not
move if U 0.25, move to the first nearest point if
0.25 = U 0.5, move to the second nearest point if
).5 = U 0.75 and move to
Hi SK,
If I understand you correctly you are trying to assign
to each point of a 10x10 2D grid a normal variable
with a certain dependency between the values at each
grid point. Is this correct?
If so, please explain the dependence structure. Do you
need something like X(i,j) = f(i,j) + Z(i,j)
Hi Nair,
If the two populations are normal the t-test gives you
the exact result for whatever the sample size is (the
sample size will affect the number of degrees of
freedom).
When the populations are not normal and the sample
size is large it is still OK to use t-test (because of
the Central
Hi Animesh,
Can you send an example of an Excel file you need to
process (and the result you wish to get)?
Regards,
Moshe.
--- Acharjee, Animesh [EMAIL PROTECTED]
wrote:
Dear All,
I am new to R. I need to read XLs
files, parse the data
and convert them into txt format for
:
Indeed, I understand what you say. The df of freedom
for the dummy example is n1+n2-2 = 8. But when I run
the t.test I get it as 5.08, am I missing something?
-Original Message-
From: Moshe Olshansky [mailto:[EMAIL PROTECTED]
Sent: Tuesday, August 07, 2007 9:05 PM
To: Nair
Well, this an explanation of what is done in the
paired t-test (and why the number of df is as it is).
I was too lazy to write all this.
It is nice that some list members are less lazy!
--- Rolf Turner [EMAIL PROTECTED] wrote:
On 9/08/2007, at 2:57 PM, Moshe Olshansky wrote:
As Thomas
PROTECTED]
[mailto:[EMAIL PROTECTED] On
Behalf Of Nair,
Murlidharan T
Sent: Thursday, August 09, 2007 9:19 AM
To: Moshe Olshansky; Rolf Turner;
r-help@stat.math.ethz.ch
Subject: Re: [R] small sample techniques
Thanks, that discussion was helpful. Well, I have
another question
I am
P(T+,D+)=P(T+|D+)*P(D+)
Thank you for your reply,
Sigalit.
On 8/13/07, Moshe Olshansky [EMAIL PROTECTED]
wrote:
Hi Sigalit,
Do you want to find the probability P(T+ = t AND
D+ =
d) for all the combinations of t and d (for ICU
and
Reg.)?
Is the probability of false detection
While inverting the matrix may be a problem, if you
need to solve an equation A*x = b you do not need to
invert A, there exist iterative methods which do need
A or inv(A) - all you need to provide is a function
that computes A*x for an arbitrary vector x.
For such a large matrix this may be slow
again,
Sigalit.
On 8/14/07, Moshe Olshansky [EMAIL PROTECTED]
wrote:
As I understand this,
P(T+ | D-)=1-P(T+ | D+)=0.05
is the probability not to detect desease for a
person
at ICU who has the desease. Correct?
What I asked was whether it is possible to
mistakenly
detect
Hi Yuchen,
First of all please notice that you may not have more
than 2^8 = 256 columns in Excel, so if you have more
than 256 time-series you can not put them all in one
Excel sheet. I also believe that you can not have more
than 2^16 = 65536 rows.
If you do not have more than 256 time-series, I
)
?choose
choose(200,2)
19900
choose(200,100)
9.054851e+58
N=200; k=100; m=50; p=.6; q=.95
choose(N,k)*p^k*(1-p)^(N-k)*choose(k,m)*q^m*(1-q)^(k-m)
6.554505e-41
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf
Of Moshe Olshansky
Sent
This does not work in the general case. To see this,
try:
rownames(a}[5] - a
and see what happens then.
--- François Pinard [EMAIL PROTECTED] wrote:
[Gianni Burgin]
let say something like this
a=matrix(1:25, nrow=5)
rownames(a)=letters[1:5]
colnames(a)=rep(A, 5)
a
A A A A A
A not very good solution is as below:
If your array's dimensions were KxMxN and the linear
index is i then
n - ceiling(i/(K*M))
i1 - i - (n-1)*(K*M)
m - ceiling(i1/K)
k - i1 - (m-1)*K
and your index is (k,m,n)
I am almost sure that there is a function in R which
does this (it exists in Matlab).
Note: forwarded message attached.
Mistakenly I did not send this to the list!
---BeginMessage---
Here is a solution which (I think) works, but it
contains a loop which most likely can be eliminated by
more knowledgable people):
Let a and b be like in your example but to make things
more
Use grep(^e,x) if you are looking for the entries
where e is the first character.
--- Vladimir Eremeev [EMAIL PROTECTED] wrote:
Hi!
seq(along=x) %in% grep(e,x)
Steve Powell-4 wrote:
I have a vector of strings
x=c(w,ex,ee)
And I want to get a logical vector showing the
Hi Bart,
I have never used image processing software in R (I
was doing this with Matlab), but here is what I would
have done algorithmically:
1) convert the picture to gray-scale
2) find a threshold value which separates the circles
from the background and convert your image to black
and white
3)
This won't work since it produces a matrix (try this).
What should work is
x[(1:nrow(x)) + nrow(x)*(v-1)]
--- Johannes Graumann [EMAIL PROTECTED]
wrote:
Thanks!
Joh
On Thursday 23 August 2007 12:01:50 you wrote:
x[cbind(1:nrow(x), the.vector)]
Patrick Burns
[EMAIL PROTECTED]
Correct, I didn't notice that the coma was inside the
cbind(). Sorry...
--- Rolf Turner [EMAIL PROTECTED] wrote:
On 24/08/2007, at 12:51 PM, Moshe Olshansky wrote:
This won't work since it produces a matrix (try
this).
On the contrary, Patrick's solution is correct. I
tried
which touches the boundaries is the
background while all other components are holes and
you can make them white (1) in the original
black-and-white image.
--- Moshe Olshansky [EMAIL PROTECTED] wrote:
Hi Bart,
I have never used image processing software in R (I
was doing this with Matlab
you mean with
morphological closing and the labeling to split the
images.
Could you please clarify this a bit?
Thanks for your support
Bart
Moshe Olshansky-2 wrote:
Hi Bart,
One more comment:
You do not really need the morphological closing
to
close the holes
As far as I understand, changing the format changes
the way data is displayed by Excel but this does not
change the data itself - if while reading the data
Excel decided that it was a date, it is being
converted to an integer (the number of days since
January 1, 1900 - and they mistakenly think
Below is one way to do this:
Loc1.alleles - c(1,5,6,7,8)
Loc1.Freq- c(0.35, 0.15, 0.05, 0.10, 0.35)
Loc1 - cbind( Loc1.alleles,Loc1.Freq)
X- data.frame(Loc1)
Loc2.alleles - c(1,4,6,8)
Loc2.Freq - c(0.35, 0.35, 0.10, 0.20)
Loc2 -
Hi,
Getting your e is not a problem:
a- list(c(1,3),c(1,2),c(2,3))
b-c(10,20,30)
aa-matrix(unlist(a),nrow=2)
ee-aa+rbind(b,b)
e-apply(ee,2,sum)
e
[1] 24 43 65
But this will not work if different list members have
different number of elements.
--- Yongwan Chun [EMAIL PROTECTED] wrote:
The simlest answer is that X and Y are two vectors of
length n then (un-normalized) cross correlation
between X and Y is sum(i=1 to n)
{X(i)-Xbar)*(Y(i)-Ybar)} where Xbar and Ybar are the
means of X and Y respectively.
You may need something different, so could you be more
specific? What do you
Hi Thomas,
On my computer the solution is (0,0,1,0,0,0,0) (within
machine accuracy) and it satisfies the constraints.
--- [EMAIL PROTECTED] wrote:
Hi everybody,
I'm using Windows XP Prof, R 2.5.1 and a Pentium 4
Processor.
Now, I want to solve a quadratic optimization
program (Portfolio
If your data is in data.txt file you can do the
following:
x - read.table(file=data.txt,header=TRUE)
t-ISOdatetime(x[,1],x[,2],x[,3],x[,4],x[,5],x[,6])
secs - as.numeric(t)
So now secs represents your time in seconds and you
can use any type of interpolation you wish to
interpolate co2obs.
---
Below is one possibility:
If you knew MA you would get a regular linear
least-squares for parameters A,B and constant which
can be easily solved. So now you can define a function
f(MA) which returns that value. Now you must minimize
that f - a function of one argument. It can have
several local
observation %in% ID
--- Takatsugu Kobayashi [EMAIL PROTECTED] wrote:
Hi RUsers,
I am wonder if I can search observations whose IDs
matches any of the
values in another vector, such as in MySQL. While I
am learing MySQL for
future database management, I appreciate if anyone
could give
unlist(strsplit(bA531F16-rep,\\-))[1]
[1] bA531F16
--- Carlos Morales [EMAIL PROTECTED]
wrote:
Hello,
I would like to know what can I do if I use
strplit with a string and I want to use the middle
left,I mean I have this:
strsplit(bA531F16-rep,\\-)
[[1]]
[1] bA531F16 rep
Hi Jonathan,
What exactly do you mean by replication?
Do you want to keep a1,b1,c1,... unchanged but have 30
different sets of random numbers?
Regards,
Moshe.
--- VTLT1999 [EMAIL PROTECTED] wrote:
Hello All,
I have searched many help forums, message boards,
etc. and I just can't
Either
min(diff(sort(x))[diff(sort(x))0])
or
min(diff(sort(unique(x
--- dxc13 [EMAIL PROTECTED] wrote:
useRs,
I am looking to find the minimum positive value of
some data I have.
Currently, I am able to find the minimum of data
after I apply some other
functions to it:
x
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