On 16/10/2008, at 11:07 AM, Prof Brian Ripley wrote:
[[ ]] works on vectors!
So it does. My bad.
letters[[3]]
[1] "c"
See help("[["). But strictly V4 is a factor and hence not a
vector: [[ ]]
also works on factors.
Yes. I had forgotten that not everyone sets options
(stringsAsFactors=FALSE)
the way that all right thinking people do. So I was thinking in
terms of
V4 being a character vector. Silly me. Of *course* you'd want it to
be a
factor .... (Huh?)
So as.character() is required, single brackets or double brackets.
cheers,
Rolf
On Thu, 16 Oct 2008, Rolf Turner wrote:
On 16/10/2008, at 10:03 AM, jim holtman wrote:
try putting as.character in the call:
x = read.csv(as.character(V4[[i]]), header = FALSE
No. This won't help. V4 is a column of the data frame optdata,
and hence is a vector. Not a list! Use single brackets --- V4[i]
---
and all will be well.
cheers,
Rolf
On Wed, Oct 15, 2008 at 4:46 PM, Ted Byers
<[EMAIL PROTECTED]> wrote:
Here is what I tried:
optdata =
read.csv("K:\\MerchantData\\RiskModel\\AutomatedRiskModel\
\soptions.dat",
header = FALSE, na.strings="")
optdata
attach(optdata)
for (i in 1:length(V4) ) { x = read.csv(V4[[i]], header = FALSE,
na.strings="");x }
And here is the outcome (just a few of the 60 records
successfully read):
optdata =
read.csv("K:\\MerchantData\\RiskModel\\AutomatedRiskModel\
\soptions.dat",
header = FALSE, na.strings="")
optdata
V1 V2 V3 V4
1 251 2008 18 Plus_Shipping.2008.18.dat
2 251 2008 19 Plus_Shipping.2008.19.dat
3 251 2008 20 Plus_Shipping.2008.20.dat
4 251 2008 22 Plus_Shipping.2008.22.dat
5 251 2008 23 Plus_Shipping.2008.23.dat
6 251 2008 24 Plus_Shipping.2008.24.dat
I can see the data has been correctly read. But for some reason
that
isn't
clear, read.csv doesn't like the data in the last column.
attach(optdata)
for (i in 1:length(V4) ) { x = read.csv(V4[[i]], header = FALSE,
na.strings="");x }
Error in read.table(file = file, header = header, sep = sep,
quote =
quote,
:
'file' must be a character string or connection
V4[[1]]
[1] Plus_Shipping.2008.18.dat
60 Levels: Easyway.2008.17.dat Easyway.2008.18.dat Easyway.
2008.19.dat
Easyway.2008.20.dat ... Secured_Pay.2008.31.dat
The last column is comprised of valid Windows filenames (and no
whitespace,
so as not to confuse things).
I see in the docuentation "`[[...]]' is the operator used to
select a
single
element, whereas `[...]' is a general subscripting operator.",
so I assume
V4[[i]] is the correct way to get the ith value from V4. So why
does
read.csv complain that "'file' must be a character string or
connection"?
It seems obvious that the value in V4[[i]i] is a string. V4
[[1]] does
give
me the right value, although that is followed by output I didn't
ask for.
In the loop above, I was going to replace the output obtained by
'x' with
output from fitdistr(x,"exponential"), but I can't proceed with
that until
I
can get the data in these files read.
What have I missed?
Thanks
Ted
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PLEASE do read the posting guide
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
______________________________________________
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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