> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ali Salekfard
> Sent: Wednesday, December 29, 2010 6:25 AM
> To: r-help@r-project.org
> Subject: Re: [R] Removing rows with earlier dates
>
> Than
David,
Thanks alot. Your code is worked fine on the whole dataset (no memory error
as I had with the other ideas). I do like the style - especialy the fact
that it is all in one line - , but for large datasets it takes longer than
what I wrote. I ran it on the same machine with the same set of rul
On Dec 29, 2010, at 11:03 AM, Ali Salekfard wrote:
> David,
>
> Thanks alot. Your code is worked fine on the whole dataset (no
> memory error as I had with the other ideas). I do like the style -
> especialy the fact that it is all in one line - , but for large
> datasets it takes longer th
On Dec 29, 2010, at 9:24 AM, Ali Salekfard wrote:
Thanks to everyone. Joshua's response seemed the most concise one,
but it
used up so much memory that my R just gave error. I checked the other
replies and all in all I came up with this, and thought to share it
with
others and get comments
> David Winsemius
> on Fri, 24 Dec 2010 11:47:05 -0500 writes:
> On Dec 24, 2010, at 11:04 AM, David Winsemius wrote:
>>
>> On Dec 24, 2010, at 8:45 AM, Ali Salekfard wrote:
>>
>>> Hi all,
>>>
>>> I'm new to the list but have benfited from it quite exte
Thanks to everyone. Joshua's response seemed the most concise one, but it
used up so much memory that my R just gave error. I checked the other
replies and all in all I came up with this, and thought to share it with
others and get comments.
My structure was as follows:
ACCOUNT RULE DATE
A1
Whenever a task calls for breaking a data object into pieces, operate on the
pieces, then put it back together, then think about using the plyr package.
Sent from my iPod
On Dec 24, 2010, at 6:58 AM, "Ali Salekfard" wrote:
> Hi all,
>
> I'm new to the list but have benfited from it quite ext
> with(YourDataFrame, tapply(`Effective Date`, `RULE COLUMNS`,
> function(x) x[which.max(x)]))
David pointed out that this will just return a table of dates. One
work around is:
do.call("rbind", by(DataFrame, DataFrame[, "RULE COLUMNS"],
function(x) x[which.max(x[, "Effective Date"]), ]))
bu
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ali Salekfard
> Sent: Friday, December 24, 2010 5:46 AM
> To: r-help@r-project.org
> Subject: [R] Removing rows with earlier dates
>
> Hi all,
>
>
Hi,
On Fri, Dec 24, 2010 at 5:45 AM, Ali Salekfard wrote:
[snip]
> I have a data frame that contains mapping rules in this way:
>
> ACCOUNT, RULE COLUMNS, Effective Date
>
>
> The dataframe comes from a database that stores all dates. What I would like
> to do is to create a data frame with only
On Dec 24, 2010, at 11:04 AM, David Winsemius wrote:
On Dec 24, 2010, at 8:45 AM, Ali Salekfard wrote:
Hi all,
I'm new to the list but have benfited from it quite extensively.
Straight to
my rather strange question:
I have a data frame that contains mapping rules in this way:
ACCOUNT,
On Dec 24, 2010, at 8:45 AM, Ali Salekfard wrote:
Hi all,
I'm new to the list but have benfited from it quite extensively.
Straight to
my rather strange question:
I have a data frame that contains mapping rules in this way:
ACCOUNT, RULE COLUMNS, Effective Date
The dataframe comes from
Hi all,
I'm new to the list but have benfited from it quite extensively. Straight to
my rather strange question:
I have a data frame that contains mapping rules in this way:
ACCOUNT, RULE COLUMNS, Effective Date
The dataframe comes from a database that stores all dates. What I would like
to do
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