[R] Automatic creation of columns in zoo object

[Gabor Grothendieck]

That's not really in the spirit of R.  Normally one works with
whole objects at a time.  You might wish to rethink your
entire approach to this.

On Tue, Feb 3, 2009 at 9:11 AM, Sergey Goriatchev <sergeyg at gmail.com> wrote:
> Dear Gabor,
>
> Yes, these extra columns are of value as I later write code that fills
> in those columns row by row conditional on some event taking place,
> and when there is no event there should be zero in a particular cell.
> BIG thanks for the quick reply. I will try the code out right away!
>
> Kind Regards,
> Sergey
>
>
> On Tue, Feb 3, 2009 at 15:05, Gabor Grothendieck
> <ggrothendieck at gmail.com> wrote:
>> Not sure why you need to have all these extra columns if they are only
>> zero anyways.  Could you not append them when you get data for them?
>> Are these placeholders really of any value?  At any rate its done using
>> cbind or merge like this:
>>
>> library(zoo)
>> library(chron)
>>
>> nms <- sub(".Comdty", "tr", me.la.tickers)
>> zeromat <- matrix(0, nrow(me.la), length(nms), dimnames = list(NULL, nms))
>> cbind(me.la, zeromat)
>>
>> In this case zeromat and me.la have the same dimensions so we could
>> alternately reduce it to this slightly briefer code:
>>
>> zeromat <- 0 * me.la
>> colnames(zeromat) <- sub(".Comdty", "tr", me.la.tickers)
>> cbind(me.la, zeromat)
>>
>>
>> On Tue, Feb 3, 2009 at 8:44 AM, Sergey Goriatchev <sergeyg at gmail.com> wrote:
>>> Hello, everyone
>>>
>>> I have a question.
>>>
>>> Assume I have the following zoo object:
>>>
>>> me.la <- structure(c(1524.75, 1554.5, 1532.25, 1587.5, 1575.25, 1535.5,
>>> 1550, 1493.5, 1492.5, 1472.25, 1457.5, 1442.75, 1399, 1535.75,
>>> 1565.25, 1543.5, 1598.5, 1586.5, 1547, 1561.5, 1504.75, 1503.75,
>>> 1483.75, 1468.75, 1453.75, 1410, 1546.75, 1575.25, 1554, 1609,
>>> 1597.5, 1558.5, 1573, 1516.25, 1515.5, 1495, 1480, 1465, 1421.25,
>>> 1561.5, 1590, 1568.75, 1623.5, 1612, 1573, 1587.5, 1530.5, 1530,
>>> 1509.75, 1494.5, 1479.5, 1435.75, 1573.5, 1601.5, 1580.25, 1635,
>>> 1623.5, 1584.5, 1599, 1541.75, 1541.5, 1521.5, 1506, 1491, 1447.25,
>>> 1585.5, 1613, 1591.75, 1646, 1634.5, 1595.5, 1610, 1552.75, 1552.75,
>>> 1532.75, 1517, 1502, 1458.25, 1600, 1627.5, 1606.5, 1660, 1649,
>>> 1609.75, 1624.25, 1567, 1567, 1547, 1531, 1516, 1472.25, 1612,
>>> 1639.5, 1618.25, 1671.5, 1661, 1621.5, 1635.75, 1578, 1578, 1558,
>>> 1542, 1527, 1483.5), .Dim = c(13L, 8L), index = structure(c(14245,
>>> 14246, 14249, 14250, 14251, 14252, 14253, 14256, 14257, 14258,
>>> 14259, 14260, 14263), format = "m/d/y", origin = structure(c(1,
>>> 1, 1970), .Names = c("month", "day", "year")), class = c("dates",
>>> "times")), class = "zoo", .Dimnames = list(NULL, c("LA1 COMDTY",
>>> "LA2 COMDTY", "LA3 COMDTY", "LA4 COMDTY", "LA5 COMDTY", "LA6 COMDTY",
>>> "LA7 COMDTY", "LA8 COMDTY")))
>>>
>>> I also have a following variable (used in RBloomberg) in my environment:
>>>
>>> me.la.tickers <- c("LA1 Comdty", "LA2 Comdty", "LA3 Comdty", "LA4
>>> Comdty", "LA5 Comdty", "LA6 Comdty", "LA7 Comdty", "LA8 Comdty")
>>>
>>> What I need to do is to automate the following manual piece of code:
>>>
>>> me.la$LA2tr <- 0
>>> me.la$LA3tr <- 0
>>> me.la$LA4tr <- 0
>>> me.la$LA5tr <- 0
>>> me.la$LA6tr <- 0
>>> me.la$LA7tr <- 0
>>> me.la$LA8tr <- 0
>>>
>>> Basically, I need to automatically create new columns in futures
>>> object taking first part of names in me.la.tickers.
>>>
>>> I tried with paste() and assign() combination, but could not get.
>>>
>>> I could do the manual part, of course, it does not take much time, but:
>>>
>>> 1) I want to learn how to automatically create new columns in zoo objects
>>> 2) I have many more variables that have to be treated similarly (that
>>> is, I have me.lp with corresponding me.lp.tickers, me.qc with
>>> me.qc.tickers, etc. Then I have a bunch of variables starting with
>>> "en", like en.co and en.cl, and corresponding ticker vectors, then
>>> "ag" variables and "so" variables, as well). I have all in all 24 zoo
>>> variables with 24 corresponding ticker vectors, and for each a
>>> corresponding ticker vector, and for each zoo variable I need to
>>> create 7 extra columns. That would take much time to do manually, and
>>> a lot of code.
>>>
>>> How would I do this automatically, please?
>>>
>>> Thank you in advance for your help!
>>>
>>> Regards,
>>> Sergey
>>>
>>> ______________________________________________
>>> R-help at r-project.org 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, reproducible code.
>>>
>>
>
>
>
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