[R] better code?

[Spencer Graves]

	  OK.  How about the following:

R> x1.df$d. <- as.Date(as.character(x1.df$d), format="%m/%d/%Y")
R> sapply(x1.df, class)
         d         x         y        d.
  "factor" "numeric" "numeric"    "Date"
R>
R> (byLH <- by(x1.df[2:3], x1.df$d., var))
x1.df$d.: 2005-01-01
           x         y
x  1.272688 -0.783719
y -0.783719  0.884866
------------------------------------------------------------
x1.df$d.: 2005-01-08
           x         y
x 1.4893139 0.2845754
y 0.2845754 0.6273839
------------------------------------------------------------
x1.df$d.: 2005-01-15
           x         y
x 0.9992097 0.8719848
y 0.8719848 0.7753879
------------------------------------------------------------
x1.df$d.: 2005-01-22
            x          y
x  1.2269211 -0.8694323
y -0.8694323  0.6161053
------------------------------------------------------------
x1.df$d.: 2005-01-29
             x           y
x 0.072721473 0.006551483
y 0.006551483 1.003923812

spencer graves
##########################################
Hi again!

The solution is good except that the dates are in alpha and not date order.
But this is way better.

Thanks

############################
Hi, Laura:

	  How about the following:

> sapply(x1.df, class)
         d         x         y
  "factor" "numeric" "numeric"
> (byLH <- by(x1.df[-1], x1.df$d, var))
x1.df$d: 1/1/2005
           x         y
x  1.272688 -0.783719
y -0.783719  0.884866
-------------------------------------------------------------
x1.df$d: 1/15/2005
           x         y
x 0.9992097 0.8719848
y 0.8719848 0.7753879
-------------------------------------------------------------
x1.df$d: 1/22/2005
            x          y
x  1.2269211 -0.8694323
y -0.8694323  0.6161053
-------------------------------------------------------------
x1.df$d: 1/29/2005
             x           y
x 0.072721473 0.006551483
y 0.006551483 1.003923812
-------------------------------------------------------------
x1.df$d: 1/8/2005
           x         y
x 1.4893139 0.2845754
y 0.2845754 0.6273839

	  spencer graves

Laura Holt wrote:

> Hi R!
> 
> I have a data.frame with dates in the first column and numeric values in 
> columns 2 -3.
> 
> I want to have a covariance matrix for each date.  The following code 
> works fine:
> 
>> x1.df
> 
>          d           x          y
> 1  01/01/05  1.06014788 0.72595670
> 2  01/01/05 -1.56330741 2.44930531
> 3  01/01/05 -0.58001696 0.48626682
> 4  01/01/05  0.27653308 0.52676239
> 5  01/08/05  1.91849382 1.72136239
> 6  01/08/05 -0.74774661 0.16657346
> 7  01/08/05  0.18505727 1.33570129
> 8  01/08/05  1.47015895 1.59666054
> 9  01/08/05 -0.67119562 2.31980255
> 10 01/15/05  0.39728456 2.48849586
> 11 01/15/05 -0.96484152 1.47565372
> 12 01/15/05 -1.55109398 0.73436620
> 13 01/22/05 -1.36172373 1.12015635
> 14 01/22/05  0.20475072 0.01010656
> 15 01/29/05 -0.49909855 0.15583279
> 16 01/29/05 -0.07834782 0.99454434
> 17 01/29/05  0.12845272 0.09605443
> 18 01/29/05 -0.44926053 1.48959860
> 19 01/29/05 -0.07033900 2.50253296
> 
>> for(i in 1:5) {
> 
> + yy <- x1.df$d == u3[i]
> + zz <- cov(x1.df[yy,2:3])
> + print(dates(u3[i]))
> + prmatrix(zz)
> + }
> [1] 01/01/05
>          x         y
> x  1.272688 -0.783719
> y -0.783719  0.884866
> [1] 01/08/05
>          x         y
> x 1.4893139 0.2845754
> y 0.2845754 0.6273839
> [1] 01/15/05
>          x         y
> x 0.9992097 0.8719848
> y 0.8719848 0.7753879
> [1] 01/22/05
>           x          y
> x  1.2269211 -0.8694323
> y -0.8694323  0.6161053
> [1] 01/29/05
>            x           y
> x 0.072721473 0.006551485
> y 0.006551485 1.003923812
> 
>>
> However, I'm sure that there is a MUCH better way to accomplish this task.
> 
> Any suggestions, please?
> 
> R V 2.1.0 Windows
> 
> Thank you in advance!
> Sincerely,
> Laura Holt
> mailto: lauraholt_983 at hotmail.com
> 
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-- 
Spencer Graves, PhD
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