performanceanalytics - R: using apply.fromstart to calculate returns and standard deviation -

i find total return has made on period standard deviation of returns. have found function, apply.fromstart in performanceanalytics package, looks promising. am, however, having difficulty implementing it.

here have: dataframe containing various data, including return per period:
hourlydata

        time            position   2014-08-01 01:00:00      1.01     2014-08-01 02:00:00      0.99   2014-08-01 03:00:00      1.01   2014-08-01 04:00:00      1.02   

i find total return in each period, follows:

period                totalreturn 2014-08-01 01:00:00      1.01     2014-08-01 02:00:00      1.01*0.99 2014-08-01 03:00:00      1.01*.099*1.01 2014-08-01 04:00:00      1.01*.099*1.01*1.02 

my code reads:
apply.fromstart(hourlydata[,2,drop = false],fun="*",width=1)

i find standard deviation of returns. code part reads follows:
apply.fromstart(hourlydata[,2,drop = false],fun="sd",width=1)

the data type of hourlydata$position "zoo"

i getting following error: in zoo(na, order.by = as.date(time(r))) : methods “zoo” objects not work if index entries in ‘order.by’ not unique

i have checked, , there no duplicate in row names

here result running dput(hourlydata):

> structure(list(period = structure(c(1406844000, 1406847600, > 1406851200,  1406854800, 1406858400, 1406862000), class = c("posixct", > "posixt" )), login = c(173908l, 173908l, 173908l, 173908l, 173908l, > 173908l ), netexposureusd = c(2188640, 2188730, 2189230, 2189000, > 2188310,  2187710), equityusd = c(9303.51, 9237.82, 8582.18, 9074.76, > 9929.96,  > 10743.57), unrealizedprofitusd = c(-31.64, -97.33, -752.97, -260.39,  > 594.81, 1408.42), depositwithdrawal = c(0, 0, 0, 0, 0, 0), laggedequity = structure(c(0,  > 9303.51, 9237.82, 8582.18, 9074.76, 9929.96), index = 1:6, class = "zoo"),  >     return = structure(c(0, -0.00706077598669755, -0.0709734547761268,  >     0.0573956733603816, 0.0942394068823857, 0.0819348718423841 >     ), index = 1:6, class = "zoo"), position = structure(c(1,  >     0.992939224013302, 0.929026545223873, 1.05739567336038, 1.09423940688239,  >     1.08193487184238), index = 1:6, class = "zoo"), sd = c(na,  >     na, 0.0390979887599392, 0.0641847560185966, 0.0867288719859795,  >     0.0187573743033249)), .names = c("period", "login", "netexposureusd",  "equityusd", "unrealizedprofitusd", > "depositwithdrawal", "laggedequity",  "return", "position", "sd"), > row.names = c("2014-08-01 01:00:00",  "2014-08-01 02:00:00", > "2014-08-01 03:00:00", "2014-08-01 04:00:00",  "2014-08-01 05:00:00", > "2014-08-01 06:00:00"), class = "data.frame") >                                  period  login netexposureusd equityusd unrealizedprofitusd depositwithdrawal laggedequity       > return  position         sd 2014-08-01 01:00:00 2014-08-01 01:00:00 > 173908        2188640   9303.51              -31.64                 0  > 0.00  0.000000000 1.0000000         na 2014-08-01 02:00:00 2014-08-01 02:00:00 173908        2188730   9237.82              -97.33           > 0      9303.51 -0.007060776 0.9929392         na 2014-08-01 03:00:00 > 2014-08-01 03:00:00 173908        2189230   8582.18             > -752.97                 0      9237.82 -0.070973455 0.9290265 0.03909799 2014-08-01 04:00:00 2014-08-01 04:00:00 173908        2189000   9074.76             -260.39                 0      8582.18  > 0.057395673 1.0573957 0.06418476 2014-08-01 05:00:00 2014-08-01 05:00:00 173908        2188310   9929.96              594.81           > 0      9074.76  0.094239407 1.0942394 0.08672887 2014-08-01 06:00:00 > 2014-08-01 06:00:00 173908        2187710  10743.57             > 1408.42                 0      9929.96  0.081934872 1.0819349 0.01875737 

use efficient vectorized base r function cumprod first desired result. while second result achieved (less efficiently) using simple *apply loop

if want keep zoo class,

cumprod(hourlydata$position) #         1         2         3         4         5         6  # 1.0000000 0.9929392 0.9224669 0.9754125 1.0673348 1.1547867 

otherwise

cumprod(as.numeric(hourlydata$position)) ## [1] 1.0000000 0.9929392 0.9224669 0.9754125 1.0673348 1.1547867 

for sd (as proposed @akrun) (used vapply instead of sapply in order "squeeze" maximum performance out of it)

vapply(seq_len(nrow(hourlydata)), function(i) sd(hourlydata$position[1:i]), fun.value = double(1)) # [1] na 0.004992723 0.039097989 0.052519398 0.063598345 0.063156702