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Home/ Questions/Q 9236339
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T07:16:07+00:00 2026-06-18T07:16:07+00:00

Reproducible code: library(quantmod) getSymbols(‘SPY’, from = ‘1950-01-01’) Y <- Cl(to.monthly(SPY)) start <- c(1993, 1)

  • 0

Reproducible code:

library(quantmod)
getSymbols('SPY', from = '1950-01-01')
Y <- Cl(to.monthly(SPY))
start <- c(1993, 1)
end <- c(2013, 2)
y <- ts(log(Y), start = start, end = end, frequency = 12)

I would like start and end are set equal to start(Y) and end(Y) without creating them “manually”.

Is it possible?

This is what I would like to get without specifying start and end variables:

> y
          Jan      Feb      Mar      Apr      May      Jun      Jul      Aug
1993 3.782825 3.793465 3.810876 3.784871 3.811539 3.807995 3.803101 3.840742
1994 3.875774 3.846097 3.797510 3.808660 3.824502 3.794815 3.826683 3.864092
1995 3.852061 3.892228 3.914221 3.943328 3.982295 3.996548 4.028205 4.032646
1996 4.153713 4.157006 4.169607 4.180369 4.202750 4.206333 4.160288 4.179451
1997 4.361951 4.371471 4.322409 4.383151 4.444532 4.480853 4.557135 4.503912
1998 4.588126 4.655103 4.699935 4.712589 4.691623 4.730127 4.716533 4.564348
1999 4.849370 4.816727 4.854995 4.892227 4.869072 4.919981 4.888468 4.883256
2000 4.938495 4.923187 5.013165 4.977354 4.961515 4.978663 4.962845 5.026115
2001 4.920127 4.819878 4.759521 4.841506 4.835885 4.808927 4.798679 4.737513
2002 4.728979 4.710881 4.740749 4.680834 4.674883 4.594716 4.512616 4.519394
2003 4.455045 4.441474 4.439588 4.520810 4.574195 4.581185 4.599052 4.619467
2004 4.731627 4.745106 4.728272 4.709170 4.726148 4.740837 4.708088 4.710521
2005 4.772040 4.792728 4.770346 4.751433 4.783149 4.780635 4.818183 4.808764
2006 4.848116 4.853826 4.866226 4.878779 4.848195 4.846389 4.850858 4.872445
2007 4.968076 4.948263 4.955827 4.999170 5.032527 5.013498 4.981687 4.994438
2008 4.922678 4.896496 4.882575 4.929136 4.944139 4.851874 4.842848 4.858183
2009 4.416790 4.303119 4.376009 4.470724 4.527533 4.521245 4.593199 4.629472
2010 4.676467 4.707185 4.762174 4.777526 4.694737 4.636863 4.702932 4.656908
2011 4.857329 4.891476 4.887262 4.915812 4.904534 4.882575 4.862367 4.805823
2012 4.877637 4.920127 4.947411 4.940713 4.878779 4.913390 4.925150 4.949894
2013 5.008633 5.007564                                                      
          Sep      Oct      Nov      Dec
1993 3.827336 3.846738 3.836006 3.841386
1994 3.832330 3.860309 3.819688 3.819030
1995 4.068685 4.065774 4.109397 4.118712
1996 4.228584 4.260424 4.330996 4.301901
1997 4.547223 4.522441 4.560382 4.575329
1998 4.622519 4.700480 4.754624 4.814702
1999 4.857873 4.919981 4.936486 4.989616
2000 4.967241 4.962495 4.884921 4.876647
2001 4.648613 4.661551 4.736637 4.738827
2002 4.404155 4.483229 4.543082 4.479947
2003 4.604670 4.656813 4.667675 4.712050
2004 4.716354 4.729156 4.769752 4.794716
2005 4.812510 4.788574 4.831588 4.824386
2006 4.894701 4.925731 4.945421 4.953147
2007 5.027689 5.041164 5.001662 4.985044
2008 4.753504 4.572957 4.500809 4.502473
2009 4.659564 4.640151 4.699935 4.713486
2010 4.737338 4.774829 4.774829 4.834296
2011 4.728714 4.832306 4.828234 4.832306
2012 4.969605 4.951239 4.956883 4.958710
2013  

I’m adding these lines of text because SO tells me I’m not using a sufficient number of explanation text words.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-18T07:16:08+00:00Added an answer on June 18, 2026 at 7:16 am

    Try the following:

    as.ts(log(Y), start = head(index(Y), 1), end = tail(index(Y), 1))
    

    This utilizes the dates present in the index of an xts object.

    As @Arun mentions in the comments, the following options are even more direct:

    ts(log(Y), start = start(Y), end = end(Y), frequency = 12)
    as.ts(log(Y), start = start(Y), end = end(Y))
    

    I would suggest the ts(log(Y)... approach since the attributes are also retained. Compare:

    ytest1 <- as.ts(log(Y), start = start(Y), end = end(Y))
    ytest2 <- ts(log(Y), start = start(Y), end = end(Y), frequency = 12)
    
    identical(y, ytest1)
    # [1] FALSE
    all.equal(y, ytest1, check.attributes = FALSE)
    # [1] TRUE
    identical(y, ytest2)
    # [1] TRUE
    
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