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Home/ Questions/Q 8901939
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T01:29:13+00:00 2026-06-15T01:29:13+00:00

How can I parse this date format? Should I change this colon to dot

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How can I parse this date format? Should I change this colon to dot or maybe someone know better solution?

> x <- "2012.01.15 09:00:02:002"
> strptime(x, "%Y.%m.%d %H:%M:%S:%OS") 
[1] "2012-01-15 09:00:02"
> strptime(x, "%Y.%m.%d %H:%M:%OS")
[1] "2012-01-15 09:00:02"
> x <- "2012.01.15 09:00:02.002"
> strptime(x, "%Y.%m.%d %H:%M:%OS")
[1] "2012-01-15 09:00:02.001"
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  1. Editorial Team
    Editorial Team
    2026-06-15T01:29:15+00:00Added an answer on June 15, 2026 at 1:29 am

    There’s a subtle distinction here that may be throwing you off.
    As ?strptime notes:

    for ‘strptime’ ‘%OS’ will input seconds including fractional seconds.

    To emphasize that a bit, %OS represents the seconds including fractional seconds — not just the fractional part of the seconds: if the seconds value is 44.234, %OS or %OS3 represents 44.234, not .234

    So the solution is indeed to substitute a . for that final :.

    Here’s one way you might do that:

    x <- "2012.01.15 09:00:02:002"
    strptime(gsub(":", ".", x), "%Y.%m.%d %H.%M.%OS") 
    
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