Hi!
10.06.2015, 13:20, khatri wrote:
My date column is in following format : %m/%d H:M:S
There is not mention of year in the data.So how can I read this using
strptime function.
I have tried strptime(dates,%m/%d H:M:S) but this is returning NA.
Thanks
How about:
strptime(dates,%m/%d
Hey sorry for my typing.
I have actually used the same format with % that is
dd-c(21/01 11:11:11)
strptime(dd,%d/%m H:M:S)
[1] NA
it is giving NA.
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Hi khatri,
strptime(6/20 21:56:32,%m/%d %H:%M:%S)
[1] 2015-06-20 21:56:32 AEST
You forgot the % signs in the format for H:M:S
Jim
On Wed, Jun 10, 2015 at 8:20 PM, khatri dheeraj.kha...@abzooba.com wrote:
My date column is in following format : %m/%d H:M:S
There is not mention of year in the
Your example is not the same as the answers you have received. Look again.
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Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
No really, what Jim said: you need the % for H:M:S as well.
dd-c(21/01 11:11:11)
strptime(dd,%d/%m %H:%M:%S)
[1] 2015-01-21 11:11:11 EST
On Wed, Jun 10, 2015 at 8:03 AM, khatri dheeraj.kha...@abzooba.com wrote:
Hey sorry for my typing.
I have actually used the same format with % that is
Hello,
Inline.
Em 10-06-2015 13:03, khatri escreveu:
Hey sorry for my typing.
I have actually used the same format with % that is
No you have not, like others allready have told you.
You need %H:%M:%S not H:M:S.
strptime(dd,%d/%m %H:%M:%S)
[1] 2015-01-21 11:11:11 GMT
Hope this helps,
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