如何从strptime中提取年份? [英] How can I extract year from strptime?
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问题描述
我有这样的格式:
t < - 4/2/2004 12:45
我可以使用这两行来提取日期和月份:
strptime(t,%m /%d /%Y%H:%M,tz =UTC )$ day
strptime(t,%m /%d /%Y%H:%M,tz =UTC)$ mday
但是,这并不是我想要的岁月。我给了我这个世纪的岁月:
strptime(t,%m /%d /%Y%H:%M ,tz =UTC)$ year
像1987年,2012年等等,而不是87和12。
如果还有其他功能用于此,我也可以这样做。
解决方案
添加1900.请参阅?POSIXlt
详细信息部分> for,um,details。
> tm< - 4/2/2004 12:45
/ pre>
> strptime(tm,%m /%d /%Y%H:%M,tz =UTC)$ year
[1] 104
> strptime(tm,%m /%d /%Y%H:%M,tz =UTC)$ year + 1900
[1] 2004
I have times in this format:
t <- "4/2/2004 12:45" I can extract day and month using these two lines: strptime(t, "%m/%d/%Y %H:%M", tz="UTC")$day strptime(t, "%m/%d/%Y %H:%M", tz="UTC")$mday
But this doesn't give me the years I want. I gives me year without the century:
strptime(t, "%m/%d/%Y %H:%M", tz="UTC")$year
How can I get the year exactly? Like 1987, 2012 etc. And not 87 and 12.
If there is any other function to use for this, I am OK with that as well.
解决方案Add 1900. See the Details section of
?POSIXlt
for, um, details.> tm <- "4/2/2004 12:45" > strptime(tm, "%m/%d/%Y %H:%M", tz="UTC")$year [1] 104 > strptime(tm, "%m/%d/%Y %H:%M", tz="UTC")$year + 1900 [1] 2004
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