blocks|key|405358|text|您可以使用format|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|405359|library(zoo)
x+<-+as.yearmon(Sys.time())
format(x,"%25b")
[1]+"Mar"
format(x,"%25Y")
[1]+"2012"|code-block|syntax|javascript|405360|entityMap^0|5|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>format</code>:

<pre><code>library(zoo)
x &lt;- as.yearmon(Sys.time())
format(x,"%b")
[1] "Mar"
format(x,"%Y")
[1] "2012"
</code></pre>

blocks|key|405418|text|对类"yearmon"的对象使用format()方法。这是您的示例日期(正确创建！)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|405419|date1+<-+as.yearmon("Mar+2012",+"%25b+%25Y")|code-block|syntax|javascript|405420|然后我们可以根据需要提取日期部分：|405421|>+format(date1,+"%25b")+##+Month,+char,+abbreviated
[1]+"Mar"
>+format(date1,+"%25Y")+##+Year+with+century
[1]+"2012"
>+format(date1,+"%25m")+##+numeric+month
[1]+"03"|405422|这些将作为字符返回。在适当的情况下，如果希望年或数字月作为数值变量，则在as.numeric()中换行，例如|405423|>+as.numeric(format(date1,+"%25m"))
[1]+3
>+as.numeric(format(date1,+"%25Y"))
[1]+2012|405424|有关详细信息，请参阅?yearmon和?strftime+-后者解释了可以使用的占位符字符。|405425|entityMap^0|2|9|G|8|0|0|0|0|10|C|0|0|A|8|J|9|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|V|$]]

Use the <code>format()</code> method for objects of class <code>"yearmon"</code>. Here is your example date (properly created!)

<pre><code>date1 &lt;- as.yearmon("Mar 2012", "%b %Y")
</code></pre>

Then we can extract the date parts as required:

<pre><code>&gt; format(date1, "%b") ## Month, char, abbreviated
[1] "Mar"
&gt; format(date1, "%Y") ## Year with century
[1] "2012"
&gt; format(date1, "%m") ## numeric month
[1] "03"
</code></pre>

These are returned as characters. Where appropriate, wrap in <code>as.numeric()</code> if you want the year or numeric month as a numeric variable, e.g.

<pre><code>&gt; as.numeric(format(date1, "%m"))
[1] 3
&gt; as.numeric(format(date1, "%Y"))
[1] 2012
</code></pre>

See <code>?yearmon</code> and <code>?strftime</code> for details - the latter explains the placeholder characters you can use.

blocks|key|621190|text|对于这类事情，lubridate+package非常棒：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|621191|>+require(lubridate)
>+month(date1)
[1]+3
>+year(date1)
[1]+2012|code-block|syntax|javascript|621192|entityMap|0|LINK|mutability|MUTABLE|url|http://www.r-statistics.com/2012/03/do-more-with-dates-and-times-in-r-with-lubridate-1-1-0/^0|7|H|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

The <a href="http://www.r-statistics.com/2012/03/do-more-with-dates-and-times-in-r-with-lubridate-1-1-0/" rel="noreferrer">lubridate package</a> is amazing for this kind of thing:

<pre><code>&gt; require(lubridate)
&gt; month(date1)
[1] 3
&gt; year(date1)
[1] 2012
</code></pre>

blocks|key|621264|text|我知道这里的OP使用的是zoo，但我发现这个线程正在搜索解决相同问题的标准ts解决方案。所以我想我也应该为ts添加一个zoo-free回答。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|621265|#+create+an+example+Date+
date_1+<-+as.Date("1990-01-01")
#+extract+year
as.numeric(format(date_1,+"%25Y"))
#+extract+month
as.numeric(format(date_1,+"%25m"))|code-block|syntax|javascript|621266|entityMap^0|C|3|11|2|1H|2|1N|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

I know the OP is using <code>zoo</code> here, but I found this thread googling for a standard <code>ts</code> solution for the same problem. So I thought I'd add a <code>zoo</code>-free answer for <code>ts</code> as well. 

<pre><code># create an example Date 
date_1 &lt;- as.Date("1990-01-01")
# extract year
as.numeric(format(date_1, "%Y"))
# extract month
as.numeric(format(date_1, "%m"))
</code></pre>

blocks|key|621307|text|对于较大的向量：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|621308|y+=+as.POSIXlt(date1)$year+%2B+1900++++#+x$year+:+years+since+1900
m+=+as.POSIXlt(date1)$mon+%2B+1++++++++#+x$mon+:+0–11|code-block|syntax|javascript|621309|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For large vectors:

<pre><code>y = as.POSIXlt(date1)$year + 1900 # x$year : years since 1900
m = as.POSIXlt(date1)$mon + 1 # x$mon : 0–11
</code></pre>

blocks|key|405715|text|根据注释，结果应该是月份数字(1月=+1)和4位数的年份，所以假设我们刚刚运行了问题中的代码，我们有以下代码。除了问题中已经使用的包之外，它不使用任何额外的包，非常简短，并且比任何其他解决方案都要快得多(请参阅下面的基准测试部分)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|405716|cycle(date1)
##+[1]+3
as.integer(date1)
##+[1]+2012|code-block|syntax|javascript|405717|基准测试|405718|在长度为1000的一年对象上，上面的解决方案比其他任何一年快1000倍，比其他任何一个月快200倍。|405719|library(zoo)
library(microbenchmark)
library(lubridate)

ym+<-+as.yearmon(rep(2000,+1000))

microbenchmark(
++as.integer(ym),
++as.numeric(format(ym,+"%25y")),
++as.POSIXlt(ym)$year+%2B+1900,
++year(ym)
)

Unit:+microseconds
+++++++++++++++++++++++++expr+++++min+++++++lq+++++mean+++median+++++++uq+++++max+neval+cld
+++++++++++++++as.integer(ym)++++18.2++++27.90++++28.93++++29.15++++31.15++++51.2+++100+a++
+as.numeric(format(ym,+"%25y"))+46515.8+47090.05+48122.28+47525.00+48080.25+69967.6+++100+++c
+++as.POSIXlt(ym)$year+%2B+1900+40874.4+41223.65+41798.60+41747.30+42171.25+44381.2+++100++b+
+++++++++++++++++++++year(ym)+40793.2+41167.70+42003.07+41742.40+42140.30+65203.3+++100++b+
+
microbenchmark(
++cycle(ym),
++as.numeric(format(ym,+"%25m")),
++as.POSIXlt(ym)$mon+%2B+1,
++month(ym)
)

Unit:+microseconds
+++++++++++++++++++++++++expr+++++min++++++lq++++++mean+++median+++++++uq+++++max+neval+cld
++++++++++++++++++++cycle(ym)+++138.1+++166.0+++173.893+++172.95+++181.45+++344.0+++100+a++
+as.numeric(format(ym,+"%25m"))+46637.1+46954.8+47632.307+47325.90+47672.40+67690.1+++100+++c
+++++++as.POSIXlt(ym)$mon+%2B+1+40923.3+41339.1+41976.836+41689.95+42078.15+65786.4+++100++b+
++++++++++++++++++++month(ym)+41056.4+41408.9+42082.975+41743.35+42164.95+66651.0+++100++b+|405720|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Based on comments the result should be the month number (January = 1) and the 4 digit year so assuming that we have just run the code in the question we have the following. This uses no extra packages other than what is already used in the question, is very short and is much faster than any of the other solutions (see Benchmark section below).
<pre><code>cycle(date1)
## [1] 3
as.integer(date1)
## [1] 2012
</code></pre>
<h2>Benchmark</h2>
On a yearmon object of length 1000 the solution above is about 1000x faster than any of the others for year and 200x faster for month.
<pre><code>library(zoo)
library(microbenchmark)
library(lubridate)

ym &lt;- as.yearmon(rep(2000, 1000))

microbenchmark(
 as.integer(ym),
 as.numeric(format(ym, &quot;%y&quot;)),
 as.POSIXlt(ym)$year + 1900,
 year(ym)
)

Unit: microseconds
 expr min lq mean median uq max neval cld
 as.integer(ym) 18.2 27.90 28.93 29.15 31.15 51.2 100 a 
 as.numeric(format(ym, &quot;%y&quot;)) 46515.8 47090.05 48122.28 47525.00 48080.25 69967.6 100 c
 as.POSIXlt(ym)$year + 1900 40874.4 41223.65 41798.60 41747.30 42171.25 44381.2 100 b 
 year(ym) 40793.2 41167.70 42003.07 41742.40 42140.30 65203.3 100 b 
 
microbenchmark(
 cycle(ym),
 as.numeric(format(ym, &quot;%m&quot;)),
 as.POSIXlt(ym)$mon + 1,
 month(ym)
)

Unit: microseconds
 expr min lq mean median uq max neval cld
 cycle(ym) 138.1 166.0 173.893 172.95 181.45 344.0 100 a 
 as.numeric(format(ym, &quot;%m&quot;)) 46637.1 46954.8 47632.307 47325.90 47672.40 67690.1 100 c
 as.POSIXlt(ym)$mon + 1 40923.3 41339.1 41976.836 41689.95 42078.15 65786.4 100 b 
 month(ym) 41056.4 41408.9 42082.975 41743.35 42164.95 66651.0 100 b 
</code></pre>

blocks|key|621386|text|从1800年到现在，我遇到了类似的数据问题，这对我来说很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|621387|data2$date=as.character(data2$date)+
lct+<-+Sys.getlocale("LC_TIME");+
Sys.setlocale("LC_TIME","C")
data2$date<-+as.Date(data2$date,+format+=+"%25Y+%25m+%25d")+#+and+it+works|code-block|syntax|javascript|621388|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Having had a similar problem with data from 1800 to now, this worked for me:

<pre><code>data2$date=as.character(data2$date) 
lct &lt;- Sys.getlocale("LC_TIME"); 
Sys.setlocale("LC_TIME","C")
data2$date&lt;- as.Date(data2$date, format = "%Y %m %d") # and it works
</code></pre>

I have a <code>yearmon</code> object:

<pre><code>require(zoo)
date1 &lt;- as.yearmon("Mar 2012", "%b %Y")
class(date1)
# [1] "yearmon"
</code></pre>

How can I extract the month and year from this?

<pre><code>month1 &lt;- fn(date1)
year1 &lt;- fn(date1)
</code></pre>

What function should I use in place of <code>fn()</code>

Extract month and year from a zoo::yearmon object

我有一个yearmon对象：require(zoo)date1 <- as.yearmon("Mar 2012", "%b %Y")class(date1)# [1] "yearmon"我如何从中提取月份和年份？month1 <- fn(date1)year1 <- fn(date1)我应该用什么函数来代替fn()

问从zoo::yearmon对象中提取月份和年份
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