blocks|key|2339986|text|您是否希望获得一天%2B一天的一个数字部分作为输出？如果是这样的话，像这样的东西将会起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2339987|test+<-+strptime(x1,"%25m/%25d/%25Y+%25H:%25M")

(test$yday%2B1)+%2B+(test$hour/24)+%2B+(test$min/(24*60))
#[1]+125.083333+365.248611+++2.684028|code-block|syntax|javascript|2339988|虽然这符合您的要求，但我认为删除%2B1可能更有意义：|offset|length|style|CODE|2339989|(test$yday)+%2B+(test$hour/24)+%2B+(test$min/(24*60))
#[1]+124.083333+364.248611+++1.684028|2339990|尽管我的感觉有点麻木，Dirk将会出现，并向我展示如何使用POSIXct日期/时间表示来实现这一点。|2339991|下面是使用基函数来尝试这样一个答案：|2339992|mapply(julian,+as.POSIXct(test),+paste(format(test,"%25Y"),"01","01",sep="-"))
#[1]+124.083333+364.248611+++1.684028|2339993|entityMap^0|0|0|G|2|0|0|T|7|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@$I|13|J|14|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

Are you hoping to get the day + a numerical part of a day as output? If so, something like this will work:

<pre><code>test &lt;- strptime(x1,"%m/%d/%Y %H:%M")

(test$yday+1) + (test$hour/24) + (test$min/(24*60))
#[1] 125.083333 365.248611 2.684028
</code></pre>

Although this matches what you ask for, I think removing the <code>+1</code> might make more sense:

<pre><code>(test$yday) + (test$hour/24) + (test$min/(24*60))
#[1] 124.083333 364.248611 1.684028
</code></pre>

Though my spidey senses are tingling that Dirk is going to show up and show me how to do this with a <code>POSIXct</code> date/time representation.

Here is an attempt of such an answer using base functions:

<pre><code>mapply(julian, as.POSIXct(test), paste(format(test,"%Y"),"01","01",sep="-"))
#[1] 124.083333 364.248611 1.684028
</code></pre>

blocks|key|2340024|text|您还可以使用POSIXct和POSIXlt表示以及来自xts的firstof函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2340025|x1+<-+c("5/5/2007+02:00",+"12/31/2007+05:58",+"1/2/2008+16:25")
x1
##+[1]+"5/5/2007+02:00"+++"12/31/2007+05:58"+"1/2/2008+16:25"++


y+<-+as.POSIXlt(x1,+format+=+"%25m/%25d/%25Y+%25H:%25M")

result+<-+mapply(julian,+x+=+as.POSIXct(y),+origin+=+firstof(y$year+%2B+1900))

result
##+[1]+124.083333+364.248611+++1.684028|code-block|syntax|javascript|2340026|如果不想使用xts，那么可以使用下面这样的代码|2340027|result+<-+mapply(julian,+
+++++++++++++++++x+=+as.POSIXct(x1,+format+=+"%25m/%25d/%25Y+%25H:%25M",+tz+=+"GMT"),
+++++++++++++++++origin+=+as.Date(paste0(gsub(".*([0-9]{4}).*",+"\\1",+x1),+
+++++++++++++++++++++++++++++++++++++++++"-01-01"),
++++++++++++++++++++++++++++++++++tz+=+"GMT"))

result
##+[1]+124.083333+364.248611+++1.684028|2340028|entityMap^0|6|7|E|7|R|3|V|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

You can also use <code>POSIXct</code> and <code>POSIXlt</code> representations along with <code>firstof</code> function from <code>xts</code>.

<pre><code>x1 &lt;- c("5/5/2007 02:00", "12/31/2007 05:58", "1/2/2008 16:25")
x1
## [1] "5/5/2007 02:00" "12/31/2007 05:58" "1/2/2008 16:25" 


y &lt;- as.POSIXlt(x1, format = "%m/%d/%Y %H:%M")

result &lt;- mapply(julian, x = as.POSIXct(y), origin = firstof(y$year + 1900))

result
## [1] 124.083333 364.248611 1.684028
</code></pre>

if you don't want to use xts then perhaps something like this

<pre><code>result &lt;- mapply(julian, 
 x = as.POSIXct(x1, format = "%m/%d/%Y %H:%M", tz = "GMT"),
 origin = as.Date(paste0(gsub(".*([0-9]{4}).*", "\\1", x1), 
 "-01-01"),
 tz = "GMT"))

result
## [1] 124.083333 364.248611 1.684028
</code></pre>

blocks|key|931028|text|如果你想做相反的事情(将一年中的某一天转换为日期和时间)，你可以使用这个小函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|931029|931030|+doy2date+=函数(Mydoy){|blockquote|931031|931032|mydate+=+as.Date(mydoy，origin+=+"2008-01-01+00:00:00"，tz+=+"GMT")|931033|dech+=+(mydoy+-+as.integer(mydoy))+*+24|931034|myh+=+as.integer(dech)|931035|mym+=+as.integer(+(dech+-+as.integer(dech))+*+60)|931036|mys+=圆形(i(dech-+as.integer(dech))+*+60)+-+mym)+*+60)，digits=0+)|931037|posixdate+=as.POSIXct(粘贴(mydate，“"，myh，":"，mym，":"，mys，sep="")，tz+=+"GMT")|931038|return(posixdate)|931039|}|931040|931041|举个例子，如果你尝试：|931042|931043|+doy2date(117.6364)+|931044|931045|该函数将返回"2008-04-27+15:16:25+GMT“作为POSIXct。|931046|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|19|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1B|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|1C|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|1D|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1E|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1F|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1G|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1H|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1I|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1K|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|10|3|11|5|E|7|1M|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]|$1|13|3|14|5|6|7|1O|8|@]|9|@]|A|$]]|$1|15|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|16|$]]

If you want to do the other way around (convert day of the year to date and time), you can use this little function:

<blockquote>
 doy2date = function(mydoy){ 
 mydate = as.Date(mydoy, origin = "2008-01-01 00:00:00", tz = "GMT") 
 dech = (mydoy - as.integer(mydoy)) * 24 
 myh = as.integer(dech) 
 mym = as.integer( (dech - as.integer(dech)) * 60) 
 mys = round(I( (((dech - as.integer(dech)) * 60) - mym) * 60), digits=0 ) 
 posixdate = as.POSIXct(paste(mydate, " ", myh,":",mym,":",mys, sep=""), tz = "GMT") 
 return(posixdate) 
 }
</blockquote>

As an example, if you try:

<blockquote>
 doy2date(117.6364) 
</blockquote>

The function will return "2008-04-27 15:16:25 GMT" as a POSIXct.

I have data for more than 3 years. For each year I want to find the day corresponding to Jaunary 1 of that year. For example: 

<pre><code>&gt; x &lt;- c('5/5/2007','12/31/2007','1/2/2008')
&gt; #Convert to day of year (julian date) –
&gt; strptime(x,"%m/%d/%Y")$yday+1
[1] 125 365 2
</code></pre>

I want to know how to do the same thing but with time added. But I still get the day not time. Can anyone suggest what is the better way to find the julian date with date and time ?

<pre><code>&gt; x1 &lt;- c('5/5/2007 02:00','12/31/2007 05:58','1/2/2008 16:25')
&gt; #Convert to day of year (julian date) –
&gt; strptime(x1,"%m/%d/%Y %H:%M")$yday+1
[1] 125 365 2
</code></pre>

Rather than this result, I want the output in decimal days. For example the first example would be <code>125.0833333</code> and so on. 

Thank you so much.

Convert to the day and time of the year in R

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有超过3年的数据。对于每一年，我希望找到与该年的Jaunary 1相对应的日期。例如：> x <- c('5/5/2007','12/31/2007','1/2/2008')> #Convert to day of year (julian date) –> strptime(x,"%m/%d/%Y")$yday+1...

问转换为R中一年中的日期和时间
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问转换为R中一年中的日期和时间EN