blocks|key|2938780|text|我很确定这是在某个地方复制的，但我在前三次搜索中失败了，下面是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2938781|tapply(+dat$long,+list(dt+=+format(+as.POSIXct(dat$date,+"%25d-%25m-%25Y+%25H:%25M"),+
++++++++++++++++++++++++++++++++++++"%25d-%25m-%25Y"),
++++++++++++++++++++++++grp=dat$apps+),
++++++++++++++++++sum)|code-block|syntax|javascript|2938782|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|$]]

I'm fairly sure this is duplicated somewhere but I failed in my first three searches so here it is:

<pre><code>tapply( dat$long, list(dt = format( as.POSIXct(dat$date, "%d-%m-%Y %H:%M"), 
 "%d-%m-%Y"),
 grp=dat$apps ),
 sum)
</code></pre>

blocks|key|2245663|text|首先，我假设您的data.frame名为dd。在这里，它是以拷贝/可压纸的形式出现的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2245664|dd+<-+structure(list(date+=+structure(c(1L,+2L,+2L,+3L,+4L,+5L,+5L,+
5L,+6L,+7L,+7L,+8L,+9L,+9L),+.Label+=+c("10/22/2013+23:51",+"10/22/2013+23:52",+
"10/23/2013+7:03",+"10/23/2013+7:13",+"10/23/2013+7:31",+"10/24/2013+0:54",+
"10/24/2013+1:16",+"10/24/2013+3:27",+"10/24/2013+7:30"),+class+=+"factor"),+
++++apps+=+structure(c(1L,+2L,+3L,+3L,+1L,+2L,+1L,+2L,+2L,+3L,+
++++3L,+1L,+1L,+1L),+.Label+=+c("A",+"B",+"C"),+class+=+"factor"),+
++++long+=+c(2L,+3L,+1L,+5L,+1L,+4L,+5L,+2L,+3L,+2L,+1L,+2L,+
++++3L,+1L)),+.Names+=+c("date",+"apps",+"long"),+class+=+"data.frame",+row.names+=+c(NA,+
-14L))|code-block|syntax|javascript|2245665|您应该将日期转换为适当的日期值。|2245666|dd$date+<-+as.POSIXct(as.character(dd$date),+format="%25m/%25d/%25Y+%25H:%25M",+tz="GMT")|2245667|然后，您可以使用aggregate创建一个很好的aggregate，这里使用as.Date来节省时间。|2245668|aggregate(long+~+as.Date(date)+%2B+apps,+dd,+FUN=sum)|2245669|这会返回|2245670|++as.Date(date)+apps+long
1++++2013-10-22++++A++++2
2++++2013-10-23++++A++++6
3++++2013-10-24++++A++++6
4++++2013-10-22++++B++++3
5++++2013-10-23++++B++++6
6++++2013-10-24++++B++++3
7++++2013-10-22++++C++++1
8++++2013-10-23++++C++++5
9++++2013-10-24++++C++++3|2245671|entityMap^0|K|2|0|0|0|0|8|9|O|9|12|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$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]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1C|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|X|$]]

First, i'm assuming your data.frame is called <code>dd</code>. Here it is in a copy/pasteable form

<pre><code>dd &lt;- structure(list(date = structure(c(1L, 2L, 2L, 3L, 4L, 5L, 5L, 
5L, 6L, 7L, 7L, 8L, 9L, 9L), .Label = c("10/22/2013 23:51", "10/22/2013 23:52", 
"10/23/2013 7:03", "10/23/2013 7:13", "10/23/2013 7:31", "10/24/2013 0:54", 
"10/24/2013 1:16", "10/24/2013 3:27", "10/24/2013 7:30"), class = "factor"), 
 apps = structure(c(1L, 2L, 3L, 3L, 1L, 2L, 1L, 2L, 2L, 3L, 
 3L, 1L, 1L, 1L), .Label = c("A", "B", "C"), class = "factor"), 
 long = c(2L, 3L, 1L, 5L, 1L, 4L, 5L, 2L, 3L, 2L, 1L, 2L, 
 3L, 1L)), .Names = c("date", "apps", "long"), class = "data.frame", row.names = c(NA, 
-14L))
</code></pre>

You should convert your dates to a proper date value

<pre><code>dd$date &lt;- as.POSIXct(as.character(dd$date), format="%m/%d/%Y %H:%M", tz="GMT")
</code></pre>

Then you can create a nice data.frame with <code>aggregate</code>, here using <code>as.Date</code> to strip off time

<pre><code>aggregate(long ~ as.Date(date) + apps, dd, FUN=sum)
</code></pre>

This returns

<pre><code> as.Date(date) apps long
1 2013-10-22 A 2
2 2013-10-23 A 6
3 2013-10-24 A 6
4 2013-10-22 B 3
5 2013-10-23 B 6
6 2013-10-24 B 3
7 2013-10-22 C 1
8 2013-10-23 C 5
9 2013-10-24 C 3
</code></pre>

blocks|key|2938861|text|在Using先生的dplyr上使用dd|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2938862|library(dplyr)
dd%25>%25+
group_by(apps,+date=gsub("\\s%2B.*","",date))%25>%25
summarize(long=sum(long))
#++++++apps+++++++date+long
#+1++++A+10/22/2013++++2
#+2++++A+10/23/2013++++6
#+3++++A+10/24/2013++++6
#+4++++B+10/22/2013++++3
#+5++++B+10/23/2013++++6
#+6++++B+10/24/2013++++3
#+7++++C+10/22/2013++++1
#+8++++C+10/23/2013++++5
#+9++++C+10/24/2013++++3|code-block|syntax|javascript|2938863|entityMap^0|9|5|H|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Using <code>dplyr</code> on Mr.Flick's <code>dd</code>

<pre><code>library(dplyr)
dd%&gt;% 
group_by(apps, date=gsub("\\s+.*","",date))%&gt;%
summarize(long=sum(long))
# apps date long
# 1 A 10/22/2013 2
# 2 A 10/23/2013 6
# 3 A 10/24/2013 6
# 4 B 10/22/2013 3
# 5 B 10/23/2013 6
# 6 B 10/24/2013 3
# 7 C 10/22/2013 1
# 8 C 10/23/2013 5
# 9 C 10/24/2013 3
</code></pre>

I have dataframe like this 

<pre><code>date apps long
10/22/2013 23:51 A 2
10/22/2013 23:52 B 3
10/22/2013 23:52 C 1
10/23/2013 7:03 C 5
10/23/2013 7:13 A 1
10/23/2013 7:31 B 4
10/23/2013 7:31 A 5
10/23/2013 7:31 B 2
10/24/2013 0:54 B 3
10/24/2013 1:16 C 2
10/24/2013 1:16 C 1
10/24/2013 3:27 A 2
10/24/2013 7:30 A 3
10/24/2013 7:30 A 1
</code></pre>

The problems that i have is :
I want to sum how long A, B, C apps spent time for each days. so the output will looks like:

<pre><code>A 10/22/2013 2
A 10/23/2013 6
A 10/24/2013 6
etc...
</code></pre>

I've tried some syntax but did not work, thank you

Sum by condition in two columns data frame 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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我有这样的数据date            apps    long10/22/2013 23:51    A   210/22/2013 23:52    B   310/22/2013 23:52    C   110/23/2013 7:03     C   510/23/2013 7:13     A   1...

问两列数据帧R中按条件求和
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问两列数据帧R中按条件求和EN