blocks|key|3848321|text|您可以使用lubridate和dplyr实现这一点。使用year()根据年份和国家/地区获取年份和组。最后一步是条件总结：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3848322|library(dplyr)
library(lubridate)

mydf+%25>%25
++group_by(year+=+year(Date),+Country)+%25>%25
++summarise(p+=+sum(between(People,+15,+18)))|code-block|syntax|javascript|3848323|​|3848324|3848325|这可能会导致|3848326|3848327|+++year+Country+++++++p
++<dbl>+<fct>+++++<int>
1+2010.+Australia++++22
2+2010.+Canada+++++++34
3+2010.+France+++++++26
4+2011.+Australia++++21
5+2011.+Canada+++++++30
6+2011.+France+++++++13
7+2012.+Australia++++28
8+2012.+Canada+++++++31
9+2012.+France+++++++23|3848328|entityMap^0|5|9|F|5|S|6|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$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]]|$1|K|3|L|5|6|7|12|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|13|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|14|8|@]|D|@]|E|$]]|$1|P|3|L|5|6|7|15|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|T|$]]

You could achieve it with <code>lubridate</code> and <code>dplyr</code>. Use <code>year()</code> to grab the year and group by the year and the country. The last step is a conditional summing up:

<pre><code>library(dplyr)
library(lubridate)

mydf %&gt;%
 group_by(year = year(Date), Country) %&gt;%
 summarise(p = sum(between(People, 15, 18)))
</code></pre>

<hr>
This could yield

<pre><code> year Country p
 &lt;dbl&gt; &lt;fct&gt; &lt;int&gt;
1 2010. Australia 22
2 2010. Canada 34
3 2010. France 26
4 2011. Australia 21
5 2011. Canada 30
6 2011. France 13
7 2012. Australia 28
8 2012. Canada 31
9 2012. France 23
</code></pre>

blocks|key|1362662|text|对于data.table解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1362663|library(data.table)
setDT(mydf)[,(People=sum(between(People,+15,+18))),+by+=+.(year(Date),+Country)]

+++year+++Country+V1
1:+2010++++Canada+22
2:+2010+Australia+17
3:+2010++++France+22
4:+2011++++Canada+23
5:+2011++++France+22
6:+2011+Australia+26
7:+2012++++Canada+21
8:+2012++++France+29
9:+2012+Australia+26|code-block|syntax|javascript|1362664|entityMap^0|2|A|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|$]]

For a <code>data.table</code> solution:

<pre><code>library(data.table)
setDT(mydf)[,(People=sum(between(People, 15, 18))), by = .(year(Date), Country)]

 year Country V1
1: 2010 Canada 22
2: 2010 Australia 17
3: 2010 France 22
4: 2011 Canada 23
5: 2011 France 22
6: 2011 Australia 26
7: 2012 Canada 21
8: 2012 France 29
9: 2012 Australia 26
</code></pre>

blocks|key|3848355|text|考虑基数R聚合：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3848356|mydf$Year+<-+format(mydf$Date,+"%25Y")
mydf$NumberTime15_18+<-+ifelse(mydf$People+>=+15+&+mydf$People+<=+18,+1,+0)

aggregate(NumberTime15_18+~+Country+%2B+Year,+mydf,+sum)

#+++++Country+Year+NumberTime15_18
#+1+Australia+2010++++++++++++++22
#+2++++Canada+2010++++++++++++++17
#+3++++France+2010++++++++++++++28
#+4+Australia+2011++++++++++++++26
#+5++++Canada+2011++++++++++++++24
#+6++++France+2011++++++++++++++20
#+7+Australia+2012++++++++++++++16
#+8++++Canada+2012++++++++++++++27
#+9++++France+2012++++++++++++++21|code-block|syntax|javascript|3848357|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|$]]

Consider base R aggregation:

<pre><code>mydf$Year &lt;- format(mydf$Date, "%Y")
mydf$NumberTime15_18 &lt;- ifelse(mydf$People &gt;= 15 &amp; mydf$People &lt;= 18, 1, 0)

aggregate(NumberTime15_18 ~ Country + Year, mydf, sum)

# Country Year NumberTime15_18
# 1 Australia 2010 22
# 2 Canada 2010 17
# 3 France 2010 28
# 4 Australia 2011 26
# 5 Canada 2011 24
# 6 France 2011 20
# 7 Australia 2012 16
# 8 Canada 2012 27
# 9 France 2012 21
</code></pre>

blocks|key|1362707|text|下面是必需的基本解决方案。要点:使用format.Date将日期转换为字符年份值，按分组需要是list-object：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1362708|aggregate(+mydf['People'],+list(mydf[['Country']],+format(mydf$Date,+"%25Y")+),+
+++++++++++++++++++++++++++FUN=function(d)+sum(+d+>=15+&+d+<=18)+)
++++Group.1+Group.2+People
1+Australia++++2010+++++25
2++++Canada++++2010+++++22
3++++France++++2010+++++24
4+Australia++++2011+++++27
5++++Canada++++2011+++++19
6++++France++++2011+++++33
7+Australia++++2012+++++19
8++++Canada++++2012+++++33
9++++France++++2012+++++24|code-block|syntax|javascript|1362709|如果您希望结果数据帧具有不同的列名，则将这些列名添加到按组定义中的列表中：|1362710|+aggregate(+mydf['People'],+list(Cntry=mydf[['Country']],+Yr=format(mydf$Date,+"%25Y")+),+
++++++++++++++++function(d)+sum(+d+>=15+&+d+<=18)+)
++++++Cntry+++Yr+People
1+Australia+2010+++++25
2++++Canada+2010+++++22
3++++France+2010+++++24
4+Australia+2011+++++27
5++++Canada+2011+++++19
6++++France+2011+++++33
7+Australia+2012+++++19
8++++Canada+2012+++++33
9++++France+2012+++++24|1362711|entityMap^0|I|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

And here is the requisite base solution. Key points: convert dates to character year values with <code>format.Date</code> and the by-grouping needs to be a list-object:

<pre><code>aggregate( mydf['People'], list(mydf[['Country']], format(mydf$Date, "%Y") ), 
 FUN=function(d) sum( d &gt;=15 &amp; d &lt;=18) )
 Group.1 Group.2 People
1 Australia 2010 25
2 Canada 2010 22
3 France 2010 24
4 Australia 2011 27
5 Canada 2011 19
6 France 2011 33
7 Australia 2012 19
8 Canada 2012 33
9 France 2012 24
</code></pre>

If you want the resulting dataframe to have different column names then add those to the list inside the by-group definition:

<pre><code> aggregate( mydf['People'], list(Cntry=mydf[['Country']], Yr=format(mydf$Date, "%Y") ), 
 function(d) sum( d &gt;=15 &amp; d &lt;=18) )
 Cntry Yr People
1 Australia 2010 25
2 Canada 2010 22
3 France 2010 24
4 Australia 2011 27
5 Canada 2011 19
6 France 2011 33
7 Australia 2012 19
8 Canada 2012 33
9 France 2012 24
</code></pre>

I have daily data and I'd like to count the number of times per year the daily measurement is within a certain range. The data is also grouped by factor, so I need to know number of days per year for each factor that is within a certain range (e.g. 15 to 18)

I have a large dataset of over 100 years but here is some data just for this example:

<pre><code>Date &lt;- seq(as.Date("2010/01/01"), by = "day", length.out = 1095)
People &lt;- sample.int(20, 1095, replace = TRUE)
Country &lt;- sample(x = c("Australia", "Canada", "France"), size = 1095, replace = TRUE)

mydf &lt;- data.frame(Date, People, Country)
</code></pre>

I would like to know the number of times per year the value of "People" is between 15 and 18 for each country.

So my output would be a new dataframe, something like:

<pre><code>myDate People Country
2010 45 Australia
2010 10 Canada
2010 24 France
2011 33 Australia
2011 100 Canada
2011 4 France
2012 21 Australia
2012 66 Canada
2012 211 France
</code></pre>

Any help would be greatly appreciated as I'm struggling with this and have looked for answers but I can't find a solution that involves the date and factor.

R - counting number of daily values within a specific range per factor and summarising by year

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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我有每天的数据，我想统计一下每天的测量在一定范围内的次数。数据也是按因子分组的，因此我需要知道在某个范围内(例如15到18)的每个因子每年的天数我有一个超过100年的大型数据集，但这里有一些仅用于此示例的数据：Date <- seq(as.Date("2010/01/01"), by = "day", length.o...

问R-统计每个因子在特定范围内的日值数量，并按年汇总
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问R-统计每个因子在特定范围内的日值数量，并按年汇总EN