Question

我有每日数据，我想计算每日测量在一定范围内的每年的次数。数据也按因子分组，因此我需要知道每个因子在一定范围内（例如15到18）每年的天数

我有一个超过100年的大型数据集，但这里只是一些数据：

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

mydf <- data.frame(Date, People, Country)

我想知道＆＃34; People＆＃34;每年的价值。每个国家的年龄在15到18之间。

所以我的输出将是一个新的数据框，如：

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

任何帮助都会受到高度赞赏，因为我正在努力解决这个问题并寻找答案，但我无法找到涉及日期和因素的解决方案。

Answer 1

您可以使用lubridate和dplyr来实现这一目标。使用year()按年份和国家/地区获取年份和组。最后一步是有条件的总结：

library(dplyr)
library(lubridate)

mydf %>%
  group_by(year = year(Date), Country) %>%
  summarise(p = sum(between(People, 15, 18)))

<小时/> 这可能会产生

   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

Answer 2

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

aggregate( mydf['People'], list(mydf[['Country']], format(mydf$Date, "%Y") ), 
                           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

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

 aggregate( mydf['People'], list(Cntry=mydf[['Country']], Yr=format(mydf$Date, "%Y") ), 
                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

Answer 3

对于data.table解决方案：

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

Answer 4

考虑基础R聚合：

mydf$Year <- format(mydf$Date, "%Y")
mydf$NumberTime15_18 <- ifelse(mydf$People >= 15 & mydf$People <= 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

R - 计算每个因子在特定范围内的每日值的数量，并按年度总结

4 个答案: