Question

我有一个带有年份索引和val索引的数据框。

我想创建每n行val的平均值并保留相应的年份索引。

基本上，输出将是（对于n = 2）

year val
1990 Mean(row1,row2)
1992 Mean(row3,row4)
1994 Mean(row5,row6)
1996 Mean(row7,row8)

我该怎么做？

structure(list(year = c(1990, 1991, 1992, 1993, 1994, 1995, 1996, 
1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 
2008, 2009, 2010, 2011, 2012, 2013), val = c(84L, 67L, 72L, 138L, 
111L, 100L, 221L, 108L, 204L, 125L, 82L, 157L, 175L, 252L, 261L, 
185L, 146L, 183L, 245L, 172L, 98L, 216L, 89L, 144L)), .Names = c("year", 
"val"), row.names = 13:36, class = "data.frame")

Answer 1

使用data.table的简短单行解决方案：

library(data.table)

setDT(df)[,.(val=mean(val)), year-0:1]
#    year   val
# 1: 1990  75.5
# 2: 1992 105.0
# 3: 1994 105.5
# 4: 1996 164.5
# 5: 1998 164.5
# 6: 2000 119.5
# 7: 2002 213.5
# 8: 2004 223.0
# 9: 2006 164.5
#10: 2008 208.5
#11: 2010 157.0
#12: 2012 116.5

Answer 2

您可以使用rep创建分组变量：

n = 2
dd$group <- rep(1:(nrow(dd)/n), each = n)

然后，您可以使用您选择的库进行group_wise操作。我使用过data.table。

library(data.table)
setDT(dd)

#Getting the result is then trivial    
res <- dd[, .(year = min(year), mean_val = mean(val)), by = group]

Answer 3

val解决方案 - 添加分组变量（1,1,2,2,3,3等），然后计算组内year的均值，并使用最小> require(dplyr) > d %>% group_by(G=trunc(2:(n()+1)/2)) %>% summarise(mean=mean(val),year=min(year)) %>% select(-G) Source: local data frame [12 x 2] mean year 1 75.5 1990 2 105.0 1992 3 105.5 1994 4 164.5 1996 5 164.5 1998 6 119.5 2000 7 213.5 2002 8 223.0 2004 9 164.5 2006 10 208.5 2008 11 157.0 2010 12 116.5 2012在组内，然后删除分组变量：

广义为meanN = function(df, n){ df %>% group_by(G=(0:(n()-1))%/%n) %>% summarise(mean=mean(val),year=min(year)) %>% select(-G) } > meanN(d, 2) Source: local data table [12 x 2] mean year 1 75.5 1990 2 105.0 1992 3 105.5 1994 4 164.5 1996 5 164.5 1998 6 119.5 2000 7 213.5 2002 8 223.0 2004 9 164.5 2006 10 208.5 2008 11 157.0 2010 12 116.5 2012 > meanN(d, 12) Source: local data table [2 x 2] mean year 1 122.4167 1990 2 180.5000 2002的函数，并使用更简洁的方法计算分组变量：

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Answer 4

使用zoo package中的rollapply

> library(zoo)
> res <- rollapply(df, width=2, by=2, FUN=mean)
> res[,1] <- floor(res[,1])
> res
      year   val
 [1,] 1990  75.5
 [2,] 1992 105.0
 [3,] 1994 105.5
 [4,] 1996 164.5
 [5,] 1998 164.5
 [6,] 2000 119.5
 [7,] 2002 213.5
 [8,] 2004 223.0
 [9,] 2006 164.5
[10,] 2008 208.5
[11,] 2010 157.0
[12,] 2012 116.5

或者：

rollapply(df, width=2, by=2, FUN=function(x) c(min(x), mean(x)))[, c(1,4)]

Answer 5

您可以使用aggregate，对已舍入的年份值进行分组：

setNames(aggregate(val~I(2*floor((year-min(year))/2)+min(year)), data=dat, mean),
         c("year", "val"))
#    year   val
# 1  1990  75.5
# 2  1992 105.0
# 3  1994 105.5
# 4  1996 164.5
# 5  1998 164.5
# 6  2000 119.5
# 7  2002 213.5
# 8  2004 223.0
# 9  2006 164.5
# 10 2008 208.5
# 11 2010 157.0
# 12 2012 116.5

Answer 6

您可以将seq与colMeans功能

一起使用

data.frame(Year = df[seq(1, length(df$year), 2), ]$year, Mean = colMeans(matrix(df$val, nrow=2)))

#   Year  Mean
# 1  1990  75.5
# 2  1992 105.0
# 3  1994 105.5
# 4  1996 164.5
# 5  1998 164.5
# 6  2000 119.5
# 7  2002 213.5
# 8  2004 223.0
# 9  2006 164.5
# 10 2008 208.5
# 11 2010 157.0
# 12 2012 116.5

Answer 7

试试这个单行：

> t(sapply(split(dat,rep(seq(1,nrow(dat),2),each=2)),colMeans))
     year   val
1  1990.5  75.5
3  1992.5 105.0
5  1994.5 105.5
7  1996.5 164.5
9  1998.5 164.5
11 2000.5 119.5
13 2002.5 213.5
15 2004.5 223.0
17 2006.5 164.5
19 2008.5 208.5
21 2010.5 157.0
23 2012.5 116.5

如果需要，您可以在这一年中完成。

如何获得每n行的平均值并保留日期索引？

7 个答案: