R 平均每年前5个数据点
我试图平均每个“年”的前5个“var1”数据点。我的数据如下。每个年度数据的长度不一样。非常感谢你的帮助!:) 像这样的R 平均每年前5个数据点,r,aggregate,mean,R,Aggregate,Mean,我试图平均每个“年”的前5个“var1”数据点。我的数据如下。每个年度数据的长度不一样。非常感谢你的帮助!:) 像这样的 t <- read.csv("t.txt", sep="") ## Read data myMean <- function(x) ifelse(length(x)<5, mean(x), mean(x[1:5])) ans <- aggregate(var1 ~ year, data = t, FUN = myMean) ans year var
t <- read.csv("t.txt", sep="") ## Read data
myMean <- function(x) ifelse(length(x)<5, mean(x), mean(x[1:5]))
ans <- aggregate(var1 ~ year, data = t, FUN = myMean)
ans
year var1
1 2008 14
2 2009 13
3 2010 12
像这样的事
t <- read.csv("t.txt", sep="") ## Read data
myMean <- function(x) ifelse(length(x)<5, mean(x), mean(x[1:5]))
ans <- aggregate(var1 ~ year, data = t, FUN = myMean)
ans
year var1
1 2008 14
2 2009 13
3 2010 12
t使用data.table,我们将'data.frame'转换为'data.table'(setDT(df1)
),按'year'分组,我们得到'var1'的前5个值(带head
),并得到平均值
library(data.table)
setDT(df1)[, list(var1=mean(head(var1,5))), year]
# year var1
#1: 2010 2.4
#2: 2009 2.6
#3: 2008 2.8
使用data.table,我们将'data.frame'转换为'data.table'(setDT(df1)
),按'year'分组,得到'var1'的前5个值(带head
),并得到平均值
library(data.table)
setDT(df1)[, list(var1=mean(head(var1,5))), year]
# year var1
#1: 2010 2.4
#2: 2009 2.6
#3: 2008 2.8
这里还有一个选项,使用split
和lappy
:
sapply( split(X$var1,X$year), function(x) ifelse(length(x)<5, mean(x), mean(x[1:5])) )
akrun
是数据表,PoChoi
是数据帧,mra68
是命名向量:
> akrun
year var1
1: 2011 5.5
2: 2010 2.4
3: 2009 2.6
4: 2008 2.8
> PoChoi.1
year var1
1 2008 2.8
2 2009 2.6
3 2010 2.4
4 2011 5.5
> PoChoi.2
year var1
1 2008 2.8
2 2009 2.6
3 2010 2.4
4 2011 5.5
> mra68.1
2008 2009 2010 2011
2.8 2.6 2.4 5.5
> mra68.2
2008 2009 2010 2011
2.8 2.6 2.4 5.5
一个更大的例子:
library(microbenchmark)
library(data.table)
set.seed(1)
X <- data.frame( year = sample( 1500:2015, 10000, replace=TRUE ),
var1 = sample( 1:10, 10000, replace=TRUE ) )
myMean <- function(x) ifelse(length(x)<5, mean(x), mean(x[1:5]))
microbenchmark(
akrun = setDT(X)[, list(var1=mean(head(var1,5))), year],
PoChoi.1 = aggregate(var1 ~ year, data = X, FUN = myMean),
PoChoi.2 = aggregate(var1 ~ year, data = X, FUN = function(x) ifelse(length(x)<5, mean(x), mean(x[1:5]))),
mra68.1 = sapply( split(X$var1,X$year), myMean ),
mra68.2 = sapply( split(X$var1,X$year), function(x) ifelse(length(x)<5, mean(x), mean(x[1:5])) ),
times = 1000
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# akrun 15.44811 23.50436 36.81674 43.12405 44.22435 69.62202 1000
# PoChoi.1 33.96411 51.52858 83.29682 95.53486 99.60884 241.59967 1000
# PoChoi.2 33.64844 51.70747 83.47835 95.07223 99.44127 247.55881 1000
# mra68.1 11.05145 17.33191 27.21526 31.41954 32.34819 126.89461 1000
# mra68.2 11.05054 17.16615 26.96236 31.25061 32.14054 85.44422 1000
这里还有一个选项,使用split
和lappy
:
sapply( split(X$var1,X$year), function(x) ifelse(length(x)<5, mean(x), mean(x[1:5])) )
akrun
是数据表,PoChoi
是数据帧,mra68
是命名向量:
> akrun
year var1
1: 2011 5.5
2: 2010 2.4
3: 2009 2.6
4: 2008 2.8
> PoChoi.1
year var1
1 2008 2.8
2 2009 2.6
3 2010 2.4
4 2011 5.5
> PoChoi.2
year var1
1 2008 2.8
2 2009 2.6
3 2010 2.4
4 2011 5.5
> mra68.1
2008 2009 2010 2011
2.8 2.6 2.4 5.5
> mra68.2
2008 2009 2010 2011
2.8 2.6 2.4 5.5
一个更大的例子:
library(microbenchmark)
library(data.table)
set.seed(1)
X <- data.frame( year = sample( 1500:2015, 10000, replace=TRUE ),
var1 = sample( 1:10, 10000, replace=TRUE ) )
myMean <- function(x) ifelse(length(x)<5, mean(x), mean(x[1:5]))
microbenchmark(
akrun = setDT(X)[, list(var1=mean(head(var1,5))), year],
PoChoi.1 = aggregate(var1 ~ year, data = X, FUN = myMean),
PoChoi.2 = aggregate(var1 ~ year, data = X, FUN = function(x) ifelse(length(x)<5, mean(x), mean(x[1:5]))),
mra68.1 = sapply( split(X$var1,X$year), myMean ),
mra68.2 = sapply( split(X$var1,X$year), function(x) ifelse(length(x)<5, mean(x), mean(x[1:5])) ),
times = 1000
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# akrun 15.44811 23.50436 36.81674 43.12405 44.22435 69.62202 1000
# PoChoi.1 33.96411 51.52858 83.29682 95.53486 99.60884 241.59967 1000
# PoChoi.2 33.64844 51.70747 83.47835 95.07223 99.44127 247.55881 1000
# mra68.1 11.05145 17.33191 27.21526 31.41954 32.34819 126.89461 1000
# mra68.2 11.05054 17.16615 26.96236 31.25061 32.14054 85.44422 1000
您是否在中进行了搜索和研究?RAlso已经回答了很多类似的问题,请记下您对2008年和2009年的预期答案,因为此答案将根据数据的排序方式而有所不同。可能重复的问题谢谢您的评论!:)您是否在中进行了搜索和研究?RAlso已经回答了很多类似的问题,请记下您对2008年和2009年的预期答案,因为此答案将根据数据的排序方式而有所不同。可能重复的问题谢谢您的评论!:)我喜欢这个答案,假设OP想要什么,因为它考虑了少于5个案例的情况。这个简单的代码很棒,这正是我想要的!非常感谢!:)我喜欢这个答案,假设OP想要什么,因为它考虑了少于5个案例的情况。这个简单的代码很棒,这正是我想要的!非常感谢!:)使用aggregate
公式var1~year
对新的R用户来说更直观,但是,相比之下,它的缺点是速度较慢。使用aggregate
公式var1~year
对新的R用户来说更直观,但相比之下,它的缺点是速度较慢。非常感谢,谢谢你的建议。下面的代码非常适合我测试和学习:)非常感谢阿克伦的建议。下面的代码非常适合我测试和学习:)