如何计算R中某些连续数据值的平均值

Question

I have a data set for example

a<-c(1,2,3,4,5,6,7,8,9)

I want to calculate average of every three consecutive data values. Say, data values

1:3,4:6,7:9

What command should I use?

Answer 1

This is another way:

Make another vector that contains different levels for 1:3, 4:6, 7:9

a<-c(1,2,3,4,5,6,7,8,9)
b<-rep(1:3,each=3)
x<-ave(a, b, FUN=mean)  #use ave to find the means
x
#[1] 2 2 2 5 5 5 8 8 8  - gives this output

x[seq(1, length(x), 3)]  #this will output every 3rd element, giving:
#[1] 2 5 8

and if you wanted it on one row:

ave(a, rep(1:3,each=3), FUN=mean)[seq(1, length(a), 3)]

And an additional way - use some rolling mean function (eg from ZOO package or TTR package) and select the 3rd element of each:

library(TTR)
runMean(a,3)[seq(3, length(a), 3)]
#[1] 2 5 8

and of course this principle could be extended to the base way of calculating rolling averages:

filter(a, rep(1/3,3), sides=1)[seq(3, length(a), 3)]

Answer 2

Here's a possible RcppRoll approach

library(RcppRoll)
n <- 3 # The summing range
a <- 1:9 # Your vector
roll_mean(a, n)[seq_len(length(a) - n + 1) %% n == 1]
## [1] 2 5 8

Answer 3

1) rollapply Try this:

library(zoo)
a <- 1:9
rollapply(a, 3, mean, by = 3, align = "left", partial = TRUE)
## [1] 2 5 8

It also works if the length of a is not a multiple of 3 in which case it still averages the small portion at the end. If you want any small portion at the end to be dropped then omit the partial=TRUE argument. If you know that the length of a is always a multiple of 3 then the partial = TRUE argument can be omitted since it has no effect in that case.

2) tapply Here is a second alternative approach. c(gl(n, 3, n)) creates a grouping vector c(1, 1, 1, 2, 2, 2, ...)) of length n and then tapply applies mean to the values of a in each group:

n <- length(a)
tapply(a, c(gl(n, 3, n)), mean)
## 1 2 3 
## 2 5 8 

3) aggregate Similar to tapply but gives a data frame as output:

n <- length(a)
group <- gl(n, 3, n)
aggregate(a ~ group, FUN = mean)
##   group a
## 1     1 2
## 2     2 5
## 3     3 8

Answer 4

This worked for me as well:

v  <- 1:9  # a given vector
gr <- 3    # consider a sequence of 3 consecutive elements
length(v) <- prod(dim(matrix(v, nrow=gr))) # will stretch the vector with NA-s if needed
colMeans(matrix(v, nrow=gr), na.rm=TRUE)
[1] 2 5 8

Need to pay attention to recycling when converting from vector to matrix. For example:

v  <- 1:11
gr <- 3 
length(v) <- prod(dim(matrix(v, nrow=gr))); v
[1]  1  2  3  4  5  6  7  8  9 10 11 NA

# Will warn about the recycling 
# Warning message:
# In matrix(v, nrow = gr) :
#  data length [11] is not a sub-multiple or multiple of the number of rows [3]
# But the conversion will take place considering the NA-s:

m <- matrix(v, nrow=gr); m
     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    2    5    8   11
[3,]    3    6    9   NA
colMeans(m, na.rm=TRUE)
[1]  2.0  5.0  8.0 10.5

An option with data.table

dt <- data.table(1:11, rep(1:3,each=3))
dt
    V1 V2
 1:  1  1
 2:  2  1
 3:  3  1
 4:  4  2
 5:  5  2
 6:  6  2
 7:  7  3
 8:  8  3
 9:  9  3
10: 10  1
11: 11  1
dt[, mean(V1), by = rleid(V2)]$V1
[1]  2.0  5.0  8.0 10.5