如何在R中标准化时间序列数据？

Question

I have the matrix below. How do I divide each row with its mean?

 TAXA   1992    1993    1994     1995   
 Aba    1        0      0.01     0  
 Abr    2      0.084    0.1      3  
 Amp    7         6     4        2

Answer 1

I think you want either of these -

For a data frame:

cbind(df[1], df[-1] / rowMeans(df[-1]))
#   TAXA    X1992      X1993      X1994     X1995
# 1  Aba 3.960396 0.00000000 0.03960396 0.0000000
# 2  Abr 1.543210 0.06481481 0.07716049 2.3148148
# 3  Amp 1.473684 1.26315789 0.84210526 0.4210526

For a matrix:

m / rowMeans(m)
#         1992       1993       1994      1995
# Aba 3.960396 0.00000000 0.03960396 0.0000000
# Abr 1.543210 0.06481481 0.07716049 2.3148148
# Amp 1.473684 1.26315789 0.84210526 0.4210526

This finds the mean of each row then divides each row by its corresponding mean. The first assumes the first column in your example is actually a column, while the second assumes it is row names in a matrix.

Data:

df <- structure(list(TAXA = structure(1:3, .Label = c("Aba", "Abr", 
"Amp"), class = "factor"), X1992 = c(1L, 2L, 7L), X1993 = c(0, 
0.084, 6), X1994 = c(0.01, 0.1, 4), X1995 = c(0L, 3L, 2L)), .Names = c("TAXA", 
"X1992", "X1993", "X1994", "X1995"), class = "data.frame", row.names = c(NA, 
-3L))

m <- structure(c(1, 2, 7, 0, 0.084, 6, 0.01, 0.1, 4, 0, 3, 2), .Dim = 3:4, .Dimnames = list(
    c("Aba", "Abr", "Amp"), c("1992", "1993", "1994", "1995"
    )))

Answer 2

Using the 'tidy data' approach (I copied the data from question to clipboard):

t <- read.table("clipboard", sep=" ", header=T)

library(tidyr)
library(dplyr)
t %>% 
  gather(year, value, -TAXA) %>% 
  group_by(TAXA) %>% 
  mutate(value=value / mean(value)) %>% 
  spread(year, value)

You get:

Source: local data frame [3 x 5]

  TAXA    X1992      X1993      X1994     X1995
1  Aba 3.960396 0.00000000 0.03960396 0.0000000
2  Abr 1.543210 0.06481481 0.07716049 2.3148148
3  Amp 1.473684 1.26315789 0.84210526 0.4210526

It gathers the values from many columns into one. (They're getting the same treatment, they should be in one column.) Then it calculates the mean for each TAXA separately, and reformats the data back into the wide format.