The output of a time-series looks like a data frame:
时间序列的输出看起来像数据框：
ts(rnorm(12*5, 17, 8), start=c(1981,1), frequency = 12)
Jan Feb Mar Apr May Jun Jul ...
1981 14.064085 21.664250 14.800249 -5.773095 16.477470 1.129674 16.747669 ...
1982 23.973620 17.851890 21.387944 28.451552 24.177141 25.212271 19.123179 ...
1983 19.801210 11.523906 8.103132 9.382778 4.614325 21.751529 9.540851 ...
1984 15.394517 21.021790 23.115453 12.685093 -2.209352 28.318686 10.159940 ...
1985 20.708447 13.095117 32.815273 9.393895 19.551045 24.847337 18.703991 ...
It would be handy to transform it into a data frame with columns Jan, Feb, Mar... and rows 1981, 1982, ... and then back. What's the most elegant way to do this?
将它转换为包含Jan，Feb，Mar ......和1981,1982行......然后返回的数据框将是很方便的。最优雅的方法是什么？
1 个解决方案
#1
17
Here are two ways. The first way creates dimnames for the matrix about to be created and then strings out the data into a matrix, transposes it and converts it to data frame. The second way creates a by list consisting of year and month variables and uses tapply on that later converting to data frame and adding names.
这有两种方式。第一种方法为即将创建的矩阵创建dimnames，然后将数据串行化为矩阵，转置它并将其转换为数据帧。第二种方法创建一个由年份和月份变量组成的按列表，并使用tapply，稍后转换为数据框并添加名称。
# create test data
set.seed(123)
tt <- ts(rnorm(12*5, 17, 8), start=c(1981,1), frequency = 12)
1) matrix. This solution requires that we have whole consecutive years
1）矩阵。这个解决方案要求我们连续几年
dmn <- list(month.abb, unique(floor(time(tt))))
as.data.frame(t(matrix(tt, 12, dimnames = dmn)))
If we don't care about the nice names it is just as.data.frame(t(matrix(tt, 12))) .
如果我们不关心好名字，那就是as.data.frame（t（matrix（tt，12）））。
We could replace the dmn<- line with the following simpler line using @thelatemail's comment:
我们可以使用@thelatemail的注释用以下更简单的行替换dmn < - 行：
dmn <- dimnames(.preformat.ts(tt))
2) tapply. A more general solution using tapply is:
2）tapply。使用tapply的更通用的解决方案是：
tapply(tt, list(year = floor(time(tt)), month = month.abb[cycle(tt)]), c)
If we don't care about names we can replace month.abb[cycle(tt)] with just cycle(tt).
如果我们不关心名称，我们可以用cycle（tt）替换month.abb [cycle（tt）]。
Note: To invert this suppose DF is any of the data frame solutions above. Then try:
注意：要反转这一点，假设DF是上面的任何数据帧解决方案。然后尝试：
ts(c(t(DF)), start = 1981, freq = 12)
#1
17
Here are two ways. The first way creates dimnames for the matrix about to be created and then strings out the data into a matrix, transposes it and converts it to data frame. The second way creates a by list consisting of year and month variables and uses tapply on that later converting to data frame and adding names.
这有两种方式。第一种方法为即将创建的矩阵创建dimnames，然后将数据串行化为矩阵，转置它并将其转换为数据帧。第二种方法创建一个由年份和月份变量组成的按列表，并使用tapply，稍后转换为数据框并添加名称。
# create test data
set.seed(123)
tt <- ts(rnorm(12*5, 17, 8), start=c(1981,1), frequency = 12)
1) matrix. This solution requires that we have whole consecutive years
1）矩阵。这个解决方案要求我们连续几年
dmn <- list(month.abb, unique(floor(time(tt))))
as.data.frame(t(matrix(tt, 12, dimnames = dmn)))
If we don't care about the nice names it is just as.data.frame(t(matrix(tt, 12))) .
如果我们不关心好名字，那就是as.data.frame（t（matrix（tt，12）））。
We could replace the dmn<- line with the following simpler line using @thelatemail's comment:
我们可以使用@thelatemail的注释用以下更简单的行替换dmn < - 行：
dmn <- dimnames(.preformat.ts(tt))
2) tapply. A more general solution using tapply is:
2）tapply。使用tapply的更通用的解决方案是：
tapply(tt, list(year = floor(time(tt)), month = month.abb[cycle(tt)]), c)
If we don't care about names we can replace month.abb[cycle(tt)] with just cycle(tt).
如果我们不关心名称，我们可以用cycle（tt）替换month.abb [cycle（tt）]。
Note: To invert this suppose DF is any of the data frame solutions above. Then try:
注意：要反转这一点，假设DF是上面的任何数据帧解决方案。然后尝试：
ts(c(t(DF)), start = 1981, freq = 12)