将数据帧转换为每月时间序列 [英] converting a data frame to monthly time series
问题描述
我有一个数据框架，其中包含每月100年（1200个数据点）的数据，其中月份以列为单位，行以年份为单位。我想将其转换为每月的时间序列，并且尝试了几种方法，但没有一种方法可以创建正确的时间结构。
I have a data frame of a monthly data for 100 yrs (1200 data points) with the months in columns and years in the rows. I want to convert it into a monthly time series and I have tried several ways, none of which create the correct "temporal" structure.
问题出在R考虑数据框为100个观测值（年）的12个变量（月）。这是我最新尝试的可复制代码：
The problem lies with R considering the data frame as a 100 observations (years) of 12 variables (the months). Here is a reproducible code for my latest try:
set.seed(12)
dummy.df <- as.data.frame(matrix(round(rnorm(1200),digits=2),nrow=100,ncol=12))
rownames(dummy.df) <- seq(from=1901, to=2000)
colnames(dummy.df) <- c("jan","feb","mar","apr","may","jun","jul","aug","sep","oct","nov","dec")
dummy.df.ts <- ts(as.vector(as.matrix(dummy.df)), start=c(1901,1), end=c(2000,12), frequency=12)
在 dummy.df .ts对象，将切换行和列，而不是逐列观察，而是将所有1月2月等依次堆叠在一起。我如何获得正确的时间结构？
In the "dummy.df.ts" object, the rows and columns are switched and instead of sequential observations in columns, all the januarys februarys etc are stacked together one after the other. How can I get to the correct temporal structure?
我的数据示例：这些是1901年至1905年的每月温度值
An example of my data: these are monthly temperature values from 1901 - 1905
fr.monthly.temp.sample
JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
1901 2.7 0.4 4.7 10.0 13.0 16.9 19.2 18.3 15.7 10.6 4.9 3.5
1902 4.1 3.2 7.5 10.3 10.0 15.1 18.2 17.4 15.0 10.2 6.3 3.5
1903 3.8 5.9 7.6 7.1 12.9 14.9 17.6 17.3 15.5 12.1 6.9 2.7
1904 3.0 4.6 5.5 10.3 13.6 16.3 20.2 18.5 13.9 11.2 5.4 4.8
1905 1.7 4.0 7.4 9.3 11.9 16.5 20.0 17.6 14.7 8.4 5.5 3.8
并使用以下ts（）调用：
And by using this ts() call:
fr.monthly.temp.sample.ts <- ts(as.vector(as.matrix(fr.monthly.temp.sample)), start=c(1901,1), end=c(1905,12), frequency=12)
这是我为时间序列对象得到的输出：
This is the output I get for the time series object:
fr.monthly.temp.sample.ts
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
1901 2.7 4.1 3.8 3.0 1.7 0.4 3.2 5.9 4.6 4.0 4.7 7.5
1902 7.6 5.5 7.4 10.0 10.3 7.1 10.3 9.3 13.0 10.0 12.9 13.6
1903 11.9 16.9 15.1 14.9 16.3 16.5 19.2 18.2 17.6 20.2 20.0 18.3
1904 17.4 17.3 18.5 17.6 15.7 15.0 15.5 13.9 14.7 10.6 10.2 12.1
1905 11.2 8.4 4.9 6.3 6.9 5.4 5.5 3.5 3.5 2.7 4.8 3.8
-请注意已更改的时间结构（列中的值现在在行中..）-
--Note the changed temporal structure (values from the columns are now in the rows..)--
谢谢。
推荐答案
解决方案1
在对矩阵进行矢量化之前，可以转置（函数 t（））：
You could transpose (function t()) the matrix before vectorizing it:
set.seed(12)
dummy.df <- as.data.frame(matrix(round(rnorm(1200), digits = 2),
nrow = 100, ncol = 12))
rownames(dummy.df) <- seq(1901, 2000)
colnames(dummy.df) <- month.abb
dummy.df.ts <- ts(as.vector(t(as.matrix(dummy.df))),
start=c(1901,1), end=c(2000,12), frequency=12)
解决方案2
您可以融化数据，按日期排序，然后应用ts（）函数。
You could melt the data, order by date, then apply the ts() function.
这是数据设置。 如果您的语言设置为英语，则可以使用month.abb来保存一些代码，但这对其他语言区域设置并不可靠。
set.seed(12)
dummy.df <- as.data.frame(matrix(round(rnorm(1200),digits=2),nrow=100,ncol=12))
months <- format(seq.Date(as.Date("2013-01-01"), as.Date("2013-12-01"),
by = "month"), format = "%b")
colnames(dummy.df) <- months
dummy.df$Year <- seq(1901, 2000) # set as variable, not as rownames
融合数据，这样您就有一个包含1200行的数据框，每行代表一个观察值：
Melt the data so you have a data frame with 1200 rows, each representing an observation:
library("reshape2")
dummy.df <- melt(dummy.df, id.vars = "Year")
按日期排序观测值：
dummy.df$Date <- as.Date(paste(dummy.df$Year, dummy.df$variable, "01", sep = "-"),
format = ("%Y-%b-%d"))
dummy.df <- dummy.df[order(dummy.df$Date), ]
然后您可以应用类似的ts （）调用，并且ts对象显示所需的顺序：
Then you can apply a similar ts() call, with the ts object showing the desired order:
dummy.df.ts <- ts(dummy.df$value, start=c(1901,1), end=c(2000,12), frequency=12)
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