尝试在 R 中使用 stl 和分解函数时出错 [英] Error when trying to use stl and decompose functions in R
问题描述
我制作了一个简单的时间序列，我在 sin 函数中添加了一点噪音，并尝试使用 R 中的stl"和decompose"函数对其进行分解，而我的序列肯定有超过 2 个周期并且是定期，R 给我两个函数的以下错误:
I have made a simple time-series, i added a little noise to a sin function and tried to decompose it using the "stl" and "decompose" function in R, while my series definitely has more than 2 period and is periodic, R gives me the following error for both functions:
x
[1] 1.4537365796 2.7185844368 2.8394728999 3.8926989923 4.3405508086 5.1959080871
[7] 5.6602505790 5.4829985648 5.6357660330 4.6084976233 4.6617322922 4.0286486832
[13] 3.3641752333 1.7408063182 0.8815147612 0.2895139342 -0.5402768515 -1.5612641107
[19] -2.1584502547 -2.9878043526 -3.5545638149 -4.0530074199 -4.0748538612 -4.7581704662
[25] -4.6555349052 -4.0726206240 -3.1646413472 -2.6934453823 -2.2364605277 -1.2643569882
[31] -0.1202011946 1.1136371449 2.2504199271 3.0313528996 3.5384449109 4.5176211013
[37] 5.4013172839 5.4252837451 5.4768196692 5.8979709077 5.6698285659 4.5133489450
[43] 4.2702602998 3.5180837069 2.2652913344 1.1975595698 0.5412697849 -0.5966162032
[49] -1.0827728340 -1.8488242277 -3.4118061838 -3.9009752140 -3.9102671954 -4.3486102172
[55] -4.7481017993 -4.0097598695 -3.9078554267 -3.8070416888 -2.5968567322 -2.2567568949
[61] -1.1423907008 0.0002492447 0.4338279080 1.2431986797 2.3216397323 3.3235925116
[67] 4.1591487169 4.9028481873 5.4535861470 5.0579349546 5.1548777627 4.9707124992
[73] 5.4496833187 4.4563072487 4.1301372986 2.4594352788 1.7253019929 0.6961453965
[79] 0.4281167695 -1.3152944759 -1.8645880957 -2.5764132038 -3.7681528112 -4.3731672862
[85] -3.9940201611 -4.5497596299 -4.9496796983 -4.1233093447 -3.7759837204 -3.3359027749
[91] -2.3518009102 -1.7488933797 -0.7225148838 0.5395759836 1.0496249652 2.0383715782
[97] 3.2357449979 3.8028316517 5.0346866280 5.2154265148
fit<- stl(x, t.window=15, s.window="per", robust=TRUE)
Error in stl(x, t.window = 15, s.window = "per", robust = TRUE) :series is not periodic or has less than two periods
fit<- decompose(x,type="multiplicative")
Error in decompose(x, type = "multiplicative") :time series has no or less than 2 periods
有人能帮我解决这个问题吗?
Would someone help me with this problem please?
推荐答案
从错误信息中是不是很明显:
Isn't it obvious from the error message:
time series has no or less than 2 periods
?R 没有告诉您您的数据不是周期性的，只是您传递给它的数据没有指示它们是周期性的.
? R's not telling you your data aren't periodic, just that the data you passed it have no indication that they are periodic.
您的时间序列 x 从 R 的角度来看没有周期性.您似乎忘记告诉 ts() 什么是周期性，或者在告诉时出错了.由于您没有显示 x 是如何创建的，因此我们无能为力，只能告诉您返回并创建 x 以使其具有 >=2期间.
Your time series x doesn't have an periodicity from the point of view of R. It looks like you forgot to tell, or made an error in telling, ts() what the periodicity is. As you don't show how x was created, there isn't much we can do except tell you to go back and create x so that it does have >=2 periods.
这里的重点是，R 本身无法推断出每单位时间的观察频率.您必须告诉 ts() 该信息.您可以通过多种方式做到这一点:
The point here is that on it's own, R can't deduce what the frequency of observation is per unit time. You have to tell ts() that information. You can do this in a number of ways:
frequency:单位时间内的观察次数.
frequency: the number of observations per unit of time.
deltat:连续两次采样周期的分数观察；例如，每月数据的 1/12.只有其中之一应提供 frequency 或 deltat.
deltat: the fraction of the sampling period between successive
observations; e.g., 1/12 for monthly data. Only one of
frequency or deltat should be provided.
如果您不提供其中之一，ts() 使用默认值 frequency = 1, deltat = 1 这将表明每单位时间一次观察的时间序列(例如每年一次).
If you don't provide one of these, ts() uses the defaults frequency = 1, deltat = 1 which would indicate a time series of one observation per unit time (one per year for example).
stl() 需要一个 "ts" 分类对象 - 如果您不提供，它将强制输入数据到一个 "ts" 对象通过 as.ts().这个函数将使用我上面描述的默认值.
stl() requires a "ts" classed object - if you don't provide one, it will coerce the input data to a "ts" object via as.ts(). This function will use the defaults which I describe above.
我认为这里发生的事情是您没有意识到 stl() 需要一个 "ts" 类对象，也没有意识到它为您创建了一个不合适的对象当您刚刚提供观察向量时的数据.
What I think has happened here is that you didn't realise that stl() requires a "ts" class object nor that it created an inappropriate one for your data when you just supplied the vector of observations.
解决方案是通过 ts() 显式创建 "ts" 分类对象并指定 frequency 或 之一deltat.
The solution would be to explicitly create the "ts" classed object via ts() and specify one of frequency or deltat.
例如
dat <- cumsum(rnorm(12*4))
x <- ts(dat)
stl(x, "periodic")
xx <- ts(dat, frequency = 12)
stl(xx, "periodic")
这里我用 frequency = 12 表示每单位时间有 12 次观察 --- 比如每月的数据.
Here I used frequency = 12 to indicate that there are 12 observations per unit time --- such as with monthly data.
上面给了我
R> stl(x, "periodic")
Error in stl(x, "periodic") :
series is not periodic or has less than two periods
R> stl(xx, "periodic")
Call:
stl(x = xx, s.window = "periodic")
Components
seasonal trend remainder
Jan 1 -0.103529 0.55245 -0.44301
Feb 1 0.001333 0.56981 0.86135
Mar 1 -0.382075 0.58717 1.11162
Apr 1 0.010552 0.59891 -1.04966
....
对于您的数据，我怀疑您希望 frequency = 10 给出时间序列的长度；那就是说你每年有十次观察.如果该系列每年有更多观察值，例如每月数据 12 个，但您没有刚开始的最后两个月或前两个月(即没有缺失数据，NAs)(结束)在当年晚些时候(早些时候)，因此在系列的一端或两端没有完整年份的数据.
For your data I suspect you want frequency = 10 given the length of the time series; that would say that you have ten observations per year. If the series has more observations per year, say 12 for monthly data, but you don;t have the last two or first two months (i.e. there is no missing data, NAs) you just started (ended) later (earlier) in the year and hence don't have a full years worth of data at one or both ends of the series.
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