问

我如何估计大熊猫时间序列的周期性

392399224_619416 发布于 2023-02-08 19:24

import

web

get

datetime

有没有办法近似熊猫时间序列的周期性？对于R,xts对象有一个调用的方法periodicity就是为了这个目的.有没有实现的方法呢？

例如,我们可以推断出不指定频率的时间序列的频率吗？

import pandas.io.data as web
aapl = web.get_data_yahoo("AAPL")


[2010-01-04 00:00:00, ..., 2013-12-19 00:00:00]
Length: 999, Freq: None, Timezone: None

这个系列的频率可以合理地近似为每日.

更新:

我认为显示R的周期性方法实现的源代码可能会有所帮助.

function (x, ...) 
{
    if (timeBased(x) || !is.xts(x)) 
        x <- try.xts(x, error = "'x' needs to be timeBased or xtsible")
    p <- median(diff(.index(x)))
    if (is.na(p)) 
        stop("can not calculate periodicity of 1 observation")
    units <- "days"
    scale <- "yearly"
    label <- "year"
    if (p < 60) {
        units <- "secs"
        scale <- "seconds"
        label <- "second"
    }
    else if (p < 3600) {
        units <- "mins"
        scale <- "minute"
        label <- "minute"
        p <- p/60L
    }
    else if (p < 86400) {
        units <- "hours"
        scale <- "hourly"
        label <- "hour"
    }
    else if (p == 86400) {
        scale <- "daily"
        label <- "day"
    }
    else if (p <= 604800) {
        scale <- "weekly"
        label <- "week"
    }
    else if (p <= 2678400) {
        scale <- "monthly"
        label <- "month"
    }
    else if (p <= 7948800) {
        scale <- "quarterly"
        label <- "quarter"
    }
    structure(list(difftime = structure(p, units = units, class = "difftime"), 
        frequency = p, start = start(x), end = end(x), units = units, 
        scale = scale, label = label), class = "periodicity")
}

我认为这条线是关键,我不太明白 p <- median(diff(.index(x)))

1 个回答

这个时间序列会跳过周末(和假期),因此它实际上没有每日开始的频率.您可以使用asfreq它将其上采样到具有每日频率的时间序列,但是:

aapl = aapl.asfreq('D', method='ffill')

这样做会将最后观察到的值向前传播到缺少值的日期.

请注意,Pandas还有一个工作日频率,因此也可以使用以下方式上传到工作日:

aapl = aapl.asfreq('B', method='ffill')

如果您希望自动化以天为单位推断中位数频率的过程,那么您可以这样做:

import pandas as pd
import numpy as np
import pandas.io.data as web
aapl = web.get_data_yahoo("AAPL")
f  = np.median(np.diff(aapl.index.values))
days = f.astype('timedelta64[D]').item().days
aapl = aapl.asfreq('{}D'.format(days), method='ffill')
print(aapl)

这段代码需要测试,但也许它接近您发布的R代码:

import pandas as pd
import numpy as np
import pandas.io.data as web

def infer_freq(ts):
    med  = np.median(np.diff(ts.index.values))
    seconds = int(med.astype('timedelta64[s]').item().total_seconds())
    if seconds < 60:
        freq = '{}s'.format(seconds)
    elif seconds < 3600:
        freq = '{}T'.format(seconds//60)
    elif seconds < 86400:
        freq = '{}H'.format(seconds//3600)
    elif seconds < 604800:
        freq = '{}D'.format(seconds//86400)
    elif seconds < 2678400:
        freq = '{}W'.format(seconds//604800)
    elif seconds < 7948800:
        freq = '{}M'.format(seconds//2678400)
    else:
        freq = '{}Q'.format(seconds//7948800)
    return ts.asfreq(freq, method='ffill')

aapl = web.get_data_yahoo("AAPL")
print(infer_freq(aapl))

2023-02-08 19:27 回答

缘来是你2502855331

撰写答案

今天，你开发时遇到什么问题呢？

立即提问

热门标签