机器学习之时间序列分析（二）：Kaggle比赛Web Traff

今天才发现kaggle的Discussion和Kernel内容区别还挺大的。我原来一直在Kernel中找解决方案。其实很多都在Discussion版块给了自己解决方案描述并附加github。

Web Traffic Time Series Forcasting
该题目中提供了过去一年多时间的一些维基词语每天的访问情况，要求预测未来一年这些维基词语的访问情况。

通过对这道题各个solution的分析可以发现一个很神奇的现象：我们在前一篇文章中提到的方法ARIMA之类的并未被这些solution使用。包括facebook提供的用来做时间序列预估的库Prophet也被证明效果不好。

这里有个大家总体方案的讨论帖。Share your general approach?。
还有个兄弟的经验总结.Tips from the winning solutions .
都值得看下。

这里重点分析我们能看到的三个solution。

Baseline Solution 使用中位数

baseline solution
使用中位数预测

import numpy as np # linear algebra
import pandas as pd # data processing, CSV file I/O (e.g. pd.read_csv)


train = pd.read_csv("../input/train_1.csv")
train = train.fillna(0.)


# I'm gong to share a solution that I found interesting with you.
# The idea is to compute the median of the series in different window sizes at the end of the series,
# and the window sizes are increasing exponentially with the base of golden ratio.
# Then a median of these medians is taken as the estimate for the next 60 days.
# This code's result has the score of around 44.9 on public leaderboard, but I could get upto 44.7 by playing with it.

# r = 1.61803398875
# Windows = np.round(r**np.arange(0,9) * 7)
Windows = [6, 12, 18, 30, 48, 78, 126, 203, 329]


n = train.shape[1] - 1 #  550
Visits = np.zeros(train.shape[0])
for i, row in train.iterrows():
    M = []
    start = row[1:].nonzero()[0]
    if len(start) == 0:
        continue
    if n - start[0] < Windows[0]:
        Visits[i] = row.iloc[start[0]+1:].median()
        continue
    for W in Windows:
        if W > n-start[0]:
            break
        M.append(row.iloc[-W:].median())
    Visits[i] = np.median(M)

Visits[np.where(Visits < 1)] = 0.
train['Visits'] = Visits


test = pd.read_csv("../input/key_1.csv")
test['Page'] = test.Page.apply(lambda x: x[:-11])

test = test.merge(train[['Page','Visits']], on='Page', how='left')
test[['Id','Visits']].to_csv('sub.csv', index=False)

原理解释如下：
就是对每一行（每个词从XX年XX月XX日到YY年Y月Y日每天的访问量）求非NAN值的中位数。

中位数求法如下：

1. 对该行最后六天访问量求中位数，最后12天求中位数，最后18天求中位数。。。。按斐波那契数列的天数求出一个中位数数组。就是按数列[6, 12, 18, 30, 48, 78, 126, 203, 329]为求中位数的天数

2. 对上面求出来的中位数数组再求一次中位数

3. 以中位数值为预测结果

Solution 2 使用CNN ??(rank 2nd)

2nd Solution

Solution 3 使用RNN seq2seq (rank 1st)

1st Solution

后面两个方案，等我撸完 < hands-on machine learning with scikit-learn and tensorflow > 再来分析