如何从熊猫的时间序列数据中获取斜率？

Question

我有一个熊猫数据框，其中包含 date 和一些如下所示的值

原始数据：

 list = [('2018-10-29', 6.1925), ('2018-10-29', 6.195), ('2018-10-29', 1.95833333333333),
        ('2018-10-29', 1.785), ('2018-10-29', 3.05), ('2018-10-29', 1.30666666666667),
        ('2018-10-29', 1.6325), ('2018-10-30', 1.765), ('2018-10-30', 1.265),
        ('2018-10-30', 2.1125), ('2018-10-30', 2.16714285714286), ('2018-10-30', 1.485),
        ('2018-10-30', 1.72), ('2018-10-30', 2.754), ('2018-10-30', 1.79666666666667),
        ('2018-10-30', 1.27833333333333), ('2018-10-30', 3.48), ('2018-10-30', 6.19),
        ('2018-10-30', 6.235), ('2018-10-30', 6.11857142857143), ('2018-10-30', 6.088),
        ('2018-10-30', 4.3), ('2018-10-30', 7.80666666666667),
        ('2018-10-30', 7.78333333333333), ('2018-10-30', 10.9766666666667),
        ('2018-10-30', 2.19), ('2018-10-30', 1.88)]

加载到熊猫后

df = pd.DataFrame(list)

             0          1
0   2018-10-29   6.192500
1   2018-10-29   6.195000
2   2018-10-29   1.958333
3   2018-10-29   1.785000
4   2018-10-29   3.050000
5   2018-10-29   1.306667
6   2018-10-29   1.632500
7   2018-10-30   1.765000
8   2018-10-30   1.265000
9   2018-10-30   2.112500
10  2018-10-30   2.167143
11  2018-10-30   1.485000
12  2018-10-30   1.720000
13  2018-10-30   2.754000
14  2018-10-30   1.796667
15  2018-10-30   1.278333
16  2018-10-30   3.480000
17  2018-10-30   6.190000
18  2018-10-30   6.235000
19  2018-10-30   6.118571
20  2018-10-30   6.088000
21  2018-10-30   4.300000
22  2018-10-30   7.806667
23  2018-10-30   7.783333
24  2018-10-30  10.976667
25  2018-10-30   2.190000
26  2018-10-30   1.880000

这就是我加载数据框的方式

df = pd.DataFrame(list)
df[0] = pd.to_datetime(df[0], errors='coerce')
df.set_index(0, inplace=True)

现在我想找到 slope 。在互联网上进行研究后，我发现这是获得 slope 所需要做的

trend_coord = list(map(list, zip(df.index.strftime('%Y-%m-%d'), sm.tsa.seasonal_decompose(df.iloc[:,0].values).trend.interpolate(method='linear',axis=0).fillna(0).values)))

results = sm.OLS(np.asarray(sm.tsa.seasonal_decompose(df.iloc[:,0].values).trend.interpolate(method='linear', axis=0).fillna(0).values), sm.add_constant(np.array([i for i in range(len(trend_coord))])), missing='drop').fit()

slope = results.params[1]
print(slope)

但我收到以下错误

Traceback (most recent call last):
  File "/home/souvik/Music/UI_Server2/test35.py", line 11, in <module>
    trend_coord = list(map(list, zip(df.index.strftime('%Y-%m-%d'), sm.tsa.seasonal_decompose(df.iloc[:,0].values).trend.interpolate(method='linear',axis=0).fillna(0).values)))
  File "/home/souvik/django_test/webdev/lib/python3.5/site-packages/statsmodels/tsa/seasonal.py", line 127, in seasonal_decompose
    raise ValueError("You must specify a freq or x must be a "
ValueError: You must specify a freq or x must be a pandas object with a timeseries index with a freq not set to None

现在，如果我将 freq 参数添加到 seasonal_decompose 方法中，例如

trend_coord = list(map(list, zip(df.index.strftime('%Y-%m-%d'), sm.tsa.seasonal_decompose(df.iloc[:,0].values, freq=1).trend.interpolate(method='linear',axis=0).fillna(0).values)))

然后我得到一个错误

Traceback (most recent call last):
  File "/home/souvik/Music/UI_Server2/test35.py", line 11, in <module>
    trend_coord = list(map(list, zip(df.index.strftime('%Y-%m-%d'), sm.tsa.seasonal_decompose(df.iloc[:,0].values, freq=1).trend.interpolate(method='linear',axis=0).fillna(0).values)))
AttributeError: 'numpy.ndarray' object has no attribute 'interpolate'

但是，如果我摆脱任何细粒度的数据，例如 interpolate 等并执行如下操作

trend_coord = sm.tsa.seasonal_decompose(df.iloc[:,0].values, freq=1, model='additive').trend

results = sm.OLS(np.asarray(trend_coord),
                 sm.add_constant(np.array([i for i in range(len(trend_coord))])), missing='drop').fit()
slope = results.params[1]
print(">>>>>>>>>>>>>>>> slope", slope)

然后我得到 slope 的值 0.13668559218559242 。

但我不确定这是否是找出 slope 的正确方法，甚至这个值是否正确。

有没有更好的方法来找出 slope ？

原文由 Souvik Ray 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

我将参与 Franco 的回答，但你不需要 sklearn。你可以用 scipy 轻松地做到这一点。

 import datetime as dt
from scipy import stats

df = pd.DataFrame(list, columns=['date', 'value'])
df.date =pd.to_datetime(df.date)
df['date_ordinal'] = pd.to_datetime(df['date']).map(dt.datetime.toordinal)
slope, intercept, r_value, p_value, std_err = stats.linregress(df['date_ordinal'], df['value'])

slope
Out[114]: 0.80959404761905

原文由 Victor Jaramillo 发布，翻译遵循 CC BY-SA 4.0 许可协议