Question

我有一个python pandas数据帧@students = Student.joins(:enrollments).where(enrollments: { is_active: false })：

df

我想根据日期预测值。我想预测01-08-2016的价值。

Group   date           Value
  A     01-02-2016     16 
  A     01-03-2016     15 
  A     01-04-2016     14 
  A     01-05-2016     17 
  A     01-06-2016     19 
  A     01-07-2016     20 
  B     01-02-2016     16 
  B     01-03-2016     13 
  B     01-04-2016     13 
  C     01-02-2016     16 
  C     01-03-2016     16

我不确定我是否正确对待日期。有没有更好的办法？

Answer 1

我认为你所做的事情没有任何问题。您可以使用datetime.toordinal代替，但这会给您相同的结果（拦截将在逻辑上不同，但这是正常的。）

df['date_ordinal'] = df['Date'].apply(lambda x: x.toordinal())
model = LinearRegression()
X = df[['date_ordinal']]
y = df.shown
model.fit(X, y)

如果您认为可能存在每日/每周/每月/季节变化的情况，则可以使用1-of-K编码。例如，请参阅this question。

根据您的评论更新

你说你想为每个组得到一个等式：

In [2]:
results = {}
for (group, df_gp) in df.groupby('Group'):
    print("Dealing with group {}".format(group))
    print("----------------------")
    X=df_gp[['date_ordinal']]
    y=df_gp.Value
    model.fit(X,y)
    print("Score: {:.2f}%".format(100*model.score(X,y)))

    coefs = list(zip(X.columns, model.coef_))
    results[group] = [('intercept', model.intercept_)] + coefs

    coefs = zip(model.coef_, X.columns)

    print ("sl = %.1f + " % model.intercept_ + \
    " + ".join("%.1f %s" % coef for coef in coefs))

    print("\n")

Out[2]:
Dealing with group A
----------------------
Score: 65.22%
sl = -735950.7 + 1.0 date_ordinal


Dealing with group B
----------------------
Score: 75.00%
sl = 1103963.0 + -1.5 date_ordinal


Dealing with group C
----------------------
Score: 100.00%
sl = 16.0 + 0.0 date_ordinal

你也可以使用方便的词典：

In [3]: results
Out[3]:
{'A': [('intercept', -735950.66666666663), ('date_ordinal', 1.0)],
 'B': [('intercept', 1103962.9999999995),
  ('date_ordinal', -1.4999999999999993)],
 'C': [('intercept', 16.0), ('date_ordinal', 0.0)]}

python pandas dataframe根据日期预测值

1 个答案: