在pandas中按行滚动窗口应用自定义函数

问题描述 投票:0回答:1

我有一个函数,我想在数据帧中逐行应用,并输出一个新的结果列。通常这对于lambda函数或.map()来说是直截了当的,但我被卡住了,因为函数需要滚动min / max和一个窗口而lambda显然只会看到那一行。

这是功能:

def divergence(series_1, series_2, local_window = 5, reference_window = 15):
    min_1_local = series_1.rolling(local_window).min().iloc[-1]
    min_1_reference = series_1.rolling(reference_window).min().iloc[-1]

    min_2_local = series_1.rolling(local_window).min().iloc[-1]
    min_2_reference = series_1.rolling(reference_window).min().iloc[-1]

    max_1_local = series_1.rolling(local_window).max().iloc[-1]
    max_1_reference = series_1.rolling(reference_window).max().iloc[-1]

    max_2_local = series_1.rolling(local_window).max().iloc[-1]
    max_2_reference = series_1.rolling(reference_window).max().iloc[-1]

    if ( (min_1_local < min_1_reference) 
        & (min_2_local > min_2_reference) 
        & (series_2.iloc[-1] > series_2.iloc[-2]) ):
        return 1
    elif ( (max_1_local > max_1_reference) 
          & (max_2_local < max_2_reference) 
          & (series_2.iloc[-1] < series_2.iloc[-2]) ):
        return -1
    else:
        return 0

这是我的数据:

    Measure1    Measure2
Date        
2018-09-18 05:00:00 1912.345679 -28.291456
2018-09-18 06:00:00 1910.802469 -28.351495
2018-09-18 07:00:00 1916.666667 -27.988846
2018-09-18 08:00:00 1907.253086 -28.039686
2018-09-18 09:00:00 1907.098765 -28.091198

有任何想法吗?

python pandas lambda pandas-apply
1个回答
0
投票

你有没有尝试过熊猫rolling function

df = pd.DataFrame(np.random.rand(50,2), columns=['input1', 'input2'])
df['min'] = df['input1'].rolling(5).min()
df['min_shifted'] = df['input1'].rolling(5).min().shift(-5)

enter image description here

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