Pythonic计算pandas数据帧条纹的方法

Question

给定df

df = pd.DataFrame([[1, 5, 2, 8, 2], [2, 4, 4, 20, 2], [3, 3, 1, 20, 2], [4, 2, 2, 1, 3], [5, 1, 4, -5, -4], [1, 5, 2, 2, -20], 
              [2, 4, 4, 3, -8], [3, 3, 1, -1, -1], [4, 2, 2, 0, 12], [5, 1, 4, 20, -2]],
             columns=['A', 'B', 'C', 'D', 'E'], index=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

根据这个答案,我创建了一个计算条纹(向上,向下)的函数.

def streaks(df, column):
    #Create sign column
    df['sign'] = 0
    df.loc[df[column] > 0, 'sign'] = 1
    df.loc[df[column] < 0, 'sign'] = 0
    # Downstreak
    df['d_streak2'] = (df['sign'] == 0).cumsum()
    df['cumsum'] = np.nan
    df.loc[df['sign'] == 1, 'cumsum'] = df['d_streak2']
    df['cumsum'] = df['cumsum'].fillna(method='ffill')
    df['cumsum'] = df['cumsum'].fillna(0)
    df['d_streak'] = df['d_streak2'] - df['cumsum']
    df.drop(['d_streak2', 'cumsum'], axis=1, inplace=True)
    # Upstreak
    df['u_streak2'] = (df['sign'] == 1).cumsum()
    df['cumsum'] = np.nan
    df.loc[df['sign'] == 0, 'cumsum'] = df['u_streak2']
    df['cumsum'] = df['cumsum'].fillna(method='ffill')
    df['cumsum'] = df['cumsum'].fillna(0)
    df['u_streak'] = df['u_streak2'] - df['cumsum']
    df.drop(['u_streak2', 'cumsum'], axis=1, inplace=True)
    del df['sign']
    return df

功能很好,但很长.我确信写这个有更好的方法.我尝试了另一个答案,但效果不佳.

这是所需的输出

streaks(df, 'E')


    A   B   C    D     E    d_streak    u_streak
1   1   5   2    8     2         0.0    1.0
2   2   4   4   20     2         0.0    2.0
3   3   3   1   20     2         0.0    3.0
4   4   2   2    1     3         0.0    4.0
5   5   1   4   -5    -4         1.0    0.0
6   1   5   2    2   -20         2.0    0.0
7   2   4   4    3    -8         3.0    0.0
8   3   3   1   -1    -1         4.0    0.0
9   4   2   2    0    12         0.0    1.0
10  5   1   4   20    -2         1.0    0.0

Answer 1

您可以简化功能,如下所示:

def streaks(df, col):
    sign = np.sign(df[col])
    s = sign.groupby((sign!=sign.shift()).cumsum()).cumsum()
    return df.assign(u_streak=s.where(s>0, 0.0), d_streak=s.where(s<0, 0.0).abs())

使用它:

streaks(df, 'E')

首先,使用计算所考虑的列中存在的每个单元的符号np.sign.这些将+1分配给正数,-1分配给负数.

接下来,使用sign!=sign.shift()并确定将在分组过程中使用的累积和来识别相邻值的集合(比较当前单元格和下一个单元格).

执行将groupby这些作为键/条件,并再次获取子组元素的累积和.

最后,将正计算cumsum值ustreak和负计算值(取其模数后的绝对值)分配给dstreak.