Pythonic计算pandas数据帧条纹的方法

给定df

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

基于this answer,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',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

解决方法

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

接下来,使用sign！= sign.shift()来识别相邻值的集合(比较当前单元格和它的下一个)并获取它将在分组过程中使用的累积和.

执行groupby,将这些作为键/条件,并再次获取子组元素的累积和.

最后,将正计算的cumsum值分配给ustreak,将负的值(取其模数后的绝对值)分配给dstreak.

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