pandas 数据框 - 如何找到满足某些条件的连续行?

may*_*ull 4 python pandas

我正在尝试制作一个程序来查找满足某些条件的连续行。例如,如果有一个如下所示的数据框:

df = pd.DataFrame([1,1,2,-13,-4,-5,6,17,8,9,-10,-11,-12,-13,14,15], 
            index=[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15], 
            columns=['value'])

>>> df
    value
0       1
1       1
2       2
3     -13
4      -4
5      -5
6       6
7      17
8       8
9       9
10    -10
11    -11
12    -12
13    -13
14    -14
15     15
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我希望它返回一个数据框,显示满足以下条件的行:

1) 顺序必须是(positive rows)and (negative rows),而不是相反。

2) 每组正或负的行必须至少有 3 行

3)正负组必须彼此相邻

          posIdx,   negIdx,  posLength,  negLength
0              2          3           3          3    # (1,1,2) (-13,-4,-5)
1              9         10           4          5    # (6,17,8,9) (-10,-11,-12,-13,-14)
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有什么简单的方法可以使用 python 或 pandas 命令来做到这一点吗?

jez*_*ael 5

我创建了辅助列以方便验证解决方案:

#column for negative and positive
df['sign'] = np.where(df['value'] < 0, 'neg','pos')
#consecutive groups
df['g'] = df['sign'].ne(df['sign'].shift()).cumsum()

#removed groups with length more like 2
df = df[df['g'].map(df['g'].value_counts()).gt(2)]

#tested if order `pos-neg` of groups, if not removed groups
m1 = df['sign'].eq('pos') & df['sign'].shift(-1).eq('neg')
m2 = df['sign'].eq('neg') & df['sign'].shift().eq('pos')
groups = df.loc[m1 | m2, 'g']
df = df[df['g'].isin(groups)].copy()

df['pairs'] = (df['sign'].ne(df['sign'].shift()) & df['sign'].eq('pos')).cumsum()
print (df)
    value sign  g  pairs
0       1  pos  1      1
1       1  pos  1      1
2       2  pos  1      1
3     -13  neg  2      1
4      -4  neg  2      1
5      -5  neg  2      1
6       6  pos  3      2
7      17  pos  3      2
8       8  pos  3      2
9       9  pos  3      2
10    -10  neg  4      2
11    -11  neg  4      2
12    -12  neg  4      2
13    -13  neg  4      2
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GroupBy.first所有组和计数的最后聚合GroupBy.size和命名聚合(pandas 0.25+),对列进行排序并展平 MultiIndex,最后Idx_pos对减法正确1

df1 = (df.reset_index()
         .groupby(['pairs','g', 'sign'])
         .agg(Idx=('index','first'),  Length=('sign','size'))
         .reset_index(level=1, drop=True)
         .unstack()
         .sort_index(axis=1, level=[0,1], ascending=[True, False])
         )
df1.columns = df1.columns.map(lambda x: f'{x[0]}_{x[1]}')
df1['Idx_pos'] = df1['Idx_neg'] - 1
print (df1)
       Idx_pos  Idx_neg  Length_pos  Length_neg
pairs                                          
1            2        3           3           3
2            9       10           4           4
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  • 非常感谢你的帮助。它比我想象的更复杂!我从你的代码中学到了很多东西! (2认同)