通过熊猫进行分组计算的连续零值

时间:2018-10-11 05:23:19

标签: python pandas

我的输入看起来像下面的df。

我需要按列(A,B)分组并计算连续零的数量/计算每个组中连续零的长度,然后写入新列“ Zero_count”

Input:
A    B  DATE      hour  measure     
A10  1  1/1/2014    0   0       
A10  1  1/1/2014    1   0       
A10  1  1/1/2014    2   0       
A10  1  1/1/2014    3   0       
A10  2  1/1/2014    4   0       
A10  2  1/1/2014    5   1       
A10  2  1/1/2014    6   2       
A10  3  1/1/2014    7   0       
A11  1  1/1/2014    8   0       
A11  1  1/1/2014    9   0       
A11  1  1/1/2014    10  2       
A11  1  1/1/2014    11  0       
A11  1  1/1/2014    12  0       
A12  2  1/1/2014    13  1       
A12  2  1/1/2014    14  3       
A12  2  1/1/2014    15  0       
A12  4  1/1/2014    16  5       
A12  4  1/1/2014    17  0       
A12  6  1/1/2014    18  0       

我尝试使用“ groupby”技术来获取组,但是我一直在寻找组内连续的零计数。我尝试使用lambda函数,但是它计算零的总数,而我有兴趣重复连续的零。我希望我的输出看起来像这样:

Output
A    B  DATE      hour  measure Consec_zero_count
A10  1  1/1/2014    0   0       4
A10  1  1/1/2014    1   0       4
A10  1  1/1/2014    2   0       4
A10  1  1/1/2014    3   0       4
A10  2  1/1/2014    4   0       1
A10  2  1/1/2014    5   1       0
A10  2  1/1/2014    6   2       0
A10  3  1/1/2014    7   0       1
A11  1  1/1/2014    8   0       2
A11  1  1/1/2014    9   0       2
A11  1  1/1/2014    10  2       0
A11  1  1/1/2014    11  0       2
A11  1  1/1/2014    12  0       2
A12  2  1/1/2014    13  1       0
A12  2  1/1/2014    14  3       0
A12  2  1/1/2014    15  0       1
A12  4  1/1/2014    16  5       0
A12  4  1/1/2014    17  0       1
A12  6  1/1/2014    18  0       1

任何线索都将不胜感激。预先感谢!

2 个答案:

答案 0 :(得分:2)

通过将ne个值的shiftSeries)与cumsum进行比较,为连续值的唯一组创建助手!=。然后groupbytransformsize。仅0numpy.where的最后拟合值:

g = df['measure'].ne(df['measure'].shift()).cumsum()
counts = df.groupby(['A','B', g])['measure'].transform('size')
df['Consec_zero_count'] = np.where(df['measure'].eq(0), counts, 0)
print (df)
      A  B      DATE  hour  measure  Consec_zero_count
0   A10  1  1/1/2014     0        0                  4
1   A10  1  1/1/2014     1        0                  4
2   A10  1  1/1/2014     2        0                  4
3   A10  1  1/1/2014     3        0                  4
4   A10  2  1/1/2014     4        0                  1
5   A10  2  1/1/2014     5        1                  0
6   A10  2  1/1/2014     6        2                  0
7   A10  3  1/1/2014     7        0                  1
8   A11  1  1/1/2014     8        0                  2
9   A11  1  1/1/2014     9        0                  2
10  A11  1  1/1/2014    10        2                  0
11  A11  1  1/1/2014    11        0                  2
12  A11  1  1/1/2014    12        0                  2
13  A12  2  1/1/2014    13        1                  0
14  A12  2  1/1/2014    14        3                  0
15  A12  2  1/1/2014    15        0                  1
16  A12  4  1/1/2014    16        5                  0
17  A12  4  1/1/2014    17        0                  1
18  A12  6  1/1/2014    18        0                  1

答案 1 :(得分:0)

类似于@jezrael的答案,但逻辑略有不同:

df.loc[df.measure.eq(0), 'Consec_zero_count'] = (df.groupby(['A','B', df.measure.ne(0).cumsum()])
                                                  .measure.transform(lambda x: x[x.eq(0)].size))


df['Consec_zero_count'] = df['Consec_zero_count'].fillna(0).astype(int)

>>> df
      A  B      DATE  hour  measure  Consec_zero_count
0   A10  1  1/1/2014     0        0                  4
1   A10  1  1/1/2014     1        0                  4
2   A10  1  1/1/2014     2        0                  4
3   A10  1  1/1/2014     3        0                  4
4   A10  2  1/1/2014     4        0                  1
5   A10  2  1/1/2014     5        1                  0
6   A10  2  1/1/2014     6        2                  0
7   A10  3  1/1/2014     7        0                  1
8   A11  1  1/1/2014     8        0                  2
9   A11  1  1/1/2014     9        0                  2
10  A11  1  1/1/2014    10        2                  0
11  A11  1  1/1/2014    11        0                  2
12  A11  1  1/1/2014    12        0                  2
13  A12  2  1/1/2014    13        1                  0
14  A12  2  1/1/2014    14        3                  0
15  A12  2  1/1/2014    15        0                  1
16  A12  4  1/1/2014    16        5                  0
17  A12  4  1/1/2014    17        0                  1
18  A12  6  1/1/2014    18        0                  1
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