Question

如果T1是这样的话：

T1 = pd.DataFrame(data = {'val':['B','D','E','A','D','B','A','E','A','D','B']})

和P是这个：

P = pd.DataFrame(data = {'val': ['E','A','D','B']})

如何在T1内获得P的位置？

就min和max而言，我希望看到这个返回

min max
3   6
8   11

如果这些数据帧表示为SQL表，我可以将此SQL方法转换为pandas：

DECLARE @Items INT = (SELECT COUNT(*) FROM @P);

SELECT MIN(t.KeyCol) AS MinKey,
MAX(t.KeyCol) AS MaxKey
FROM dbo.T1 AS t
INNER JOIN @P AS p ON p.Val = t.Val
GROUP BY t.KeyCol - p.KeyCol
HAVING COUNT(*) = @Items;

此SQL解决方案来自Pesomannen对http://sqlmag.com/t-sql/identifying-subsequence-in-sequence-part-2

的回复

Answer 1

好吧，你可以随时做一个这样的解决方法：

t1 = ''.join(T1.val)
p = ''.join(P.val)
start, res = 0, []
while True:
    try:
        res.append(t1.index(p, start))
        start = res[-1] + 1
    except:
        break

获取起始索引，然后通过mathing计算结束索引，并使用iloc访问数据帧。你应该使用基于0的索引（不是基于1的，就像在示例中那样）

Answer 2

当然，这不会使用P，但可能符合您的目的。

groups = T1.groupby(T1.val).groups
pd.DataFrame({'min': [min(x) for x in groups.values()],
              'max': [max(x) for x in groups.values()]}, index=groups.keys())

产量

   max  min
E    7    2
B   10    0
D    9    1
A    8    3

[4 rows x 2 columns]

Answer 3

我认为我已经按照与SQL解决方案相同的方法解决了这个问题 - 一种关系划分（即匹配值，按键列中的差异分组并选择具有计数等于子序列的大小）：

import pandas as pd

T1 = pd.DataFrame(data = {'val':['B','D','E','A','D','B','A','E','A','D','B']})

# use the index to create a new column that's going to be the key (zero based)
T1 = T1.reset_index()

# do the same for the subsequence that we want to find within T1
P = pd.DataFrame(data = {'val': ['E','A','D','B']})
P = P.reset_index()

# join on the val column
J = T1.merge(P,on=['val'],how='inner')

# group by difference in key columns calculating the min, max and count of the T1 key
FullResult = J.groupby(J['index_x'] - J['index_y'])['index_x'].agg({min,max,'count'})

# Final result is where the count is the size of the subsequence - in this case 4
FullResult[FullResult['count'] == 4]

真的很喜欢使用熊猫！

查找序列中子序列的位置

3 个答案: