在 Pandas 数据框中查找连续段
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Finding consecutive segments in a pandas data frame
提问by languitar
I have a pandas.DataFrame with measurements taken at consecutive points in time. Along with each measurement the system under observation had a distinct state at each point in time. Hence, the DataFrame also contains a column with the state of the system at each measurement. State changes are much slower than the measurement interval. As a result, the column indicating the states might look like this (index: state):
我有一个 pandas.DataFrame ，在连续的时间点进行测量。随着每次测量，被观察系统在每个时间点都有不同的状态。因此，DataFrame 还包含一个列，其中包含每次测量时系统的状态。状态变化比测量间隔慢得多。因此，指示状态的列可能如下所示（索引：状态）：
1: 3
2: 3
3: 3
4: 3
5: 4
6: 4
7: 4
8: 4
9: 1
10: 1
11: 1
12: 1
13: 1
Is there an easy way to retrieve the indices of each segment of consecutively equal states. That means I would like to get something like this:
有没有一种简单的方法来检索每个连续相等状态段的索引。这意味着我想得到这样的东西：
[[1,2,3,4], [5,6,7,8], [9,10,11,12,13]]
The result might also be in something different than plain lists.
结果也可能与普通列表不同。
The only solution I could think of so far is manually iterating over the rows, finding segment change points and reconstructing the indices from these change points, but I have the hope that there is an easier solution.
到目前为止我能想到的唯一解决方案是手动迭代行，找到段变化点并从这些变化点重建索引，但我希望有一个更简单的解决方案。
回答by Zelazny7
One-liner:
单线：
df.reset_index().groupby('A')['index'].apply(np.array)
Code for example:
代码示例：
In [1]: import numpy as np
In [2]: from pandas import *
In [3]: df = DataFrame([3]*4+[4]*4+[1]*4, columns=['A'])
In [4]: df
Out[4]:
A
0 3
1 3
2 3
3 3
4 4
5 4
6 4
7 4
8 1
9 1
10 1
11 1
In [5]: df.reset_index().groupby('A')['index'].apply(np.array)
Out[5]:
A
1 [8, 9, 10, 11]
3 [0, 1, 2, 3]
4 [4, 5, 6, 7]
You can also directly access the information from the groupby object:
您也可以直接访问 groupby 对象中的信息：
In [1]: grp = df.groupby('A')
In [2]: grp.indices
Out[2]:
{1L: array([ 8, 9, 10, 11], dtype=int64),
3L: array([0, 1, 2, 3], dtype=int64),
4L: array([4, 5, 6, 7], dtype=int64)}
In [3]: grp.indices[3]
Out[3]: array([0, 1, 2, 3], dtype=int64)
To address the situation that DSM mentioned you could do something like:
要解决 DSM 提到的情况，您可以执行以下操作：
In [1]: df['block'] = (df.A.shift(1) != df.A).astype(int).cumsum()
In [2]: df
Out[2]:
A block
0 3 1
1 3 1
2 3 1
3 3 1
4 4 2
5 4 2
6 4 2
7 4 2
8 1 3
9 1 3
10 1 3
11 1 3
12 3 4
13 3 4
14 3 4
15 3 4
Now groupby both columns and apply the lambda function:
现在对两列进行分组并应用 lambda 函数：
In [77]: df.reset_index().groupby(['A','block'])['index'].apply(np.array)
Out[77]:
A block
1 3 [8, 9, 10, 11]
3 1 [0, 1, 2, 3]
4 [12, 13, 14, 15]
4 2 [4, 5, 6, 7]
回答by Rutger Kassies
You could use np.diff() to test where a segment starts/ends and iterate over those results. Its a very simple solution, so probably not the most performent one.
您可以使用 np.diff() 来测试段的开始/结束位置并迭代这些结果。它是一个非常简单的解决方案，所以可能不是性能最好的解决方案。
a = np.array([3,3,3,3,3,4,4,4,4,4,1,1,1,1,4,4,12,12,12])
prev = 0
splits = np.append(np.where(np.diff(a) != 0)[0],len(a)+1)+1
for split in splits:
print np.arange(1,a.size+1,1)[prev:split]
prev = split
Results in:
结果是：
[1 2 3 4 5]
[ 6 7 8 9 10]
[11 12 13 14]
[15 16]
[17 18 19]
