根据连续出现的值对数据框进行分组 [英] Grouping dataframe based on consecutive occurrence of values
问题描述
我有一个pandas数组，该数组的一列为true或false(在下面的示例中标题为"condition").我想按连续的true或false值对数组进行分组.我尝试使用pandas.groupby，但未成功使用该方法，尽管我认为这是由于我缺乏理解.数据框的示例可以在下面找到:
I have a pandas array which has one column which is either true or false (titled 'condition' in the example below). I would like to group the array by consecutive true or false values. I have tried to use pandas.groupby but haven't succeeded using that method, albeit I think that's down to my lack of understanding. An example of the dataframe can be found below:
df = pd.DataFrame(df)
print df
print df
index condition H t
0 1 2 1.1
1 1 7 1.5
2 0 1 0.9
3 0 6.5 1.6
4 1 7 1.1
5 1 9 1.8
6 1 22 2.0
理想情况下，程序的输出将类似于以下内容.我当时在考虑使用某种分组"方法来简化每个结果集的调用，但不确定这是否是最好的方法.任何帮助将不胜感激.
Ideally the output of the program would be something along the lines of what can be found below. I was thinking of using some sort of 'grouping' method to make it easier to call each set of results but not sure if this is the best method. Any help would be greatly appreciated.
index condition H t group
0 1 2 1.1 1
1 1 7 1.5 1
2 0 1 0.9 2
3 0 6.5 1.6 2
4 1 7 1.1 3
5 1 9 1.8 3
6 1 22 2.0 3
推荐答案
由于您处理的是0/1，这是使用diff + cumsum-
Since you're dealing with 0/1s, here's another alternative using diff + cumsum -
df['group'] = df.condition.diff().abs().cumsum().fillna(0).astype(int) + 1
df
condition H t group
index
0 1 2.0 1.1 1
1 1 7.0 1.5 1
2 0 1.0 0.9 2
3 0 6.5 1.6 2
4 1 7.0 1.1 3
5 1 9.0 1.8 3
6 1 22.0 2.0 3
如果您不介意浮动，则可以使其更快一点.
If you don't mind floats, this can be made a little faster.
df['group'] = df.condition.diff().abs().cumsum() + 1
df.loc[0, 'group'] = 1
df
index condition H t group
0 0 1 2.0 1.1 1.0
1 1 1 7.0 1.5 1.0
2 2 0 1.0 0.9 2.0
3 3 0 6.5 1.6 2.0
4 4 1 7.0 1.1 3.0
5 5 1 9.0 1.8 3.0
6 6 1 22.0 2.0 3.0
以下是具有numpy等效项的版本-
Here's the version with numpy equivalents -
df['group'] = 1
df.loc[1:, 'group'] = np.cumsum(np.abs(np.diff(df.condition))) + 1
df
condition H t group
index
0 1 2.0 1.1 1
1 1 7.0 1.5 1
2 0 1.0 0.9 2
3 0 6.5 1.6 2
4 1 7.0 1.1 3
5 1 9.0 1.8 3
6 1 22.0 2.0 3
在我的机器上，这是时间-
On my machine, here are the timings -
df = pd.concat([df] * 100000, ignore_index=True)
%timeit df['group'] = df.condition.diff().abs().cumsum().fillna(0).astype(int) + 1
10 loops, best of 3: 25.1 ms per loop
%%timeit
df['group'] = df.condition.diff().abs().cumsum() + 1
df.loc[0, 'group'] = 1
10 loops, best of 3: 23.4 ms per loop
%%timeit
df['group'] = 1
df.loc[1:, 'group'] = np.cumsum(np.abs(np.diff(df.condition))) + 1
10 loops, best of 3: 21.4 ms per loop
%timeit df['group'] = df['condition'].ne(df['condition'].shift()).cumsum()
100 loops, best of 3: 15.8 ms per loop
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