pandas 识别连续出现的值
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Identifying consecutive occurrences of a value
提问by Stefano Potter
I have a df like so:
我有一个像这样的 df：
Count
1
0
1
1
0
0
1
1
1
0
and I want to return a 1in a new column if there are two or more consecutive occurrences of 1in Countand a 0if there is not. So in the new column each row would get a 1based on this criteria being met in the column Count. My desired output would then be:
1如果有两个或多个连续出现的1inCount和 a0如果没有，我想在新列中返回 a 。因此，在新列中，每行都将1根据列中满足此条件获得Count。我想要的输出是：
Count New_Value
1 0
0 0
1 1
1 1
0 0
0 0
1 1
1 1
1 1
0 0
I am thinking I may need to use itertoolsbut I have been reading about it and haven't come across what I need yet. I would like to be able to use this method to count any number of consecutive occurrences, not just 2 as well. For example, sometimes I need to count 10 consecutive occurrences, I just use 2 in the example here.
我想我可能需要使用，itertools但我一直在阅读它，但还没有遇到我需要的东西。我希望能够使用这种方法来计算任意数量的连续出现次数，而不仅仅是 2 次。例如，有时我需要计算连续出现的 10 次，这里的示例中我只使用了 2。
采纳答案by Stefan
You could:
你可以：
df['consecutive'] = df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count
to get:
要得到：
Count consecutive
0 1 1
1 0 0
2 1 2
3 1 2
4 0 0
5 0 0
6 1 3
7 1 3
8 1 3
9 0 0
From here you can, for any threshold:
从这里你可以，对于任何阈值：
threshold = 2
df['consecutive'] = (df.consecutive > threshold).astype(int)
to get:
要得到：
Count consecutive
0 1 0
1 0 0
2 1 1
3 1 1
4 0 0
5 0 0
6 1 1
7 1 1
8 1 1
9 0 0
or, in a single step:
或者，在一个步骤中：
(df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count >= threshold).astype(int)
In terms of efficiency, using pandasmethods provides a significant speedup when the size of the problem grows:
在效率方面，pandas当问题规模增大时，使用方法提供了显着的加速：
df = pd.concat([df for _ in range(1000)])
%timeit (df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count >= threshold).astype(int)
1000 loops, best of 3: 1.47 ms per loop
compared to:
相比：
%%timeit
l = []
for k, g in groupby(df.Count):
size = sum(1 for _ in g)
if k == 1 and size >= 2:
l = l + [1]*size
else:
l = l + [0]*size
pd.Series(l)
10 loops, best of 3: 76.7 ms per loop
回答by Psidom
Not sure if this is optimized, but you can give it a try:
不确定这是否已优化，但您可以尝试一下：
from itertools import groupby
import pandas as pd
l = []
for k, g in groupby(df.Count):
size = sum(1 for _ in g)
if k == 1 and size >= 2:
l = l + [1]*size
else:
l = l + [0]*size
df['new_Value'] = pd.Series(l)
df
Count new_Value
0 1 0
1 0 0
2 1 1
3 1 1
4 0 0
5 0 0
6 1 1
7 1 1
8 1 1
9 0 0
