识别值的连续出现 [英] Identifying consecutive occurrences of a value

问题描述

我有这样的df:

Count
1
0
1
1
0
0
1
1
1
0

，并且如果在Count中连续出现两次或多次1，并且在没有0的情况下，我想在新列中返回1.因此，在新列中，根据Count列中满足的条件，每一行都将得到1.我想要的输出将是:

and I want to return a 1 in a new column if there are two or more consecutive occurrences of 1 in Count and a 0 if there is not. So in the new column each row would get a 1 based on this criteria being met in the column Count. My desired output would then be:

Count  New_Value
1      0 
0      0
1      1
1      1
0      0
0      0
1      1
1      1 
1      1
0      0

我认为我可能需要使用itertools，但是我一直在阅读有关它，但是还没有遇到我需要的内容.我希望能够使用此方法来计数任意数量的连续出现，而不仅仅是2.例如，有时候我需要计算10次连续出现，在这里的示例中我只使用2次.

I am thinking I may need to use itertools but 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.

推荐答案

您可以:

df['consecutive'] = df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count

获取:

   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

在这里，您可以设置任何阈值:

From here you can, for any threshold:

threshold = 2
df['consecutive'] = (df.consecutive > threshold).astype(int)

获得:

   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

或者，只需一步:

(df.Count.groupby((df.Count != df.Count.shift()).cumsum()).transform('size') * df.Count >= threshold).astype(int)

在效率方面，使用pandas方法可以在问题规模扩大时显着提高速度:

In terms of efficiency, using pandas methods provides a significant speedup when the size of the problem grows:

 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

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

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