英文:
How to count the number of zeros if they lie between ones
问题
Sure, here are the translations of the code-related parts:
我有一个数据框架
ID A B C
60 0 1 1
120 0 1 1
180 0 1 1
240 0 0 0
300 0 0 0
360 1 1 0
420 1 1 1
480 0 0 0
540 0 0 0
600 0 0 0
660 0 0 0
720 1 1 1
780 1 0 1
840 0 0 0
900 0 0 0
如何计算仅在1之间的零的数量
在这个例子中，答案应该是A中的[4]，B中的[2,4]，C中的[3,4]
请帮助我。
我已经看到这个选项，但它不适用，因为它计算一行中的所有零。我需要只计算1之间的零。请帮助。
英文:
I have dataframe
ID A B C
60 0 1 1
120 0 1 1
180 0 1 1
240 0 0 0
300 0 0 0
360 1 1 0
420 1 1 1
480 0 0 0
540 0 0 0
600 0 0 0
660 0 0 0
720 1 1 1
780 1 0 1
840 0 0 0
900 0 0 0
How to count the number of zeros only between 1
in this example, the answer should be [4] in A, [2,4] in B, [3,4] in C
Could you help me, please?
i have seen this option, but it does not fit, since it counts all zeros in a row. And I need to count zeros only between 1. Please help.
答案1
得分: 3
你可以尝试类似于bfill和ffill，然后对结果进行groupby操作。
s = (df.mask(df==0).ffill() == df.mask(df==0).bfill()) & (df==0)
s = s.apply(lambda x: x[x].groupby((~x).cumsum()).count().tolist())
ID []
A [4]
B [2, 4]
C [3, 4]
dtype: object
英文:
You can try something like bfill and ffill then groupby the result
s = (df.mask(df==0).ffill() == df.mask(df==0).bfill()) & (df==0)
s = s.apply(lambda x : x[x].groupby((~x).cumsum()).count().tolist())
ID []
A [4]
B [2, 4]
C [3, 4]
dtype: object
答案2
得分: 0
以下是翻译好的内容：
您还可以使用 itertools.groupby：
from itertools import groupby
df.apply(lambda s: [g for k,g in [(k, len(list(g)))
for k,g in groupby(s, lambda x: x==0)
][1:-1] if k])
输出：
ID []
A [4]
B [2, 4]
C [3, 4]
dtype: object
英文:
You can also use itertools.groupby:
from itertools import groupby
df.apply(lambda s: [g for k,g in [(k, len(list(g)))
for k,g in groupby(s, lambda x: x==0)
][1:-1] if k])
Output:
ID []
A [4]
B [2, 4]
C [3, 4]
dtype: object
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