英文:
Drop rows where a certain column has more than n zeros in a row
问题
我正在处理来自数据记录器的数据，该记录器以毫秒间隔收集数据，无论是否发生有趣的事情。因此，存在大段的数据，在其中某些列的值为零。我想要删除这些行，但保留其中的一些作为填充，以便进一步处理。
示例输入：
a b c
1 2 3
0 3 2
0 1 0
0 3 7
0 9 6
4 0 1
0 2 5
0 6 3
0 1 6
示例输出 - 在列a上进行筛选，将包含超过2个零行的列替换为正好有2个零行（删除哪些行可以是任意的）：
a b c
1 2 3
0 3 2
0 1 0
4 0 1
0 2 5
0 6 3
我的当前代码删除了所有列中值为零的行：
df = df[df['ColName'] != 0]
英文:
I am processing data from a datalogger which collects data at millisecond intervals, regardless of whether anything interesting is happening. As a result, there are long stretches of data where certain columns are zero. I would like to drop these rows, but leave some number of them as padding, which will help in further processing.
Example input:
a b c
1 2 3
0 3 2
0 1 0
0 3 7
0 9 6
4 0 1
0 2 5
0 6 3
0 1 6
Example output - filter on column a, replace column with more than 2 zero rows with exactly 2 zero rows (selection of which rows to drop can be arbitrary):
a b c
1 2 3
0 3 2
0 1 0
4 0 1
0 2 5
0 6 3
My current code drops all rows where a column is zero:
df= df[df['ColName'] != 0]
答案1
得分: 1
你可以使用滚动的（n+1）总和来排除你不想保留的行。例如，对于连续两个0，你可以使用：
exclude = df['a'].rolling(3).sum() != 0
然后：
output = df.loc[exclude,:]
返回：
a b c
0 1 2 3
1 0 3 2
2 0 1 0
5 4 0 1
6 0 2 5
7 0 6 3
或者，可以使用一行代码实现：
output = df[df['a'].rolling(3).sum() != 0]
你可以更改滚动函数中的数字以适应你想要检查的周期数量。
英文:
You can use a rolling (n+1) sum to exclude the rows you wouldn't like to keep. For example for two consecutive 0s you can use:
exclude = df['a'].rolling(3).sum() != 0
And then:
output = df.loc[exclude,:]
Returning:
a b c
0 1 2 3
1 0 3 2
2 0 1 0
5 4 0 1
6 0 2 5
7 0 6 3
Or, in a one-liner:
output = df[df['a'].rolling(3).sum() != 0]
You can change the number in the rolling function to fit the amount of periods you'd like to check.
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