如何从df中删除NaN值以及附近的非Nan值？

Question

I have large CSVs (~100k rows x 30 cols). Occasionally the data has sections of nan values which span sections of the df of various sizes. I need to drop the nan s but also ~3 data points either side because the non- nan data either side is borked.

One could drop any row containing a nan but this would throw away more data than needs to be.

How can I do this with python? The data has been loaded into a df .

Answer 1

Use:

df = pd.DataFrame({'col':['a','b','c', np.nan, 'd','e',np.nan, 's','r'],
                   'col1':4})

print (df)
   col  col1
0    a     4
1    b     4
2    c     4
3  NaN     4
4    d     4
5    e     4
6  NaN     4
7    s     4
8    r     4

#test at least one missing value
m = df.isna().any(axis=1)

#test row above and bellow match value by mask, chain by | for bitwise OR
#filter in inverted mask by ~ in boolean indexing
df = df[~(m | m.shift(fill_value=False) | m.shift(-1, fill_value=False))]
print (df)
  col  col1
0   a     4
1   b     4
8   r     4

Alternative solution:

m = df.notna().all(axis=1)

df = df[(m & m.shift(fill_value=True) & m.shift(-1, fill_value=True))]

Answer 2

Here is another way if the number of rows to look above an below might change.

l = 1

(df.loc[~df.isna().any(axis=1)
        .replace(False,None,method = 'ffill',limit= l)
        .replace(False,None,method = 'bfill',limit= l)])