过滤值小于 0 的 Pandas 数据帧行

Question

I have a pandas dataframe like this

df = pd.DataFrame(data=[[21, 1],[32, -4],[-4, 14],[3, 17],[-7,NaN]], columns=['a', 'b'])
df

I want to be able to remove all rows with negative values in a list of columns and conserving rows with NaN.

In my example there is only 2 columns, but I have more in my dataset, so I can't do it one by one.

Answer 1

If you want to apply it to all columns, do df[df > 0] with dropna() :

>>> df[df > 0].dropna()
    a   b
0  21   1
3   3  17

If you know what columns to apply it to, then do for only those cols with df[df[cols] > 0] :

>>> cols = ['b']
>>> df[cols] = df[df[cols] > 0][cols]
>>> df.dropna()
    a   b
0  21   1
2  -4  14
3   3  17

Answer 2

I've found you can simplify the answer by just doing this:

>>> cols = ['b']
>>> df = df[df[cols] > 0]

dropna() is not an in-place method, so you have to store the result.

>>> df = df.dropna()

Answer 3

I was looking for a solution for this that doesn't change the dtype (which will happen if NaN's are mixed in with ints as suggested in the answers that use dropna . Since the questioner already had a NaN in their data, that may not be an issue for them. I went with this solution which preserves the int64 dtype. Here it is with my sample data:

df = pd.DataFrame(data={'a':[0, 1, 2], 'b': [-1,0,1], 'c': [-2, -1, 0]})
columns = ['b', 'c']
filter_ = (df[columns] >= 0).all(axis=1)
df[filter_]


   a  b  c
2  2  1  0