[更新以适应现代pandas，这isnull是一种方法DataFrame。]
您可以使用isnullandany构建一个布尔系列并使用它来索引到您的框架中：
>>> df = pd.DataFrame([range(3), [0, np.NaN, 0], [0, 0, np.NaN], range(3), range(3)])
>>> df.isnull()
0 1 2
0 False False False
1 False True False
2 False False True
3 False False False
4 False False False
>>> df.isnull().any(axis=1)
0 False
1 True
2 True
3 False
4 False
dtype: bool
>>> df[df.isnull().any(axis=1)]
0 1 2
1 0 NaN 0
2 0 0 NaN
[对于老年人pandas：]
您可以使用函数isnull而不是方法：
In [56]: df = pd.DataFrame([range(3), [0, np.NaN, 0], [0, 0, np.NaN], range(3), range(3)])
In [57]: df
Out[57]:
0 1 2
0 0 1 2
1 0 NaN 0
2 0 0 NaN
3 0 1 2
4 0 1 2
In [58]: pd.isnull(df)
Out[58]:
0 1 2
0 False False False
1 False True False
2 False False True
3 False False False
4 False False False
In [59]: pd.isnull(df).any(axis=1)
Out[59]:
0 False
1 True
2 True
3 False
4 False
导致相当紧凑：
In [60]: df[pd.isnull(df).any(axis=1)]
Out[60]:
0 1 2
1 0 NaN 0
2 0 0 NaN
def nans(df): return df[df.isnull().any(axis=1)]
然后当你需要它时，你可以输入：
nans(your_dataframe)
如果你想通过一定数量的空值列来过滤行，你可以使用这个：
df.iloc[df[(df.isnull().sum(axis=1) >= qty_of_nuls)].index]
所以，这里是一个例子：
您的数据框：
>>> df = pd.DataFrame([range(4), [0, np.NaN, 0, np.NaN], [0, 0, np.NaN, 0], range(4), [np.NaN, 0, np.NaN, np.NaN]])
>>> df
0 1 2 3
0 0.0 1.0 2.0 3.0
1 0.0 NaN 0.0 NaN
2 0.0 0.0 NaN 0.0
3 0.0 1.0 2.0 3.0
4 NaN 0.0 NaN NaN
如果要选择具有空值的两列或多列的行，请运行以下命令：
>>> qty_of_nuls = 2
>>> df.iloc[df[(df.isnull().sum(axis=1) >=qty_of_nuls)].index]
0 1 2 3
1 0.0 NaN 0.0 NaN
4 NaN 0.0 NaN NaN
少了 4 个字符，但多了 2 毫秒
%%timeit
df.isna().T.any()
# 52.4 ms ± 352 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%%timeit
df.isna().any(axis=1)
# 50 ms ± 423 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
我可能会用axis=1
.any()并且.all()非常适合极端情况，但不是在您寻找特定数量的空值时。这是一种非常简单的方法来做我相信你问的事情。它非常冗长，但很实用。
import pandas as pd
import numpy as np
# Some test data frame
df = pd.DataFrame({'num_legs': [2, 4, np.nan, 0, np.nan],
'num_wings': [2, 0, np.nan, 0, 9],
'num_specimen_seen': [10, np.nan, 1, 8, np.nan]})
# Helper : Gets NaNs for some row
def row_nan_sums(df):
sums = []
for row in df.values:
sum = 0
for el in row:
if el != el: # np.nan is never equal to itself. This is "hacky", but complete.
sum+=1
sums.append(sum)
return sums
# Returns a list of indices for rows with k+ NaNs
def query_k_plus_sums(df, k):
sums = row_nan_sums(df)
indices = []
i = 0
for sum in sums:
if (sum >= k):
indices.append(i)
i += 1
return indices
# test
print(df)
print(query_k_plus_sums(df, 2))
输出
num_legs num_wings num_specimen_seen
0 2.0 2.0 10.0
1 4.0 0.0 NaN
2 NaN NaN 1.0
3 0.0 0.0 8.0
4 NaN 9.0 NaN
[2, 4]
然后，如果你像我一样想要清除这些行，你只需写下：
# drop the rows from the data frame
df.drop(query_k_plus_sums(df, 2),inplace=True)
# Reshuffle up data (if you don't do this, the indices won't reset)
df = df.sample(frac=1).reset_index(drop=True)
# print data frame
print(df)
输出：
num_legs num_wings num_specimen_seen
0 4.0 0.0 NaN
1 0.0 0.0 8.0
2 2.0 2.0 10.0
df1 = df[df.isna().any(axis=1)]