使用isna()方法(或者它的别名isnull()，这也兼容较旧的pandas版本< 0.21.0)，然后求和来计算NaN值。其中一列:

>>> s = pd.Series([1,2,3, np.nan, np.nan])

>>> s.isna().sum()   # or s.isnull().sum() for older pandas versions
2

对于一些列，这也适用:

>>> df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})

>>> df.isna().sum()
a    1
b    2
dtype: int64

自从pandas 0.14.1以来，我的建议在value_counts方法中有一个关键字参数已经实现:

import pandas as pd
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
for col in df:
    print df[col].value_counts(dropna=False)

2     1
 1     1
NaN    1
dtype: int64
NaN    2
 1     1
dtype: int64

import pandas as pd
import numpy as np

# example DataFrame
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})

# count the NaNs in a column
num_nan_a = df.loc[ (pd.isna(df['a'])) , 'a' ].shape[0]
num_nan_b = df.loc[ (pd.isna(df['b'])) , 'b' ].shape[0]

# summarize the num_nan_b
print(df)
print(' ')
print(f"There are {num_nan_a} NaNs in column a")
print(f"There are {num_nan_b} NaNs in column b")

给出输出:

     a    b
0  1.0  NaN
1  2.0  1.0
2  NaN  NaN

There are 1 NaNs in column a
There are 2 NaNs in column b

数零:

df[df == 0].count(axis=0)

计算NaN:

df.isnull().sum()

or

df.isna().sum()

你可以试试:

In [1]: s = pd.DataFrame('a'=[1,2,5, np.nan, np.nan,3],'b'=[1,3, np.nan, np.nan,3,np.nan])

In [4]: s.isna().sum()   
Out[4]: out = {'a'=2, 'b'=3} # the number of NaN values for each column

如果需要nan的总和:

In [5]: s.isna().sum().sum()
Out[6]: out = 5  #the inline sum of Out[4] 

对于你的任务，你可以使用pandas.DataFrame.dropna (https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.dropna.html):

import pandas as pd
import numpy as np

df = pd.DataFrame({'a': [1, 2, 3, 4, np.nan],
                   'b': [1, 2, np.nan, 4, np.nan],
                   'c': [np.nan, 2, np.nan, 4, np.nan]})
df = df.dropna(axis='columns', thresh=3)

print(df)

使用thresh参数，您可以声明DataFrame中所有列的NaN值的最大计数。

代码输出:

     a    b
0  1.0  1.0
1  2.0  2.0
2  3.0  NaN
3  4.0  4.0
4  NaN  NaN