df.isnull().sum() 
      //type: <class 'pandas.core.series.Series'>

or

df.column_name.isnull().sum()
     //type: <type 'numpy.int64'>

下面的代码将按降序打印所有Nan列。

df.isnull().sum().sort_values(ascending = False)

or

下面将按降序打印前15个Nan列。

df.isnull().sum().sort_values(ascending = False).head(15)

下面是按列计数Null值的代码:

df.isna().sum()

另一种完整的方法是使用np。带有.isna()的count_non0:

np.count_nonzero(df.isna())

%timeit np.count_nonzero(df.isna())
512 ms ± 24.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

使用1000005行× 16列的数据框架与顶部答案进行比较:

%timeit df.isna().sum()
492 ms ± 55.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit df.isnull().sum(axis = 0)
478 ms ± 34.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit count_nan = len(df) - df.count()
484 ms ± 47.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

数据:

raw_data = {'first_name': ['Jason', np.nan, 'Tina', 'Jake', 'Amy'], 
        'last_name': ['Miller', np.nan, np.nan, 'Milner', 'Cooze'], 
        'age': [22, np.nan, 23, 24, 25], 
        'sex': ['m', np.nan, 'f', 'm', 'f'], 
        'Test1_Score': [4, np.nan, 0, 0, 0],
        'Test2_Score': [25, np.nan, np.nan, 0, 0]}
results = pd.DataFrame(raw_data, columns = ['first_name', 'last_name', 'age', 'sex', 'Test1_Score', 'Test2_Score'])

# big dataframe for %timeit 
big_df = pd.DataFrame(np.random.randint(0, 100, size=(1000000, 10)), columns=list('ABCDEFGHIJ'))
df = pd.concat([big_df,results]) # 1000005 rows × 16 columns

https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.count.html#pandas.Series.count

pandas.Series.count
Series.count(level=None)[source]

返回系列中非na /null观测值的个数

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]