我使用这个循环来计算每一列的缺失值:

# check missing values
import numpy as np, pandas as pd
for col in df:
      print(col +': '+ np.str(df[col].isna().sum()))

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

import numpy as np
import pandas as pd

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'])

results 
'''
  first_name last_name   age  sex  Test1_Score  Test2_Score
0      Jason    Miller  22.0    m          4.0         25.0
1        NaN       NaN   NaN  NaN          NaN          NaN
2       Tina       NaN  23.0    f          0.0          NaN
3       Jake    Milner  24.0    m          0.0          0.0
4        Amy     Cooze  25.0    f          0.0          0.0
'''

您可以使用以下函数，它将在Dataframe中提供输出

零值 缺失值 占总额的% 总零缺失值 总零缺失值% 数据类型

只需复制和粘贴下面的函数，并通过传递你的熊猫数据帧来调用它

def missing_zero_values_table(df):
        zero_val = (df == 0.00).astype(int).sum(axis=0)
        mis_val = df.isnull().sum()
        mis_val_percent = 100 * df.isnull().sum() / len(df)
        mz_table = pd.concat([zero_val, mis_val, mis_val_percent], axis=1)
        mz_table = mz_table.rename(
        columns = {0 : 'Zero Values', 1 : 'Missing Values', 2 : '% of Total Values'})
        mz_table['Total Zero Missing Values'] = mz_table['Zero Values'] + mz_table['Missing Values']
        mz_table['% Total Zero Missing Values'] = 100 * mz_table['Total Zero Missing Values'] / len(df)
        mz_table['Data Type'] = df.dtypes
        mz_table = mz_table[
            mz_table.iloc[:,1] != 0].sort_values(
        '% of Total Values', ascending=False).round(1)
        print ("Your selected dataframe has " + str(df.shape[1]) + " columns and " + str(df.shape[0]) + " Rows.\n"      
            "There are " + str(mz_table.shape[0]) +
              " columns that have missing values.")
#         mz_table.to_excel('D:/sampledata/missing_and_zero_values.xlsx', freeze_panes=(1,0), index = False)
        return mz_table

missing_zero_values_table(results)

输出

Your selected dataframe has 6 columns and 5 Rows.
There are 6 columns that have missing values.

             Zero Values  Missing Values  % of Total Values  Total Zero Missing Values  % Total Zero Missing Values Data Type
last_name              0               2               40.0                          2                         40.0    object
Test2_Score            2               2               40.0                          4                         80.0   float64
first_name             0               1               20.0                          1                         20.0    object
age                    0               1               20.0                          1                         20.0   float64
sex                    0               1               20.0                          1                         20.0    object
Test1_Score            3               1               20.0                          4                         80.0   float64

如果你想保持简单，那么你可以使用下面的函数来获取%中缺失的值

def missing(dff):
    print (round((dff.isnull().sum() * 100/ len(dff)),2).sort_values(ascending=False))


missing(results)
'''
Test2_Score    40.0
last_name      40.0
Test1_Score    20.0
sex            20.0
age            20.0
first_name     20.0
dtype: float64
'''

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

2017年7月，Dzone有一篇不错的文章，详细介绍了总结NaN值的各种方法。点击这里查看。

我所引用的文章提供了额外的价值:(1)展示了一种方法来计算和显示每列的NaN计数，以便人们可以轻松地决定是否丢弃这些列;(2)演示了一种方法来选择那些特定的具有NaN的行，以便它们可以选择性地丢弃或估算。

这里有一个快速的例子来演示这种方法的实用性——只有几个列，也许它的有用性不明显，但我发现它对较大的数据框架很有帮助。

import pandas as pd
import numpy as np

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

# Check whether there are null values in columns
null_columns = df.columns[df.isnull().any()]
print(df[null_columns].isnull().sum())

# One can follow along further per the cited article

请使用以下方法计算特定的列数

dataframe.columnName.isnull().sum()