希望这能有所帮助，

import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan],'c':[np.nan,2,np.nan], 'd':[np.nan,np.nan,np.nan]})

df.isnull().sum()/len(df) * 100

Thres = 40
(df.isnull().sum()/len(df) * 100 ) < Thres

对于第一部分，我们有多种方法计算NaN。

方法1计数，由于计数将忽略与大小不同的NaN

print(len(df) - df.count())

方法2:isnull / isna chain with sum

print(df.isnull().sum())
#print(df.isna().sum())

方法3 describe / info:注意这将输出' notull '值计数

print(df.describe())
#print(df.info())

方法。

print(np.count_nonzero(np.isnan(df.values),axis=0))

对于问题的第二部分，如果我们想要在thresh中删除列，我们可以尝试dropna

thresh, optional要求多个非na值。

Thresh = n # no null value require, you can also get the by int(x% * len(df))
df = df.dropna(thresh = Thresh, axis = 1)

数零:

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

计算NaN:

df.isnull().sum()

or

df.isna().sum()

如果你需要得到非NA (non-None)和NA (None)计数在不同的组拉出groupby:

gdf = df.groupby(['ColumnToGroupBy'])

def countna(x):
    return (x.isna()).sum()

gdf.agg(['count', countna, 'size'])

这将返回每个组的非NA、NA和总条目数。

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

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
'''