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 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()

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

你可以从非nan值的计数中减去总长度:

count_nan = len(df) - df.count()

你应该根据你的数据计算时间。与isnull解相比，小级数的速度提高了3倍。

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。