等级NaN必须与名称相同

Question

我试图计算使用此代码在数据帧的列中出现NaN的次数:

count = enron_df.loc['salary'].count('NaN')

但每次我运行这个我得到以下错误:

KeyError: 'Level NaN must be same as name (None)'

我在网上搜索了很多试图寻找解决方案,但无济于事.

Answer 1

如果NaNs 缺少值:

enron_df = pd.DataFrame({'salary':[np.nan, np.nan, 1, 5, 7]})
print (enron_df)
   salary
0     NaN
1     NaN
2     1.0
3     5.0
4     7.0

count = enron_df['salary'].isna().sum()
#alternative
#count = enron_df['salary'].isnull().sum()
print (count)
2

如果NaNs是strings:

enron_df = pd.DataFrame({'salary':['NaN', 'NaN', 1, 5, 'NaN']})
print (enron_df)
  salary
0    NaN
1    NaN
2      1
3      5
4    NaN

count = enron_df['salary'].eq('NaN').sum()
#alternative
#count = (enron_df['salary'] == 'NaN').sum()
print (count)
3

Answer 2

根据定义,count省略NaNs而size不是.

因此,应该做一个简单的区别

count = enron_df['salary'].size - enron_df['salary'].count()