Question

我正在尝试使用此代码计算NaN在数据框列中出现的次数：

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

但每次我运行时都会出现以下错误：

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NaN是missing values：

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NaN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NaN s而size则省略count = enron_df['salary'].size - enron_df['salary'].count()。

因此，应该做一个简单的区别

{
  "crypto": {
    "BTC": {
      "name": "bitcoin",
      "current_price": "$6592.3"
    }
  }
}

Answer 3

试试这样：

count = df.loc[df['salary']=='NaN'].shape[0]

或者更好：

count = df.loc[df['salary']=='NaN', 'salary'].size

而且，走在你的道路上，你需要这样的事情：

count = df.loc[:, 'salary'].str.count('NaN').sum()

Answer 4

还有dropna参数

的值计数

import numpy as np
import pandas as pd

enron_df = pd.DataFrame({'salary':[np.nan, np.nan, 1, 5, 7]})

enron_df.salary.value_counts(dropna=False)
#NaN     2
# 7.0    1
# 5.0    1
# 1.0    1
#Name: salary, dtype: int64

如果您只想要数字，只需从值计数中选择np.NaN即可。（如果它们是字符串'NaN'，则只需将np.NaN替换为'NaN'）

enron_df.salary.value_counts(dropna=False)[np.NaN]
#2

等级NaN必须与名称相同

4 个答案: