根据另一列的运算符填充 NaN 值

Question

我有一个像这样的数据库 (pd.DataFrame):

    condition     odometer
0    new           NaN
1    bad           1100
2    excellent     110
3    NaN           200
4    NaN           2000
5    new           20
6    bad           NaN

我想根据“里程表”的值填充“条件”的NaN：

new: odometer >0 and <= 100 
excellent: odometer >100 and <= 1000
bad: odometer >1000

我尝试这样做，但它不起作用：

for i in range(len(database)): 
   if math.isnan(database['condition'][i]) == True:
      odometer = database['odometer'][i] 
      if   odometer > 0 & odometer <= 100:       value = 'new'
      elif odometer > 100 & odometer <= 1000:    value = 'excellent'
      elif odometer > 1000:                      value = 'bad'
      database['condition'][i] = value

还尝试了第一个“if”条件：

database['condition'][i] == np.nan

但效果不佳。

Answer 1

您可以使用 DataFrame.apply() 用您的函数生成一个新的条件列，然后替换它。不确定您的列是什么类型。 df['condition'].dtype 会告诉你。看起来条件可以是字符串或对象，这可能会在您的逻辑中产生错误。如果是字符串列，则需要直接比较== 'NaN'。如果它是一个对象，您可以使用 np.nan 或 math.nan。我在下面为每个案例提供了一个示例数据库。您可能还想测试里程表列的类型。

import numpy as np
import pandas as pd

# condition column as string
df = pd.DataFrame({'condition':['new','bad','excellent','NaN','NaN','new','bad'], 'odometer':np.array([np.nan, 1100, 110, 200, 2000, 20, np.nan], dtype=object)})
# condition column as object
# df = pd.DataFrame({'condition':np.array(['new','bad','excellent',np.nan,np.nan,'new','bad'], dtype=object), 'odometer':np.array([np.nan, 1100, 110, 200, 2000, 20, np.nan], dtype=object)})
def f(database):
    if database['condition'] == 'NaN':
    #if np.isnan(database['condition']):
        odometer = database['odometer'] 
        if   odometer > 0 & odometer <= 100:       value = 'new'
        elif odometer > 100 & odometer <= 1000:    value = 'excellent'
        elif odometer > 1000:                      value = 'bad'
        return value
    return database['condition']

df['condition'] = df.apply(f, axis=1)

Answer 2

我有一个很好的单线解决方案给你：

让我们创建一个示例数据框：

import pandas as pd

df = pd.DataFrame({'condition':['new','bad',None,None,None], 'odometer':[None,1100,50,500,2000]})
df
Out:    
  condition odometer
0   new     NaN
1   bad     1100.0
2   None    50.0
3   None    500.0
4   None    2000.0

解决方案：

df.condition = df.condition.fillna(df.odometer.apply(lambda number: 'new' if number in range(101) else 'excellent' if number in range(101,1000) else 'bad'))
df
Out:    
  condition  odometer
0       new        NaN
1       bad     1100.0
2       new       50.0
3 excellent      500.0
4       bad     2000.0

根据另一列的运算符填充 NaN 值

Fill NaN values based on operators from another column

python

for-loop

operators

nan

dataframe