条件NaN填充不更改列或使全部为无

Question

我有一个带有列的df，critist_Score，它有NaN值。我试图用来自同一平台的评论家的平均分数来代替它们。这个问题已经被问了好几次堆栈溢出，我使用了4个建议，没有给我想要的输出。请告诉我如何解决这个问题

这是df的一个子集：

x[['Platform','Critic_Score']].head()

Platform    Critic_Score
0   wii 76.0
1   nes NaN
2   wii 82.0
3   wii 80.0
4   gb  NaN

有关原始df的更多信息：

x.head().to_dict('list')
{'Name': ['wii sports',
  'super mario bros.',
  'mario kart wii',
  'wii sports resort',
  'pokemon red/pokemon blue'],
 'Platform': ['wii', 'nes', 'wii', 'wii', 'gb'],
 'Year_of_Release': [2006.0, 1985.0, 2008.0, 2009.0, 1996.0],
 'Genre': ['sports', 'platform', 'racing', 'sports', 'role-playing'],
 'NA_sales': [41.36, 29.08, 15.68, 15.61, 11.27],
 'EU_sales': [28.96, 3.58, 12.76, 10.93, 8.89],
 'JP_sales': [3.77, 6.81, 3.79, 3.28, 10.22],
 'Other_sales': [8.45, 0.77, 3.29, 2.95, 1.0],
 'Critic_Score': [76.0, nan, 82.0, 80.0, nan],
 'User_Score': ['8', nan, '8.3', '8', nan],
 'Rating': ['E', nan, 'E', 'E', nan]}

以下是我尝试过的语句及其输出：

一,

x['Critic_Score'] = x['Critic_Score'].fillna(x.groupby('Platform')['Critic_Score'].transform('mean'), inplace = True)

0    None
1    None
2    None
3    None
4    None
Name: Critic_Score, dtype: object

x.loc[x.Critic_Score.isnull(), 'Critic_Score'] = x.groupby('Platform').Critic_Score.transform('mean')
#no change in column
0    76.0
1     NaN
2    82.0
3    80.0
4     NaN

x['Critic_Score'] = x.groupby('Platform')['Critic_Score']\
    .transform(lambda y: y.fillna(y.mean()))
#no change in column
0    76.0
1     NaN
2    82.0
3    80.0
4     NaN
Name: Critic_Score, dtype: float64

x['Critic_Score']=x.groupby('Platform')['Critic_Score'].apply(lambda y:y.fillna(y.mean()))
​
x['Critic_Score'].head()
​

Out[73]:
0    76.0
1     NaN
2    82.0
3    80.0
4     NaN
Name: Critic_Score, dtype: float64

Answer 1

x.update(
    x.groupby('Platform').Critic_Score.transform('mean'),
    overwrite=False)

• 首先，创建一个具有相同行数但每行平台平均值的新df

• 然后用它来更新原始文件

请记住，您的样本只有一行nes和另一行gb，这两行都有nan分数，因此没有任何东西可以平均