熊猫 - 具有条件公式的Groupby

3 回答

• 7

  您可以在列表中定义条件，并使用下面的函数 group_by_condition 为每个条件创建筛选列表 . 之后，您可以使用模式匹配选择结果项：

  df = [
    {"Survived": 0, "SibSp": 1, "Parch": 0},
    {"Survived": 1, "SibSp": 1, "Parch": 0},
    {"Survived": 1, "SibSp": 0, "Parch": 0}]
  
  conditions = [
    lambda x: (x['SibSp'] > 0) or (x['Parch'] > 0),  # has family
    lambda x: (x['SibSp'] == 0) and (x['Parch'] == 0)  # no family
  ]
  
  def group_by_condition(l, conditions):
      return [[item for item in l if condition(item)] for condition in conditions]
  
  [has_family, no_family] = group_by_condition(df, conditions)
  

  回复于 2024-04-29T23:44:08+08:00
• 1

  一种简单的分组方法是使用这两列的总和 . 如果它们中的任何一个为正，则结果将大于1.并且只要长度与DataFrame的长度相同，groupby就接受任意数组，因此您不需要添加新列 .

  family = np.where((df['SibSp'] + df['Parch']) >= 1 , 'Has Family', 'No Family')
  df.groupby(family)['Survived'].mean()
  Out: 
  Has Family    0.5
  No Family     1.0
  Name: Survived, dtype: float64
  

  回复于 2024-04-29T23:44:08+08:00
• 1

  如果列 SibSp 和 Parch 中的值从不像 0 那样只使用一个条件：

  m1 = (df['SibSp'] > 0) | (df['Parch'] > 0)
  
  df = df.groupby(np.where(m1, 'Has Family', 'No Family'))['Survived'].mean()
  print (df)
  Has Family    0.5
  No Family     1.0
  Name: Survived, dtype: float64
  

  如果不可能使用首先使用两个条件：

  m1 = (df['SibSp'] > 0) | (df['Parch'] > 0)
  m2 = (df['SibSp'] == 0) & (df['Parch'] == 0)
  a = np.where(m1, 'Has Family', 
      np.where(m2, 'No Family', 'Not'))
  
  df = df.groupby(a)['Survived'].mean()
  print (df)
  Has Family    0.5
  No Family     1.0
  Name: Survived, dtype: float64
  

  回复于 2024-04-29T23:44:08+08:00