Question

   Survived  SibSp  Parch
0         0      1      0
1         1      1      0
2         1      0      0
3         1      1      0
4         0      0      1

鉴于上述数据框架，groupby有条件的优雅方式吗？我想根据以下条件将数据分成两组：

(df['SibSp'] > 0) | (df['Parch'] > 0) =   New Group -"Has Family"
 (df['SibSp'] == 0) & (df['Parch'] == 0) = New Group - "No Family"

然后采用这两个组的方法，最终输出如下：

               SurvivedMean
 Has Family    Mean
 No Family     Mean

是否可以使用groupby完成，还是必须使用上述条件语句追加新列？

谢谢！

Answer 1

分组的简单方法是使用这两列的总和。如果它们中的任何一个是正数，则结果将大于1.并且groupby接受任意数组，只要该长度与DataFrame的长度相同，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

Answer 2

如果SibSp和Parch列中的值永远不会超过0，则只使用一个条件：

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，请先使用两个条件：

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

Answer 3

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

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)

熊猫 - 具有条件公式的Groupby

3 个答案: