Pandas - Groupby with conditional formula

Question

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

Given the above dataframe, is there an elegant way to groupby with a condition? I want to split the data into two groups based on the following conditions:

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

then take the means of both of these groups and end up with an output like this:

               SurvivedMean
 Has Family    Mean
 No Family     Mean

Can it be done using groupby or would I have to append a new column using the above conditional statement?

Answer 1

An easy way to group that is to use the sum of those two columns. If either of them is positive, the result will be greater than 1. And groupby accepts an arbitrary array as long as the length is the same as the DataFrame's length so you don't need to add a new column.

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

Use only one condition if never values in columns SibSp and Parch are less as 0:

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

If is impossible use first use both conditions:

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

You could define your conditions in a list and use the function group_by_condition below to create a filtered list for each condition. Afterwards you can select the resulting items using pattern matching:

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)