Python Pandas groupby apply lambda arguments
Python Pandas groupby apply lambda arguments
The apply method itself passes each group of the groupby object as the first argument to the function. So it knows to associate Weight and Quantity to a and b based on position. (eg they are the 2nd and 3rd arguments if you count the first group argument.
df = pd.DataFrame(np.random.randint(0,11,(10,3)), columns = [num1,num2,num3])
df[category] = [a,a,a,b,b,b,b,c,c,c]
df = df[[category,num1,num2,num3]]
df
category num1 num2 num3
0 a 2 5 2
1 a 5 5 2
2 a 7 3 4
3 b 10 9 1
4 b 4 7 6
5 b 0 5 2
6 b 7 7 5
7 c 2 2 1
8 c 4 3 2
9 c 1 4 6
gb = df.groupby(category)
implicit argument is each group or in this case each category
gb.apply(lambda grp: grp.sum())
The grp is the first argument to the lambda function
notice I dont have to specify anything for it as it is already, automatically taken to be each group of the groupby object
category num1 num2 num3
category
a aaa 14 13 8
b bbbb 21 28 14
c ccc 7 9 9
So apply goes through each of these and performs a sum operation
print(gb.groups)
{a: Int64Index([0, 1, 2], dtype=int64), b: Int64Index([3, 4, 5, 6], dtype=int64), c: Int64Index([7, 8, 9], dtype=int64)}
print(1st GROUP:n, df.loc[gb.groups[a]])
1st GROUP:
category num1 num2 num3
0 a 2 5 2
1 a 5 5 2
2 a 7 3 4
print(SUM of 1st group:n, df.loc[gb.groups[a]].sum())
SUM of 1st group:
category aaa
num1 14
num2 13
num3 8
dtype: object
Notice how this is the same as the first row of our previous operation
So apply is implicitly passing each group to the function argument as the first argument.
From the docs
GroupBy.apply(func, *args, **kwargs)
args, kwargs : tuple and dict
Optional positional and keyword arguments to pass to func
Additional Args passed in *args get passed after the implict group argument.
so using your code
gb.apply(lambda df,a,b: sum(df[a] * df[b]), num1, num2)
category
a 56
b 167
c 20
dtype: int64
here num1 and num2 are being passed as additional arguments to each call of the lambda function
So apply goes through each of these and performs your lambda operation
# copy and paste your lambda function
fun = lambda df,a,b: sum(df[a] * df[b])
print(gb.groups)
{a: Int64Index([0, 1, 2], dtype=int64), b: Int64Index([3, 4, 5, 6], dtype=int64), c: Int64Index([7, 8, 9], dtype=int64)}
print(1st GROUP:n, df.loc[gb.groups[a]])
1st GROUP:
category num1 num2 num3
0 a 2 5 2
1 a 5 5 2
2 a 7 3 4
print(Output of 1st group for function fun:n,
fun(df.loc[gb.groups[a]], num1,num2))
Output of 1st group for function fun:
56