Pandas: Selecting rows for which groupby.sum() satisfies condition

Question

In pandas I have a dataframe of the form:

>>> import pandas as pd  
>>> df = pd.DataFrame({\'ID\':[51,51,51,24,24,24,31], \'x\':[0,1,0,0,1,         


        

Answer 1

Use groupby and filter

df.groupby('ID').filter(lambda s: s.x.sum()>=2)

Output:

   ID  x
3  24  0
4  24  1
5  24  1

Answer 2

df = pd.DataFrame({'ID':[51,51,51,24,24,24,31], 'x':[0,1,0,0,1,1,0]})
df.loc[df.groupby(['ID'])['x'].transform(func=sum)>=2,:]
out:
   ID  x
3  24  0
4  24  1
5  24  1

Answer 3

Using np.bincount and pd.factorize
alternative advance technique to draw better performance

f, u = df.ID.factorize()
df[np.bincount(f, df.x.values)[f] >= 2]

   ID  x
3  24  0
4  24  1
5  24  1

---

In obnoxious one-liner form

df[(lambda f, w: np.bincount(f, w)[f] >= 2)(df.ID.factorize()[0], df.x.values)]

   ID  x
3  24  0
4  24  1
5  24  1

---

np.bincount and np.unique
I could've used np.unique with the return_inverse parameter to accomplish the same exact thing. But, np.unique will sort the array and will change the time complexity of the solution.

u, f = np.unique(df.ID.values, return_inverse=True)
df[np.bincount(f, df.x.values)[f] >= 2]

---

One-liner

df[(lambda f, w: np.bincount(f, w)[f] >= 2)(np.unique(df.ID.values, return_inverse=True)[1], df.x.values)]

---

Timing

%timeit df[(lambda f, w: np.bincount(f, w)[f] >= 2)(df.ID.factorize()[0], df.x.values)]
%timeit df[(lambda f, w: np.bincount(f, w)[f] >= 2)(np.unique(df.ID.values, return_inverse=True)[1], df.x.values)]
%timeit df.groupby('ID').filter(lambda s: s.x.sum()>=2)
%timeit df.loc[df.groupby(['ID'])['x'].transform(func=sum)>=2]
%timeit df.loc[df.groupby(['ID'])['x'].transform('sum')>=2]

small data

1000 loops, best of 3: 302 µs per loop
1000 loops, best of 3: 241 µs per loop
1000 loops, best of 3: 1.52 ms per loop
1000 loops, best of 3: 1.2 ms per loop
1000 loops, best of 3: 1.21 ms per loop

large data

np.random.seed([3,1415])
df = pd.DataFrame(dict(
        ID=np.random.randint(100, size=10000),
        x=np.random.randint(2, size=10000)
    ))

1000 loops, best of 3: 528 µs per loop
1000 loops, best of 3: 847 µs per loop
10 loops, best of 3: 20.9 ms per loop
1000 loops, best of 3: 1.47 ms per loop
1000 loops, best of 3: 1.55 ms per loop

larger data

np.random.seed([3,1415])
df = pd.DataFrame(dict(
        ID=np.random.randint(100, size=100000),
        x=np.random.randint(2, size=100000)
    ))

1000 loops, best of 3: 2.01 ms per loop
100 loops, best of 3: 6.44 ms per loop
10 loops, best of 3: 29.4 ms per loop
100 loops, best of 3: 3.84 ms per loop
100 loops, best of 3: 3.74 ms per loop