Pandas: selecting rows for which groupby.sum () satisfies a condition
In pandas, I have a dataframe of a form:
>>> import pandas as pd
>>> df = pd.DataFrame({'ID':[51,51,51,24,24,24,31], 'x':[0,1,0,0,1,1,0]})
>>> df
ID x
51 0
51 1
51 0
24 0
24 1
24 1
31 0
For each "ID" the value "x" is written several times, it is equal to 0 or 1. I want to select those lines from df
that contain "ID" for which "x" is 1 at least twice.
For each "ID" I manage to count the number of times "x" is 1, by
>>> df.groupby('ID')['x'].sum()
ID
51 1
24 2
31 0
But I don't know how to proceed from here. I need the following output:
ID x
24 0
24 1
24 1
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Uses np.bincount
and pd.factorize
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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 disgusting form one-liner
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 use np.unique
with a parameter return_inverse
to do the same exact thing. But np.unique
will sort the array and 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
big 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
big 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
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