How to filter rows that fall within 1st and 3rd quartile of a particular column in pandas dataframe?

Question

I am working on a data frame in python How can I filter all the rows that have value for a particular column , say val, which fall within 1st and 3rd quartile.

Thank you.

Answer 1

low, high = df.B.quantile([0.25,0.75])
df.query('{low}<B<{high}'.format(low=low,high=high))

Answer 2

Let's create some random data with 100 rows and three columns:

import numpy as np
import pandas as pd

np.random.seed(0)

df = pd.DataFrame(np.random.randn(100, 3), columns=list('ABC'))

Now let's use loc to filter out all data in column B above and below its top and bottom quartile (retaining the middle).

lower_quantile, upper_quantile = df.B.quantile([.25, .75])

>>> df.loc[(df.B > lower_quantile) & (df.B < upper_quantile)].head()
           A         B         C
0   1.764052  0.400157  0.978738
2   0.950088 -0.151357 -0.103219
3   0.410599  0.144044  1.454274
4   0.761038  0.121675  0.443863
10  0.154947  0.378163 -0.887786

Answer 3

Using pd.Series.between() and unpacking the quantile values produced by df.A.quantile([lower, upper]), you can filter your DataFrame, here illustrated using sample data ranging 0-100:

import numpy as np
import pandas as pd

df = pd.DataFrame(data={'A': np.random.randint(0, 100, 10), 'B': np.arange(10)})

    A  B
0   4  0
1  21  1
2  96  2
3  50  3
4  82  4
5  24  5
6  93  6
7  16  7
8  14  8
9  40  9

df[df.A.between(*df.A.quantile([0.25, 0.75]).tolist())]


    A  B
1  21  1
3  50  3
5  24  5
9  40  9

On performance: .query() slows things down 2x:

df = DataFrame(data={'A': np.random.randint(0, 100, 1000), 'B': np.arange(1000)})

def query(df):
    low, high = df.B.quantile([0.25,0.75])
    df.query('{low}<B<{high}'.format(low=low,high=high))

%timeit query(df)
1000 loops, best of 3: 1.81 ms per loop

def between(df):
    df[df.A.between(*df.A.quantile([0.25, 0.75]).tolist())]

%timeit between(df)
1000 loops, best of 3: 995 µs per loop

@Alexander's solution performs identical to the one using .between().