/Python/Filter dataframe site:stackoverflow.com

Filter dataframe site:stackoverflow.com

df[df["column_name"] != 5].groupby("other_column_name")

Source: stackoverflow.com

Filter dataframe site:stackoverflow.com

df.query('D > B').query('C > B')
# equivalent to
# df.query('D > B and C > B')
# but defeats the purpose of demonstrating chaining

   A  B  C  D  E
0  0  2  7  3  8
1  7  0  6  8  6
4  3  6  7  7  4
5  5  3  7  5  9
7  6  2  6  6  5

Source: stackoverflow.com

Filter dataframe site:stackoverflow.com

df_filtered = df.pipe(lambda x: x['column'] == value)

Source: stackoverflow.com

Filter dataframe site:stackoverflow.com

df = pd.DataFrame(np.random.randn(30, 3), columns=['a','b','c'])
df_filtered = df.query('a > 0').query('0 < b < 2')

Source: stackoverflow.com

Filter dataframe site:stackoverflow.com

import pandas as pd
import numpy as np

np.random.seed([3,1415])
df = pd.DataFrame(
    np.random.randint(10, size=(10, 5)),
    columns=list('ABCDE')
)

df

   A  B  C  D  E
0  0  2  7  3  8
1  7  0  6  8  6
2  0  2  0  4  9
3  7  3  2  4  3
4  3  6  7  7  4
5  5  3  7  5  9
6  8  7  6  4  7
7  6  2  6  6  5
8  2  8  7  5  8
9  4  7  6  1  5

Source: stackoverflow.com

Filter dataframe site:stackoverflow.com

df_filtered = df.query('a > 0 and 0 < b < 2')


#If you need to refer to python variables in your query, the documentation says, "You can refer to variables in the environment by prefixing them with an ‘@’ character like @a + b". Note that the following are valid: df.query('a in list([1,2])'), s = set([1,2]); df.query('a in @s'). – 
#teichert

Source: stackoverflow.com

Filter dataframe site:stackoverflow.com

df.query('D > B')

   A  B  C  D  E
0  0  2  7  3  8
1  7  0  6  8  6
2  0  2  0  4  9
3  7  3  2  4  3
4  3  6  7  7  4
5  5  3  7  5  9
7  6  2  6  6  5

Source: stackoverflow.com

Filter dataframe site:stackoverflow.com

df_filtered = df.loc[lambda x: x['column'] == value]

Source: stackoverflow.com

Filter dataframe site:stackoverflow.com

df_filtered = df.query('a > 0 and 0 < b < 2')

Source: stackoverflow.com

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