### Question :

Most operations in `pandas`

can be accomplished with operator chaining (`groupby`

, `aggregate`

, `apply`

, etc), but the only way I’ve found to filter rows is via normal bracket indexing

```
df_filtered = df[df['column'] == value]
```

This is unappealing as it requires I assign `df`

to a variable before being able to filter on its values. Is there something more like the following?

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

##
Answer #1:

I’m not entirely sure what you want, and your last line of code does not help either, but anyway:

“Chained” filtering is done by “chaining” the criteria in the boolean index.

```
In [96]: df
Out[96]:
A B C D
a 1 4 9 1
b 4 5 0 2
c 5 5 1 0
d 1 3 9 6
In [99]: df[(df.A == 1) & (df.D == 6)]
Out[99]:
A B C D
d 1 3 9 6
```

If you want to chain methods, you can add your own mask method and use that one.

```
In [90]: def mask(df, key, value):
....: return df[df[key] == value]
....:
In [92]: pandas.DataFrame.mask = mask
In [93]: df = pandas.DataFrame(np.random.randint(0, 10, (4,4)), index=list('abcd'), columns=list('ABCD'))
In [95]: df.ix['d','A'] = df.ix['a', 'A']
In [96]: df
Out[96]:
A B C D
a 1 4 9 1
b 4 5 0 2
c 5 5 1 0
d 1 3 9 6
In [97]: df.mask('A', 1)
Out[97]:
A B C D
a 1 4 9 1
d 1 3 9 6
In [98]: df.mask('A', 1).mask('D', 6)
Out[98]:
A B C D
d 1 3 9 6
```

##
Answer #2:

Filters can be chained using a Pandas query:

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

Filters can also be combined in a single query:

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

##
Answer #3:

The answer from @lodagro is great. I would extend it by generalizing the mask function as:

```
def mask(df, f):
return df[f(df)]
```

Then you can do stuff like:

```
df.mask(lambda x: x[0] < 0).mask(lambda x: x[1] > 0)
```

##
Answer #4:

Since version 0.18.1 the `.loc`

method accepts a callable for selection. Together with lambda functions you can create very flexible chainable filters:

```
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0,100,size=(100, 4)), columns=list('ABCD'))
df.loc[lambda df: df.A == 80] # equivalent to df[df.A == 80] but chainable
df.sort_values('A').loc[lambda df: df.A > 80].loc[lambda df: df.B > df.A]
```

If all you’re doing is filtering, you can also omit the `.loc`

.

##
Answer #5:

I offer this for additional examples. This is the same answer as https://stackoverflow.com/a/28159296/

I’ll add other edits to make this post more useful.

`pandas.DataFrame.query`

`query`

was made for exactly this purpose. Consider the dataframe `df`

```
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
```

Let’s use `query`

to filter all rows where `D > B`

```
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
```

Which we chain

```
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
```

##
Answer #6:

pandas provides two alternatives to Wouter Overmeire’s answer which do not require any overriding. One is `.loc[.]`

with a callable, as in

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

the other is `.pipe()`

, as in

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

##
Answer #7:

I had the same question except that I wanted to combine the criteria into an OR condition. The format given by Wouter Overmeire combines the criteria into an AND condition such that both must be satisfied:

```
In [96]: df
Out[96]:
A B C D
a 1 4 9 1
b 4 5 0 2
c 5 5 1 0
d 1 3 9 6
In [99]: df[(df.A == 1) & (df.D == 6)]
Out[99]:
A B C D
d 1 3 9 6
```

But I found that, if you wrap each condition in `(... == True)`

and join the criteria with a pipe, the criteria are combined in an OR condition, satisfied whenever either of them is true:

```
df[((df.A==1) == True) | ((df.D==6) == True)]
```

##
Answer #8:

My answer is similar to the others. If you do not want to create a new function you can use what pandas has defined for you already. Use the pipe method.

```
df.pipe(lambda d: d[d['column'] == value])
```