Ignoring NaNs with str.contains

There's a flag for that:

In [11]: df = pd.DataFrame([["foo1"], ["foo2"], ["bar"], [np.nan]], columns=['a'])

In [12]: df.a.str.contains("foo")
Out[12]:
0     True
1     True
2    False
3      NaN
Name: a, dtype: object

In [13]: df.a.str.contains("foo", na=False)
Out[13]:
0     True
1     True
2    False
3    False
Name: a, dtype: bool

See the str.replace docs:

na : default NaN, fill value for missing values.

So you can do the following:

In [21]: df.loc[df.a.str.contains("foo", na=False)]
Out[21]:
      a
0  foo1
1  foo2

In addition to the above answers, I would say for columns having no single word name, you may use:-

df[df['Product ID'].str.contains("foo") == True]

Hope this helps.

df[df.col.str.contains("foo").fillna(False)]

I'm not 100% on why (actually came here to search for the answer), but this also works, and doesn't require replacing all nan values.

import pandas as pd
import numpy as np

df = pd.DataFrame([["foo1"], ["foo2"], ["bar"], [np.nan]], columns=['a'])

newdf = df.loc[df['a'].str.contains('foo') == True]

Works with or without .loc.

I have no idea why this works, as I understand it when you're indexing with brackets pandas evaluates whatever's inside the bracket as either True or False. I can't tell why making the phrase inside the brackets 'extra boolean' has any effect at all.