How to filter rows in pandas by regex

Question

Use contains instead:

In [10]: df.b.str.contains('^f')
Out[10]: 
0    False
1     True
2     True
3    False
Name: b, dtype: bool

There is already a string handling function Series.str.startswith(). You should try foo[foo.b.str.startswith('f')].

Result:

    a   b
1   2   foo
2   3   fat

I think what you expect.

Alternatively you can use contains with regex option. For example:

foo[foo.b.str.contains('oo', regex= True, na=False)]

Result:

    a   b
1   2   foo

na=False is to prevent Errors in case there is nan, null etc. values

It may be a bit late, but this is now easier to do in Pandas by calling Series.str.match. The docs explain the difference between match, fullmatch and contains.

Note that in order to use the results for indexing, set the na=False argument (or True if you want to include NANs in the results).

Multiple column search with dataframe:

frame[frame.filename.str.match('*.'+MetaData+'.*') & frame.file_path.str.match('C:\test\test.txt')]

Building off of the great answer by user3136169, here is an example of how that might be done also removing NoneType values.

def regex_filter(val):
    if val:
        mo = re.search(regex,val)
        if mo:
            return True
        else:
            return False
    else:
        return False

df_filtered = df[df['col'].apply(regex_filter)]

You can also add regex as an arg:

def regex_filter(val,myregex):
    ...

df_filtered = df[df['col'].apply(regex_filter,regex=myregex)]

Write a Boolean function that checks the regex and use apply on the column

foo[foo['b'].apply(regex_function)]