Count occurrences of certain words in pandas dataframe

Update: Original answer counts those rows which contain a substring.

To count all the occurrences of a substring you can use .str.count:

In [21]: df = pd.DataFrame(['hello', 'world', 'hehe'], columns=['words'])

In [22]: df.words.str.count("he|wo")
Out[22]:
0    1
1    1
2    2
Name: words, dtype: int64

In [23]: df.words.str.count("he|wo").sum()
Out[23]: 4

---

The str.contains method accepts a regular expression:

Definition: df.words.str.contains(self, pat, case=True, flags=0, na=nan)
Docstring:
Check whether given pattern is contained in each string in the array

Parameters
----------
pat : string
    Character sequence or regular expression
case : boolean, default True
    If True, case sensitive
flags : int, default 0 (no flags)
    re module flags, e.g. re.IGNORECASE
na : default NaN, fill value for missing values.

For example:

In [11]: df = pd.DataFrame(['hello', 'world'], columns=['words'])

In [12]: df
Out[12]:
   words
0  hello
1  world

In [13]: df.words.str.contains(r'[hw]')
Out[13]:
0    True
1    True
Name: words, dtype: bool

In [14]: df.words.str.contains(r'he|wo')
Out[14]:
0    True
1    True
Name: words, dtype: bool

To count the occurences you can just sum this boolean Series:

In [15]: df.words.str.contains(r'he|wo').sum()
Out[15]: 2

In [16]: df.words.str.contains(r'he').sum()
Out[16]: 1

To count the total number of matches, use s.str.match(...).str.get(0).count().

If your regex will be matching several unique words, to be tallied individually, use s.str.match(...).str.get(0).groupby(lambda x: x).count()

It works like this:

In [12]: s
Out[12]: 
0    ax
1    ay
2    bx
3    by
4    bz
dtype: object

The match string method handles regular expressions...

In [13]: s.str.match('(b[x-y]+)')
Out[13]: 
0       []
1       []
2    (bx,)
3    (by,)
4       []
dtype: object

...but the results, as given, are not very convenient. The string method get takes the matches as strings and converts empty results to NaNs...

In [14]: s.str.match('(b[x-y]+)').str.get(0)
Out[14]: 
0    NaN
1    NaN
2     bx
3     by
4    NaN
dtype: object

...which are not counted.

In [15]: s.str.match('(b[x-y]+)').str.get(0).count()
Out[15]: 2