Suppose we have simple Dataframe
df = pd.DataFrame(['one apple','banana','box of oranges','pile of fruits outside', 'one banana', 'fruits'])
df.columns = ['fruits']
how to calculate number of words in keywords, similar to:
1 word: 2
2 words: 2
3 words: 1
4 words: 1
IIUC then you can do the following:
In [89]:
count = df['fruits'].str.split().apply(len).value_counts()
count.index = count.index.astype(str) + ' words:'
count.sort_index(inplace=True)
count
Out[89]:
1 words: 2
2 words: 2
3 words: 1
4 words: 1
Name: fruits, dtype: int64
Here we use the vectorised str.split to split on spaces, and then apply len to get the count of the number of elements, we can then call value_counts to aggregate the frequency count.
We then rename the index and sort it to get the desired output
UPDATE
This can also be done using str.len rather than apply which should scale better:
In [41]:
count = df['fruits'].str.split().str.len()
count.index = count.index.astype(str) + ' words:'
count.sort_index(inplace=True)
count
Out[41]:
0 words: 2
1 words: 1
2 words: 3
3 words: 4
4 words: 2
5 words: 1
Name: fruits, dtype: int64
Timings
In [42]:
%timeit df['fruits'].str.split().apply(len).value_counts()
%timeit df['fruits'].str.split().str.len()
1000 loops, best of 3: 799 µs per loop
1000 loops, best of 3: 347 µs per loop
For a 6K df:
In [51]:
%timeit df['fruits'].str.split().apply(len).value_counts()
%timeit df['fruits'].str.split().str.len()
100 loops, best of 3: 6.3 ms per loop
100 loops, best of 3: 6 ms per loop
You could use str.count with space ' ' as delimiter.
In [1716]: count = df['fruits'].str.count(' ').add(1).value_counts(sort=False)
In [1717]: count.index = count.index.astype('str') + ' words:'
In [1718]: count
Out[1718]:
1 words: 2
2 words: 2
3 words: 1
4 words: 1
Name: fruits, dtype: int64
Timings
str.count is marginally faster
Small
In [1724]: df.shape
Out[1724]: (6, 1)
In [1725]: %timeit df['fruits'].str.count(' ').add(1).value_counts(sort=False)
1000 loops, best of 3: 649 µs per loop
In [1726]: %timeit df['fruits'].str.split().apply(len).value_counts()
1000 loops, best of 3: 840 µs per loop
Medium
In [1728]: df.shape
Out[1728]: (6000, 1)
In [1729]: %timeit df['fruits'].str.count(' ').add(1).value_counts(sort=False)
100 loops, best of 3: 6.58 ms per loop
In [1730]: %timeit df['fruits'].str.split().apply(len).value_counts()
100 loops, best of 3: 6.99 ms per loop
Large
In [1732]: df.shape
Out[1732]: (60000, 1)
In [1733]: %timeit df['fruits'].str.count(' ').add(1).value_counts(sort=False)
1 loop, best of 3: 57.6 ms per loop
In [1734]: %timeit df['fruits'].str.split().apply(len).value_counts()
1 loop, best of 3: 73.8 ms per loop
