Count distinct words from a Pandas Data Frame

Question

I've a Pandas data frame, where one column contains text. I'd like to get a list of unique words appearing across the entire column (space being the only split).

import pandas as pd

r1=['My nickname is ft.jgt','Someone is going to my place']

df=pd.DataFrame(r1,columns=['text'])

The output should look like this:

['my','nickname','is','ft.jgt','someone','going','to','place']

It wouldn't hurt to get a count as well, but it is not required.

Answer 1

Use a set to create the sequence of unique elements.

Do some clean-up on df to get the strings in lower case and split:

df['text'].str.lower().str.split()
Out[43]: 
0             [my, nickname, is, ft.jgt]
1    [someone, is, going, to, my, place]

Each list in this column can be passed to set.update function to get unique values. Use apply to do so:

results = set()
df['text'].str.lower().str.split().apply(results.update)
print(results)

set(['someone', 'ft.jgt', 'my', 'is', 'to', 'going', 'place', 'nickname'])

Or use with Counter() from comments:

from collections import Counter
results = Counter()
df['text'].str.lower().str.split().apply(results.update)
print(results)

Answer 2

Use collections.Counter:

>>> from collections import Counter
>>> r1=['My nickname is ft.jgt','Someone is going to my place']
>>> Counter(" ".join(r1).split(" ")).items()
[('Someone', 1), ('ft.jgt', 1), ('My', 1), ('is', 2), ('to', 1), ('going', 1), ('place', 1), ('my', 1), ('nickname', 1)]

Answer 3

If you want to do it from the DataFrame construct:

import pandas as pd

r1=['My nickname is ft.jgt','Someone is going to my place']

df=pd.DataFrame(r1,columns=['text'])

df.text.apply(lambda x: pd.value_counts(x.split(" "))).sum(axis = 0)

My          1
Someone     1
ft.jgt      1
going       1
is          2
my          1
nickname    1
place       1
to          1
dtype: float64

If you want a more flexible tokenization use nltk and its tokenize