pandas find strings in common among Series

Question

I have a Series of keywords extracted from a bigger DataFrame and a DataFrame with, among others, a column of strings. I would like to mask the DataFrame finding which strings contains at least one keyword. The "Keywords" Series is as follows (sorry for the weird words):

Skilful
Wilful
Somewhere
Thing
Strange

The DataFrame looks as follows:

User_ID;Tweet
01;hi all
02;see you somewhere
03;So weird
04;hi all :-)
05;next big thing
06;how can i say no?
07;so strange
08;not at all

So far I used a str.contains() function from pandas like:

mask = df['Tweet'].str.contains(str(Keywords['Keyword'][4]), case=False)

which works well finding the "Strange" string in the DataFrame and returns:

0    False
1    False
2    False
3    False
4    False
5    False
6     True
7    False
Name: Tweet, dtype: bool

What I would like to do is to mask the whole DataFrame with the all Keywords array, so I can have something like this:

0    False
1     True
2    False
3    False
4     True
5    False
6     True
7    False
Name: Tweet, dtype: bool

Is it possible without looping through the array? In my real case I have to search through millions of strings, so I'm looking for a fast method.

Thanks for your kind help.

Answer 1

Another way to achieve this is to use pd.Series.isin() with map and apply, with your sample it will be like:

df    # DataFrame

   User_ID              Tweet
0        1             hi all
1        2  see you somewhere
2        3           So weird
3        4         hi all :-)
4        5     next big thing
5        6  how can i say no?
6        7         so strange
7        8         not at all

---

w    # Series

0      Skilful
1       Wilful
2    Somewhere
3        Thing
4      Strange
dtype: object

---

# list
masked = map(lambda x: any(w.apply(str.lower).isin(x)), \                 
             df['Tweet'].apply(str.lower).apply(str.split))

df['Tweet_masked'] = masked

Results:

df
Out[13]: 
   User_ID              Tweet Tweet_masked
0        1             hi all        False
1        2  see you somewhere         True
2        3           So weird        False
3        4         hi all :-)        False
4        5     next big thing         True
5        6  how can i say no?        False
6        7         so strange         True
7        8         not at all        False

As a side note, isin only works if the whole string matches the values, in case you are only interested in str.contains, here's the variant:

masked = map(lambda x: any(_ in x for _ in w.apply(str.lower)), \
             df['Tweet'].apply(str.lower))

Updated: as @Alex pointed out, it could be even more efficient to combine both map and regexp, in fact I don't quite like map + lambda neither, here we go:

import re

r = re.compile(r'.*({}).*'.format('|'.join(w.values)), re.IGNORECASE)

masked = map(bool, map(r.match, df['Tweet']))

Answer 2

import re
df['Tweet'].str.match('.*({0}).*'.format('|'.join(phrases)))

Where phrases is an iterable of phrases whose existence you are conditioning on.

Answer 3

A simple apply can solve this. If you can weather a few seconds of processing, I think this is the simplest method available to you without venturing outside pandas.

import pandas as pd

df = pd.read_csv("dict.csv", delimiter=";")
ref = pd.read_csv("ref.csv")

kw = set([k.lower() for k in ref["Keywords"]])
print kw

boom = lambda x:True if any(w in kw for w in x.split()) else False

df["Tweet"] = df["Tweet"].apply(boom)
print df

I tested it against exactly 10,165,760 rows of made-up data and it completed in 18.9s. If that is not fast enough, then a better method is needed.

set(['somewhere', 'thing', 'strange', 'skilful', 'wilful'])
          User_ID  Tweet
0               1  False
1               2   True
2               3  False
3               4  False
4               5   True
5               6  False
6               7   True
7               8  False
8               1  False
9               2   True
10              3  False
11              4  False
12              5   True
13              6  False
14              7   True
15              8  False
16              1  False
17              2   True
18              3  False
19              4  False
20              5   True
21              6  False
22              7   True
23              8  False
24              1  False
25              2   True
26              3  False
27              4  False
28              5   True
29              6  False
...           ...    ...
10165730        3  False
10165731        4  False
10165732        5   True
10165733        6  False
10165734        7   True
10165735        8  False
10165736        1  False
10165737        2   True
10165738        3  False
10165739        4  False
10165740        5   True
10165741        6  False
10165742        7   True
10165743        8  False
10165744        1  False
10165745        2   True
10165746        3  False
10165747        4  False
10165748        5   True
10165749        6  False
10165750        7   True
10165751        8  False
10165752        1  False
10165753        2   True
10165754        3  False
10165755        4  False
10165756        5   True
10165757        6  False
10165758        7   True
10165759        8  False

[10165760 rows x 2 columns]
[Finished in 18.9s]

Hope this helps.