python - string match only whole words

Question

I have two lists - query and line . My code finds if a query such as:

["president" ,"publicly"]

Is contained in a line (order matters) such as:

["president" ,"publicly", "told"]

And this is the code I'm currently using:

if ' '.join(query) in ' '.join(line)

Problem is, I want to match whole words only. So the query below won't pass the condition statement:

["president" ,"pub"]

How can I do that?

Answer 1

You could use regexes and the \\b word boundaries:

import re
the_regex = re.compile(r'\b' + r'\b'.join(map(re.escape, ['president', 'pub'])) + r'\b')
if the_regex.search(' '.join(line)):
    print 'matching'
else:
    print 'not matching'

As an alternative you can write a function to check if a given list is a sublist of the line. Something like:

def find_sublist(sub, lst):
    if not sub:
        return 0
    cur_index = 0
    while cur_index < len(lst):
        try:
            cur_index = lst.index(sub[0], cur_index)
        except ValueError:
            break

        if lst[cur_index:cur_index + len(sub)] == sub:
            break
        lst = lst[cur_index + 1:]
    return cur_index

Which you can use as:

if find_sublist(query, line) >= 0:
    print 'matching'
else:
    print 'not matching'

Answer 2

Just use the "in" operator:

mylist = ['foo', 'bar', 'baz']

'foo' in mylist -> returns True 'bar' in mylist -> returns True 'fo' in mylist -> returns False 'ba' in mylist -> returns False

Answer 3

这是一种方式：

re.search(r'\b' + re.escape(' '.join(query)) + r'\b', ' '.join(line)) is not None

Answer 4

Just for fun you can also do:

a = ["president" ,"publicly", "told"]
b = ["president" ,"publicly"]
c = ["president" ,"pub"]
d = ["publicly", "president"]
e = ["publicly", "told"]

from itertools import izip
not [l for l,n in izip(a, b) if l != n] ## True
not [l for l,n in izip(a, c) if l != n] ## False
not [l for l,n in izip(a, d) if l != n] ## False
## to support query in the middle of the line:
try:
  query_list = a[a.index(e[0]):]
  not [l for l,n in izip(query_list, e) if l != n] ## True 
expect ValueError:
  pass

Answer 5

you can use issubset method to achieve this. Simply do:

a = ["president" ,"publicly"]
b = ["president" ,"publicly", "told"]

if set(a).issubset(b):
    #bla bla

this will return matching items in both lists.

Answer 6

You can use the all built in quantor function:

if all(word in b for word in a):
    """ all words in list"""

Note that this may not be run time efficient for long lists. Better use set type instead of list for a (list list of words to search in).

Answer 7

Here is a non-regex way of doing it. I'm sure regex would be much faster than this:

>>> query = ['president', 'publicly']
>>> line = ['president', 'publicly', 'told']
>>> any(query == line[i:i+len(query)] for i in range(len(line) - len(query)))
True
>>> query = ["president" ,"pub"]
>>> any(query == line[i:i+len(query)] for i in range(len(line) - len(query)))
False

Answer 8

Explicit is better than implicit. And as ordering matters, I would write it down like this:

query = ['president','publicly']
query_false = ['president','pub']
line = ['president','publicly','told']

query_len = len(query)
blocks = [line[i:i+query_len] for i in xrange(len(line)-query_len+1)]

blocks holds all relevant combinations to check for:

[['president', 'publicly'], ['publicly', 'told']]

Now you can simply check if your query is in that list:

print query in blocks # -> True
print query_false in blocks # -> False

The code works the way you would probably explain the straight forward solution in words, which is usually a good sign to me. If you have long lines and performance becomes a problem, you can replace the generated list by a generator.