Python检查字符串是否包含任何字典的键

Question

Given the example dictionary:

LANGUAGE_TO_ISO = {
    "en": "en",
    "eng": "en",
    "english": "en",
    "es": "es",
    "spanish": "es",
    ...
}

And a given example strings:

book_title = "The Dark Tower - english"
book_title = "The Dark Tower - eng"
book_title = "The Dark Tower 2 - english 2nd edition"

Is there a Python function I'm unaware of that would permit to search if a string contains any of the dictionary keys then return the corresponding values, without having to loop in the ISO dictionary?

This way, I could extract the ISO language from the many different ways a language could have been written.

If someone knows of a less dirty way of doing this, please share :)

UPDATE : As Willem mentioned, forgot to specify that "english", "eng", "spanish" etc would be separated by words. Either a dot, comma, hyphen, space, ...

Answer 1

I dont't know if that's optimal way, and yet I have a loop, however it's pretty compact:

def has_key_in(dictionary, string):
  return any(k in string for k in dictionary)

The advantage, if I'm not wrong, is that any stops at the first encountered True condition.

Now, the problem is that you don't have the corresponding value...

Answer 2

This should give you the common key:

set(book_title.split()).intersection(set(LANGUAGE_TO_ISO.keys()))

which you can lookup into the dictionary to get the corresponding value.

---

in response to comment from OP, including a snippet of the output on the shell:

In [4]: LANGUAGE_TO_ISO = { 
   ...:     "en": "en", 
   ...:     "eng": "en", 
   ...:     "english": "en", 
   ...:     "es": "es", 
   ...:     "spanish": "es", 
   ...: }                                                                                                                                                       

In [5]: book_title = "The Dark Tower - english"                                                                                                                 

In [6]: set(book_title.split()).intersection(set(LANGUAGE_TO_ISO.keys()))                                                                                       
Out[6]: {'english'}

Answer 3

The less complex way of doing it would be to try to replace each word of the sentence using regular expressions and try to replace the word by another one using a replacement function, defaulting to the current word if not found:

LANGUAGE_TO_ISO = {
    "en": "en",
    "eng": "en",
    "english": "en",
    "es": "es",
    "spanish": "es",
}

book_title = "The Dark Tower - english"

import re

print(re.sub(r"\b(\w+)\b",lambda m : LANGUAGE_TO_ISO.get(m.group(1),m.group(1)),book_title))

prints:

The Dark Tower - en

Answer 4

If you are only interested in the words of the string to process, we can perform a match linear in the number of characters of the dictionary, with:

filter(None, map(LANGUAGE_TO_ISO.get, book_title.split()))

This will contain a list of ISO codes for matched words (so we do not match 'en' in the word 'men' ).

For example:

>>> book_title = "The Dark Tower - eng"
>>> list(filter(None, map(LANGUAGE_TO_ISO.get, book_title.split())))
['en']

We can - if we want to - make it even more or less case sensitive (for some special cases, for example characters without a lowercase variant, this will not work) with:

filter(None, map(LANGUAGE_TO_ISO.get, book_title.lower().split()))

(given the keys in the dictionary are all lowercase).

If you however want to be able to parse substrings (like 'en' in 'men' ), then you may want to look for a parser (a parser works linear in the input as well, and acts like an annotated finite state machine).

Python splits words according to spaces, but dots, etc. will not separate the words. You can however split those with for example a regex, like:

import re

splt = re.compile('\W+')

filter(None, map(LANGUAGE_TO_ISO.get, splt.split(book_title)))

Or based on your edit:

Either a dot, comma, hyphen, space, ...

You can list the characters between square brackets:

import re

splt = re.compile('[\s.,-]+')

filter(None, map(LANGUAGE_TO_ISO.get, splt.split(book_title)))