I'm working with Python v2, and I'm trying to find out if you can tell if a word is in a string.
I have found some information about identifying if the word is in the string - using .find, but is there a way to do an IF statement. I would like to have something like the following:
if string.find(word):
print 'success'
Thanks for any help.
This small function compares all search words in given text. If all search words are found in text, returns length of search, or False otherwise.
Also supports unicode string search.
def find_words(text, search):
"""Find exact words"""
dText = text.split()
dSearch = search.split()
found_word = 0
for text_word in dText:
for search_word in dSearch:
if search_word == text_word:
found_word += 1
if found_word == len(dSearch):
return lenSearch
else:
return False
usage:
find_words('çelik güray ankara', 'güray ankara')
Using regex is a solution, but it is too complicated for that case.
You can simply split text into list of words. Use split(separator, num) method for that. It returns a list of all the words in the string, using separator as the separator. If separator is unspecified it splits on all whitespace (optionally you can limit the number of splits to num).
list_of_words = mystring.split()
if word in list_of_words:
print 'success'
This will not work for string with commas etc. For example:
mystring = "One,two and three"
# will split into ["One,two", "and", "three"]
If you also want to split on all commas etc. use separator argument like this:
# whitespace_chars = " \t\n\r\f" - space, tab, newline, return, formfeed
list_of_words = mystring.split( \t\n\r\f,.;!?'\"()")
if word in list_of_words:
print 'success'
Advanced way to check the exact word, that we need to find in a long string:
import re
text = "This text was of edited by Rock"
#try this string also
#text = "This text was officially edited by Rock"
for m in re.finditer(r"\bof\b", text):
if m.group(0):
print "Present"
else:
print "Absent"
You could just add a space before and after "word".
x = raw_input("Type your word: ")
if " word " in x:
print "Yes"
elif " word " not in x:
print "Nope"
This way it looks for the space before and after "word".
>>> Type your word: Swordsmith
>>> Nope
>>> Type your word: word
>>> Yes
If matching a sequence of characters is not sufficient and you need to match whole words, here is a simple function that gets the job done. It basically appends spaces where necessary and searches for that in the string:
def smart_find(haystack, needle):
if haystack.startswith(needle+" "):
return True
if haystack.endswith(" "+needle):
return True
if haystack.find(" "+needle+" ") != -1:
return True
return False
This assumes that commas and other punctuations have already been stripped out.
What about to split the string and strip words punctuation?
w in [ws.strip(',.?!') for ws in p.split()]
Or working the case:
w.lower() in [ws.strip(',.?!') for ws in p.lower().split()]
Maybe that way:
def wsearch(word, phrase):
# Attention about punctuation and about split characters
punctuation = ',.?!'
return word.lower() in [words.strip(punctuation) for words in phrase.lower().split()]
Sample:
print(wsearch('CAr', 'I own a caR.'))
I didn't check performance...
You can split string to the words and check the result list.
if word in string.split():
print 'success'
If you want to find out whether a whole word is in a space-separated list of words, simply use:
def contains_word(s, w):
return (' ' + w + ' ') in (' ' + s + ' ')
contains_word('the quick brown fox', 'brown') # True
contains_word('the quick brown fox', 'row') # False
This elegant method is also the fastest. Compared to Hugh Bothwell's and daSong's approaches:
>python -m timeit -s "def contains_word(s, w): return (' ' + w + ' ') in (' ' + s + ' ')" "contains_word('the quick brown fox', 'brown')"
1000000 loops, best of 3: 0.351 usec per loop
>python -m timeit -s "import re" -s "def contains_word(s, w): return re.compile(r'\b({0})\b'.format(w), flags=re.IGNORECASE).search(s)" "contains_word('the quick brown fox', 'brown')"
100000 loops, best of 3: 2.38 usec per loop
>python -m timeit -s "def contains_word(s, w): return s.startswith(w + ' ') or s.endswith(' ' + w) or s.find(' ' + w + ' ') != -1" "contains_word('the quick brown fox', 'brown')"
1000000 loops, best of 3: 1.13 usec per loop
Edit: A slight variant on this idea for Python 3.6+, equally fast:
def contains_word(s, w):
return f' {w} ' in f' {s} '
As you are asking for a word and not for a string, I would like to present a solution which is not sensitive to prefixes / suffixes and ignores case:
#!/usr/bin/env python
import re
def is_word_in_text(word, text):
"""
Check if a word is in a text.
Parameters
----------
word : str
text : str
Returns
-------
bool : True if word is in text, otherwise False.
Examples
--------
>>> is_word_in_text("Python", "python is awesome.")
True
>>> is_word_in_text("Python", "camelCase is pythonic.")
False
>>> is_word_in_text("Python", "At the end is Python")
True
"""
pattern = r'(^|[^\w]){}([^\w]|$)'.format(word)
pattern = re.compile(pattern, re.IGNORECASE)
matches = re.search(pattern, text)
return bool(matches)
if __name__ == '__main__':
import doctest
doctest.testmod()
If your words might contain regex special chars (such as +), then you need re.escape(word)
if 'seek' in 'those who seek shall find':
print('Success!')
but keep in mind that this matches a sequence of characters, not necessarily a whole word - for example, 'word' in 'swordsmith' is True. If you only want to match whole words, you ought to use regular expressions:
import re
def findWholeWord(w):
return re.compile(r'\b({0})\b'.format(w), flags=re.IGNORECASE).search
findWholeWord('seek')('those who seek shall find') # -> <match object>
findWholeWord('word')('swordsmith') # -> None
find returns an integer representing the index of where the search item was found. If it isn't found, it returns -1.
haystack = 'asdf'
haystack.find('a') # result: 0
haystack.find('s') # result: 1
haystack.find('g') # result: -1
if haystack.find(needle) >= 0:
print 'Needle found.'
else:
print 'Needle not found.'
Source: Stackoverflow.com