How to make a Python Regular Expression, have an optional start requirement?

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So I need to be able to check for the word "view" in a sentence, without pulling it out of another word, and even if its the first word.

import re


if re.search(r"view", "interview"):
    print "aw"                          #This cuts it out of the word "interview"

if re.search(r" view", "interview"):
    print "aw"                          #This wont cut it out of another word, but 
                                    #doesn't work if "view" is the first word.

if re.search(r" view", "view"):
    print "aw"                          #This just shows it wont work since its the first word.

It sounds like you are looking for word boundaries, or \b when using regular expressions:

Try using \bview\b as your regular expression. \b will match (^\w|\w$|\W\w|\w\W), so it will only match the standalone word view, and not when it is found inside another word.

Here are some examples:

rgx = r'\bview\b'

print(re.search(rgx, 'interview'))
print(re.search(rgx, 'view is the first word'))
print(re.search(rgx, ' view is after a space'))

# None
# <_sre.SRE_Match object; span=(0, 4), match='view'>
# <_sre.SRE_Match object; span=(1, 5), match='view'>

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The "view at the start of the string" case can only happen once per string so you can check that individually. That can be done by using one if check, or simpler, just add a space before the target string. For example check for "view" in str:

if re.search(" view", " " + str): print("Found!")

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The occurrence can be at beginning, end and in the middle and hence we try to capture all such occurrences while ignoring the ones where it occurs as part of other text.

re.search(r"(^view)|(\sview$)|(\sview\s)"," view interview is view")

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This RegEx captures every Word which has "view" somewhere inside it.

/([A-z]*view[A-z]*)/gi

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