Python regex to match words not having dot

Question

I want to accept only those strings having the pattern \'wild.flower\', \'pink.flower\',...i.e any word preceding \'.flower\', but the word should not contain dot. For example,

Answer 1

Here is the regex for you. ^([^\.]*)\.flower$. Example: https://regex101.com/r/cSL445/1.

Answer 2

Is this sufficient?

^\w+\.\w+$

Answer 3

You are looking for "^\w+\.flower$".

Answer 4

To match any character except a newline or a dot you could use a negated character class [^.\r\n]+ and repeat that one or more times and use anchors to assert the start ^ and the end $ of the line.

^[^.\r\n]+\.flower$

Or you could specify in a character class which characters you would allow to match followed by a dot \. and flower.

^[a-z0-9]+\.flower$

Answer 5

Your case of pink.blue.flower is unclear. There are 2 possibilities:

• Match only blue (cut off preceding dot and what was before).
• Reject this case altogether (you want to match a word preceding .flower only if it is not preceded with a dot).

In the first case accept other answers.

But if you want the second solution, use: \b(?<!\.)[a-z]+(?=\.flower).

Description:

• \b - Start from a word boundary (but it allows the "after a dot" case).
• (?<!\.) - Negative lookbehind - exclude the "after a dot" case.
• [a-z]+ - Match a sequence of letters.
• (?=\.flower) - Positive lookahead for .flower.

I assumed that you have only lower case letters, but if it is not the case, then add i (case insensitive) option.

Another remark: Other answers include \w, which matches also digits and _ or even [^\.] - any char other than a dot (including e.g. \n).

Are you happy with that? If you aren't, change to [a-z] (again, maybe with i option).