Why \S+ with boundaries does not extract non-space characters?

Asked Jul 17 '18 at 15:28

Active Jul 17 '18 at 16:02

Viewed 52 times

print(re.search(r'\b\w+\b', '\t hell0_there(0,1) \tbbb'))     # Works fine - 'hell0_there'
print(re.search(r'\S+', '  hell0_there(0,1)\t bbb'))          # Works fine - 'hell0_there(0,1)'

print(re.search(r'\b\S+\b', '  hell0_there(0,1)  '))      # Missing paranthesis - 'hell0_there(0,1'
print(re.search(r'\b\S+\b','   1.3.6.1.4.1.9848.3.2.2.3(1,0)   ')) # Missing closed paranthesis

print(re.search(r'\b\S+\b', '\t hell0_there(0,1) \tbbb'))      # Same issue
print(re.search(r'\b\S+\b','\t  1.3.6.1.4.1.9148.3.2.2.3(1,0)  \tnnn')) # Same issue

Why non-space boundary('\b\S+\b') does not extract with closed paranthesis?

edited Jul 17 '18 at 16:02

asked Jul 17 '18 at 15:28

overexchange

15,768
30
152
347

There is no word boundary between `)` and following space. – user4003407 Jul 17 '18 at 15:30
@PetSerAl `hell0_there(0,1)` is a single word. Isn't it? – overexchange Jul 17 '18 at 15:32
`hell0_there(0,1)` is not a single word as word contains only `[a-zA-Z0-9_]` characters. There is a word boundary between `1` and `)` but there no word boundary between `)` and space. – anubhava Jul 17 '18 at 15:33
@overexchange No, it is not. You should reread definition of `\w`. – user4003407 Jul 17 '18 at 15:33
@PetSerAl I meant, `hell0_there(0,1)` are one or more non-space characters(`\S+`). Isn't it? Yes `\w` is short cut for `[a-zA-Z0-9_]` – overexchange Jul 17 '18 at 15:34
1

Yes it is 1+ non-space characters but word boundary is always asserted between `\w` and `\W`. Looks like you just need to have `r'\S+'` – anubhava Jul 17 '18 at 15:35
@anubhava Query edited – overexchange Jul 17 '18 at 15:40
1

That query has already been answered I think. There is a word boundary between `1` and `)` but no word boundary between `)` and space (both `\W`). – anubhava Jul 17 '18 at 15:42
Last two don't work fine either. – revo Jul 17 '18 at 15:46
@revo you are right – overexchange Jul 17 '18 at 15:49
@anubhava, could you write your comment as answer? – Jul 17 '18 at 15:53
@anubhava what would be the regular expression to detect with `)` as well? as this is part of non-space character. `\S+` – overexchange Jul 17 '18 at 16:10
@anubhava [Answer](https://stackoverflow.com/a/1324784/3317808) says: *After the last character in the string, if the last character is a word character.* last word character is `e`. How come, `(` & `,` are within boundary? – overexchange Jul 17 '18 at 16:13
@overexchange *How come, `(` & `,` are within boundary?* because them match `\S+` and there is a boundary between `0` and `)`. – user4003407 Jul 17 '18 at 16:27

Why \S+ with boundaries does not extract non-space characters?

0 Answers0