\w not helpful for non-Roman scripts

[nathanlmiles]

BPO	1693050
Nosy	@malemburg, @loewis, @terryjreedy, @vstinner, @ezio-melotti
Superseder	bpo-24194: Make tokenize recognize Other_ID_Start and Other_ID_Continue chars

^{Note: these values reflect the state of the issue at the time it was migrated and might not reflect the current state.}

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GitHub fields:

assignee = None
closed_at = <Date 2018-03-15.00:32:39.376>
created_at = <Date 2007-04-02.15:27:11.000>
labels = ['expert-regex']
title = '\\w not helpful for non-Roman scripts'
updated_at = <Date 2018-03-15.00:32:39.375>
user = 'https://bugs.python.org/nathanlmiles'

bugs.python.org fields:

activity = <Date 2018-03-15.00:32:39.375>
actor = 'terry.reedy'
assignee = 'none'
closed = True
closed_date = <Date 2018-03-15.00:32:39.376>
closer = 'terry.reedy'
components = ['Regular Expressions']
creation = <Date 2007-04-02.15:27:11.000>
creator = 'nathanlmiles'
dependencies = []
files = []
hgrepos = []
issue_num = 1693050
keywords = []
message_count = 15.0
messages = ['31688', '31689', '76556', '76557', '81221', '190075', '190100', '190219', '190226', '190268', '190322', '190323', '190324', '190326', '313849']
nosy_count = 10.0
nosy_names = ['lemburg', 'loewis', 'terry.reedy', 'vstinner', 'nathanlmiles', 'rsc', 'timehorse', 'ezio.melotti', 'mrabarnett', 'l0nwlf']
pr_nums = []
priority = 'normal'
resolution = 'duplicate'
stage = 'resolved'
status = 'closed'
superseder = '24194'
type = None
url = 'https://bugs.python.org/issue1693050'
versions = ['Python 2.7', 'Python 3.3', 'Python 3.4']

[nathanlmiles]

When I try to use r"\w+(?u)" to find words in a unicode Devanagari text bad things happen. Words get chopped into small pieces. I think this is likely because vowel signs such as 093e are not considered to match \w.

I think that if you wish \w to be useful for Indic
scipts \w will need to be exanded to unclude unicode character categories Mc, Mn, Me.

I am using Python 2.4.4 on Windows XP SP2.

I ran the following script to see the characters which I think ought to match \w but don't

import re
import unicodedata

text = ""
for i in range(0x901,0x939): text += unichr(i)
for i in range(0x93c,0x93d): text += unichr(i)
for i in range(0x93e,0x94d): text += unichr(i)
for i in range(0x950,0x954): text += unichr(i)
for i in range(0x958,0x963): text += unichr(i)

parts = re.findall("\W(?u)", text)
for ch in parts:
print "%04x" % ord(ch), unicodedata.category(ch)

The odd character here is 0904. Its categorization seems to imply that you are using the uncode 3.0 database but perhaps later versions of Python are using the current 5.0 database.

[malemburg]

Python 2.4 is using Unicode 3.2. Python 2.5 ships with Unicode 4.1.

We're likely to ship Unicode 5.x with Python 2.6 or a later release.

Regarding the char classes: I don't think Mc, Mn and Me should be considered parts of a word. Those are marks which usually separate words.

[terryjreedy]

Vowel 'marks' are condensed vowel characters and are very much part of
words and do not separate words. Python3 properly includes Mn and Mc as
identifier characters.

http://docs.python.org/dev/3.0/reference/lexical_analysis.html#identifiers-and-keywords

For instance, the word 'hindi' has 3 consonants 'h', 'n', 'd', 2 vowels
'i' and 'ii' (long i) following 'h' and 'd', and a null vowel (virama)
after 'n'. [The null vowel is needed because no vowel mark indicates the
default vowel short a. So without it, the word would be hinadii.]
The difference between the devanagari vowel characters, used at the
beginning of words, and the vowel marks, used thereafter, is purely
graphical and not phonological. In short, in the sanskrit family,
word = syllable+
syllable = vowel | consonant + vowel mark

From a clp post asking why re does not see hindi as a word:

हिन्दी
ह DEVANAGARI LETTER HA (Lo)
ि DEVANAGARI VOWEL SIGN I (Mc)
न DEVANAGARI LETTER NA (Lo)
् DEVANAGARI SIGN VIRAMA (Mn)
द DEVANAGARI LETTER DA (Lo)
ी DEVANAGARI VOWEL SIGN II (Mc)

.isapha and possibly other unicode methods need fixing also
>>> 'हिन्दी'.isalpha()#2.x and 3.0
False

[loewis]

Unicode TR#18 defines \w as a shorthand for

\p{alpha}
\p{gc=Mark}
\p{digit}
\p{gc=Connector_Punctuation}

which would include all marks. We should recursively check whether we
follow the recommendation (e.g. \p{alpha} refers to all character having
the Alphabetic derived core property, which is Lu+Ll+Lt+Lm+Lo+Nl +
Other_Alphabetic, where Other_Alphabetic is a selected list of
additional character - all from Mn/Mc)

[mrabarnett]

In issue bpo-2636 I'm using the following:

Alpha is Ll, Lo, Lt, Lu.
Digit is Nd.
Word is Ll, Lo, Lt, Lu, Mc, Me, Mn, Nd, Nl, No, Pc.

These are what are specified at
http://www.regular-expressions.info/posixbrackets.html

[BreamoreBoy]

Am I correct in saying that this must stay open as it targets the re module but as given in msg81221 is fixed in the new regex module?

[mrabarnett]

I had to check what re does in Python 3.3:

>>> print(len(re.match(r'\w+', 'हिन्दी').group()))
1

Regex does this:

>>> print(len(regex.match(r'\w+', 'हिन्दी').group()))
6

[timehorse]

Matthew, I think that is considered a single word in Sanscrit or Thai so Python 3.x is correct. In this case you've written the Sanscrit word for Hindi.

[mrabarnett]

I'm not sure what you're saying.

The re module in Python 3.3 matches only the first codepoint, treating the second codepoint as not part of a word, whereas the regex module matches all 6 codepoints, treating them all as part of a single word.

[timehorse]

Maybe you could show us the byte-for-byte hex of the string you're testing so we can examine if it's really a code point intending word boundary or just a code point for the sake of beginning a new character.

[mrabarnett]

You could've obtained it from msg76556 or msg190100:

>>> print(ascii('हिन्दी'))
'\u0939\u093f\u0928\u094d\u0926\u0940'
>>> import re, regex
>>> print(ascii(re.match(r"\w+", '\u0939\u093f\u0928\u094d\u0926\u0940').group()))
'\u0939'
>>> print(ascii(regex.match(r"\w+", '\u0939\u093f\u0928\u094d\u0926\u0940').group()))
'\u0939\u093f\u0928\u094d\u0926\u0940'

[timehorse]

Thanks Matthew and sorry to put you through more work; I just wanted to verify exactly which unicode (UTF-16 I take it) were being used to verify if the UNICODE standard expected them to be treated as unique words or single letters within a word. Sanskrit is an alphabet, not an ideograph so each symbol is considered a letter. So I believe your implementation is correct and yes, you are right, re is at fault. There are just accenting characters and letters in that sequence so they should be interpreted as a single word of 6 letters, as you determine, and not one of the first letter. Mind you, I misinterpreted msg190100 in that I thought you were using findall in which case the answer should be 1, but as far as length of extraction, yes, 6, I totally agree. Sorry for the misunderstanding. http://www.unicode.org/charts/PDF/U0900.pdf contains the code chart for Hindi.

[mrabarnett]

UTF-16 has nothing to do with it, that's just an encoding (a pair of them actually, UTF-16LE and UTF-16BE).

And I don't know why you thought I was using findall in msg190100 when the examples were using match! :-)