The code
phrase = "".join([c if c.isalpha() else " " for c in phrase])
substitute all non-alphabetic character with spaces. It works very well with strings made up with occidental language characters.
But giving it the value:
phrase = u'इसका स्वामित्व और नियंत्रण किया। इसके'
the result is u'इसक स व म त व और न य त रण क य इसक ', while it shouldn't change, since the string is only made of alphabetic characters and spaces.
I think the reason is that some character is a surrogate pair.
Is it a bug with python's isalpha() method?
Or, if not, how can I deal properly with characters represented by surrogate pairs?
I've gotten lost in an edge case of sorts. I'm working on a conversion of some old plaintext documentation to reST/Sphinx format, with the intent of outputting to a few formats (including HTML and text) from there. Some of the documented functions are for dealing with bitstrings, and a common case within these is a sentence like the following: Starting character is the blank " " which has the value 0.
I tried writing this as an inline literal the following ways: Starting character is the blank `` `` which has the value 0. or Starting character is the blank :literal:` ` which has the value 0. but there are a few problems with how these end up working:
reST syntax objects to a whitespace immediately inside of the literal, and it doesn't get recognized.
The above can be "fixed"--it looks correct in the HTML () and plaintext (" ") output--with a non-breaking space character inside the literal, but technically this is a lie in our case, and if a user copied this character, they wouldn't be copying what they expect.
The space can be wrapped in regular quotes, which allows the literal to be properly recognized, and while the output in HTML is probably fine (" "), in plaintext it ends up double-quoted as "" "".
In both 2/3 above, if the literal falls on the wrap boundary, the plaintext writer (which uses textwrap) will gladly wrap inside the literal and trim the space because it's at the start/end of the line.
I feel like I'm missing something; is there a good way to handle this?
Try using the unicode character codes. If I understand your question, this should work.
Here is a "|space|" and a non-breaking space (|nbspc|)
.. |space| unicode:: U+0020 .. space
.. |nbspc| unicode:: U+00A0 .. non-breaking space
You should see:
Here is a “ ” and a non-breaking space ( )
I was hoping to get out of this without needing custom code to handle it, but, alas, I haven't found a way to do so. I'll wait a few more days before I accept this answer in case someone has a better idea. The code below isn't complete, nor am I sure it's "done" (will sort out exactly what it should look like during our review process) but the basics are intact.
There are two main components to the approach:
introduce a char role which expects the unicode name of a character as its argument, and which produces an inline description of the character while wrapping the character itself in an inline literal node.
modify the text-wrapper Sphinx uses so that it won't break at the space.
Here's the code:
class TextWrapperDeux(TextWrapper):
_wordsep_re = re.compile(
r'((?<!`)\s+(?!`)|' # whitespace not between backticks
r'(?<=\s)(?::[a-z-]+:)`\S+|' # interpreted text start
r'[^\s\w]*\w+[a-zA-Z]-(?=\w+[a-zA-Z])|' # hyphenated words
r'(?<=[\w\!\"\'\&\.\,\?])-{2,}(?=\w))') # em-dash
#property
def wordsep_re(self):
return self._wordsep_re
def char_role(name, rawtext, text, lineno, inliner, options={}, content=[]):
"""Describe a character given by unicode name.
e.g., :char:`SPACE` -> "char:` `(U+00020 SPACE)"
"""
try:
character = nodes.unicodedata.lookup(text)
except KeyError:
msg = inliner.reporter.error(
':char: argument %s must be valid unicode name at line %d' % (text, lineno))
prb = inliner.problematic(rawtext, rawtext, msg)
return [prb], [msg]
app = inliner.document.settings.env.app
describe_char = "(U+%05X %s)" % (ord(character), text)
char = nodes.inline("char:", "char:", nodes.literal(character, character))
char += nodes.inline(describe_char, describe_char)
return [char], []
def setup(app):
app.add_role('char', char_role)
The code above lacks some glue to actually force the use of the new TextWrapper, imports, etc. When a full version settles out I may try to find a meaningful way to republish it; if so I'll link it here.
Markup: Starting character is the :char:`SPACE` which has the value 0.
It'll produce plaintext output like this: Starting character is the char:` `(U+00020 SPACE) which has the value 0.
And HTML output like: Starting character is the <span>char:<code class="docutils literal"> </code><span>(U+00020 SPACE)</span></span> which has the value 0.
The HTML output ends up looking roughly like: Starting character is the char:(U+00020 SPACE) which has the value 0.
I would like to find a regex that will pick out all commas that fall outside quote sets.
For example:
'foo' => 'bar',
'foofoo' => 'bar,bar'
This would pick out the single comma on line 1, after 'bar',
I don't really care about single vs double quotes.
Has anyone got any thoughts? I feel like this should be possible with readaheads, but my regex fu is too weak.
This will match any string up to and including the first non-quoted ",". Is that what you are wanting?
/^([^"]|"[^"]*")*?(,)/
If you want all of them (and as a counter-example to the guy who said it wasn't possible) you could write:
/(,)(?=(?:[^"]|"[^"]*")*$)/
which will match all of them. Thus
'test, a "comma,", bob, ",sam,",here'.gsub(/(,)(?=(?:[^"]|"[^"]*")*$)/,';')
replaces all the commas not inside quotes with semicolons, and produces:
'test; a "comma,"; bob; ",sam,";here'
If you need it to work across line breaks just add the m (multiline) flag.
The below regexes would match all the comma's which are present outside the double quotes,
,(?=(?:[^"]*"[^"]*")*[^"]*$)
DEMO
OR(PCRE only)
"[^"]*"(*SKIP)(*F)|,
"[^"]*" matches all the double quoted block. That is, in this buz,"bar,foo" input, this regex would match "bar,foo" only. Now the following (*SKIP)(*F) makes the match to fail. Then it moves on to the pattern which was next to | symbol and tries to match characters from the remaining string. That is, in our output , next to pattern | will match only the comma which was just after to buz . Note that this won't match the comma which was present inside double quotes, because we already make the double quoted part to skip.
DEMO
The below regex would match all the comma's which are present inside the double quotes,
,(?!(?:[^"]*"[^"]*")*[^"]*$)
DEMO
While it's possible to hack it with a regex (and I enjoy abusing regexes as much as the next guy), you'll get in trouble sooner or later trying to handle substrings without a more advanced parser. Possible ways to get in trouble include mixed quotes, and escaped quotes.
This function will split a string on commas, but not those commas that are within a single- or double-quoted string. It can be easily extended with additional characters to use as quotes (though character pairs like « » would need a few more lines of code) and will even tell you if you forgot to close a quote in your data:
function splitNotStrings(str){
var parse=[], inString=false, escape=0, end=0
for(var i=0, c; c=str[i]; i++){ // looping over the characters in str
if(c==='\\'){ escape^=1; continue} // 1 when odd number of consecutive \
if(c===','){
if(!inString){
parse.push(str.slice(end, i))
end=i+1
}
}
else if(splitNotStrings.quotes.indexOf(c)>-1 && !escape){
if(c===inString) inString=false
else if(!inString) inString=c
}
escape=0
}
// now we finished parsing, strings should be closed
if(inString) throw SyntaxError('expected matching '+inString)
if(end<i) parse.push(str.slice(end, i))
return parse
}
splitNotStrings.quotes="'\"" // add other (symmetrical) quotes here
Try this regular expression:
(?:"(?:[^\\"]+|\\(?:\\\\)*[\\"])*"|'(?:[^\\']+|\\(?:\\\\)*[\\'])*')\s*=>\s*(?:"(?:[^\\"]+|\\(?:\\\\)*[\\"])*"|'(?:[^\\']+|\\(?:\\\\)*[\\'])*')\s*,
This does also allow strings like “'foo\'bar' => 'bar\\',”.
MarkusQ's answer worked great for me for about a year, until it didn't. I just got a stack overflow error on a line with about 120 commas and 3682 characters total. In Java, like this:
String[] cells = line.split("[\t,](?=(?:[^\"]|\"[^\"]*\")*$)", -1);
Here's my extremely inelegant replacement that doesn't stack overflow:
private String[] extractCellsFromLine(String line) {
List<String> cellList = new ArrayList<String>();
while (true) {
String[] firstCellAndRest;
if (line.startsWith("\"")) {
firstCellAndRest = line.split("([\t,])(?=(?:[^\"]|\"[^\"]*\")*$)", 2);
}
else {
firstCellAndRest = line.split("[\t,]", 2);
}
cellList.add(firstCellAndRest[0]);
if (firstCellAndRest.length == 1) {
break;
}
line = firstCellAndRest[1];
}
return cellList.toArray(new String[cellList.size()]);
}
#SocialCensus, The example you gave in the comment to MarkusQ, where you throw in ' alongside the ", doesn't work with the example MarkusQ gave right above that if we change sam to sam's: (test, a "comma,", bob, ",sam's,",here) has no match against (,)(?=(?:[^"']|["|'][^"']")$). In fact, the problem itself, "I don't really care about single vs double quotes", is ambiguous. You have to be clear what you mean by quoting either with " or with '. For example, is nesting allowed or not? If so, to how many levels? If only 1 nested level, what happens to a comma outside the inner nested quotation but inside the outer nesting quotation? You should also consider that single quotes happen by themselves as apostrophes (ie, like the counter-example I gave earlier with sam's). Finally, the regex you made doesn't really treat single quotes on par with double quotes since it assumes the last type of quotation mark is necessarily a double quote -- and replacing that last double quote with ['|"] also has a problem if the text doesn't come with correct quoting (or if apostrophes are used), though, I suppose we probably could assume all quotes are correctly delineated.
MarkusQ's regexp answers the question: find all commas that have an even number of double quotes after it (ie, are outside double quotes) and disregard all commas that have an odd number of double quotes after it (ie, are inside double quotes). This is generally the same solution as what you probably want, but let's look at a few anomalies. First, if someone leaves off a quotation mark at the end, then this regexp finds all the wrong commas rather than finding the desired ones or failing to match any. Of course, if a double quote is missing, all bets are off since it might not be clear if the missing one belongs at the end or instead belongs at the beginning; however, there is a case that is legitimate and where the regex could conceivably fail (this is the second "anomaly"). If you adjust the regexp to go across text lines, then you should be aware that quoting multiple consecutive paragraphs requires that you place a single double quote at the beginning of each paragraph and leave out the quote at the end of each paragraph except for at the end of the very last paragraph. This means that over the space of those paragraphs, the regex will fail in some places and succeed in others.
Examples and brief discussions of paragraph quoting and of nested quoting can be found here http://en.wikipedia.org/wiki/Quotation_mark .
I am trying to insert spaces into a string of IPA characters, e.g. to turn ɔ̃wɔ̃tɨ into ɔ̃ w ɔ̃ t ɨ. Using split/join was my first thought:
s = ɔ̃w̃ɔtɨ
s.split('').join(' ') #=> ̃ ɔ w ̃ ɔ p t ɨ
As I discovered by examining the results, letters with diacritics are in fact encoded as two characters. After some research I found the UnicodeUtils module, and used the each_grapheme method:
UnicodeUtils.each_grapheme(s) {|g| g + ' '} #=> ɔ ̃w ̃ɔ p t ɨ
This worked fine, except for the inverted breve mark. The code changes ̑a into ̑ a. I tried normalization (UnicodeUtils.nfc, UnicodeUtils.nfd), but to no avail. I don't know why the each_grapheme method has a problem with this particular diacritic mark, but I noticed that in gedit, the breve is also treated as a separate character, as opposed to tildes, accents etc. So my question is as follows: is there a straightforward method of normalization, i.e. turning the combination of Latin Small Letter A and Combining Inverted Breve into Latin Small Letter A With Inverted Breve?
I understand your question concerns Ruby but I suppose the problem is about the same as with Python. A simple solution is to test the combining diacritical marks explicitly :
import unicodedata
liste=[]
s = u"ɔ̃w̃ɔtɨ"
comb=False
prec=u""
for char in s:
if unicodedata.combining(char):
liste.append(prec+char)
prec=""
else:
liste.append(prec)
prec=char
liste.append(prec)
print " ".join(liste)
>>>> ɔ̃ w̃ ɔ t ɨ
I have the following string:
"h3. My Title Goes Here"
I basically want to remove the first four characters from the string so that I just get back:
"My Title Goes Here".
The thing is I am iterating over an array of strings and not all have the h3. part in front so I can't just ditch the first four characters blindly.
I checked the docs and the closest thing I could find was chomp, but that only works for the end of a string.
Right now I am doing this:
"h3. My Title Goes Here".reverse.chomp(" .3h").reverse
This gives me my desired output, but there has to be a better way. I don't want to reverse a string twice for no reason. Is there another method that will work?
To alter the original string, use sub!, e.g.:
my_strings = [ "h3. My Title Goes Here", "No h3. at the start of this line" ]
my_strings.each { |s| s.sub!(/^h3\. /, '') }
To not alter the original and only return the result, remove the exclamation point, i.e. use sub. In the general case you may have regular expressions that you can and want to match more than one instance of, in that case use gsub! and gsub—without the g only the first match is replaced (as you want here, and in any case the ^ can only match once to the start of the string).
You can use sub with a regular expression:
s = 'h3. foo'
s.sub!(/^h[0-9]+\. /, '')
puts s
Output:
foo
The regular expression should be understood as follows:
^ Match from the start of the string.
h A literal "h".
[0-9] A digit from 0-9.
+ One or more of the previous (i.e. one or more digits)
\. A literal period.
A space (yes, spaces are significant by default in regular expressions!)
You can modify the regular expression to suit your needs. See a regular expression tutorial or syntax guide, for example here.
A standard approach would be to use regular expressions:
"h3. My Title Goes Here".gsub /^h3\. /, '' #=> "My Title Goes Here"
gsub means globally substitute and it replaces a pattern by a string, in this case an empty string.
The regular expression is enclosed in / and constitutes of:
^ means beginning of the string
h3 is matched literally, so it means h3
\. - a dot normally means any character so we escape it with a backslash
is matched literally