My objective is to extract strings from numbered/bulleted lists in multiple Microsoft Word documents, then to organize those strings into a single, one-line string where each string is ordered in the following manner: 1.string1 2.string2 3.string3 etc. I refer to these one-line strings as procedures, consisting of 'steps' 1., 2., 3., etc.
The reason it has to be in this format is because the procedure strings are being put into a database, the database is used to create Excel spreadsheet outputs, a formatting macro is used on the spreadsheets, and the procedure strings in question have to be in this format in order for that macro to work properly.
The numbered/bulleted lists in MSword are all similar in format, but some use numbers, some use bullets, and some have extra line spaces before the first point, or extra line spaces after the last point.
The following text shows three different examples of how the Word documents are formatted:
Paragraph Keyword 1: arbitrary text
1. Step 1
2. Step 2
3. Step 3
Paragraph Keyword 2: arbitrary text
Paragraph Keyword 3: arbitrary text
• Step 1
• Step 2
• Step 3
Paragraph Keyword 4: arbitrary text
Paragraph Keyword 5: arbitrary text
Step 1
Step 2
Step 3
Paragraph Keyword 6: arbitrary text
(For some reason the first two lists didn't get indented in the formatting of the post, but in my word document all the indentation is the same)
When the numbered/bulleted list is formatted without line extra spaces, my code works fine, e.g. between "paragraph keyword 1:" and "paragraph keyword 2:".
I was trying to use isspace() to isolate the instances where there are extra line spaces that aren't part of the list that I want to include in my procedure strings.
Here is my code:
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
def extractStrings(file):
doc = file
for i in range(len(doc.paragraphs)):
str1 = doc.paragraphs[i].text
if "Paragraph Keyword 1:" in str1:
start1=i
if "Paragraph Keyword 2:" in str1:
finish1=i
if "Paragraph Keyword 3:" in str1:
start2=i
if "Paragraph Keyword 4:" in str1:
finish2=i
if "Paragraph Keyword 5:" in str1:
start3=i
if "Paragraph Keyword 6:" in str1:
finish3=i
print("----------------------------")
procedure1 = ""
y=1
for x in range(start1 + 1, finish1):
temp = str((doc.paragraphs[x].text))
print(temp)
if not temp.isspace():
if y > 1:
procedure1 = (procedure1 + " " + str(y) + "." + temp)
else:
procedure1 = (procedure1 + str(y) + "." + temp)
y=y+1
print(procedure1)
print("----------------------------")
procedure2 = ""
y=1
for x in range(start2 + 1, finish2):
temp = str((doc.paragraphs[x].text))
print(temp)
if not temp.isspace():
if y > 1:
procedure2 = (procedure2 + " " + str(y) + "." + temp)
else:
procedure2 = (procedure2 + str(y) + "." + temp)
y=y+1
print(procedure2)
print("----------------------------")
procedure3 = ""
y=1
for x in range(start3 + 1, finish3):
temp = str((doc.paragraphs[x].text))
print(temp)
if not temp.isspace():
if y > 1:
procedure3 = (procedure3 + " " + str(y) + "." + temp)
else:
procedure3 = (procedure3 + str(y) + "." + temp)
y=y+1
print(procedure3)
print("----------------------------")
del doc
''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
import docx
doc1 = docx.Document("docx_isspace_experiment_042420.docx")
extractStrings(doc1)
del doc1
Unfortunately I have no way of putting the output into this post, but the problem is that whenever there is a blank line in the word doc, isspace() returns false, and a number "x." is assigned to empty space, so I end up with something like: 1. 2.Step 1 3.Step 2 4.Step 3 5. 6. (that's the last iteration of print(procedure3) from the code)
The problem is that isspace() is returning false even when my python console output shows that the string is just a blank line.
Am I using isspace() incorrectly? Is there something in the string I am not detecting that is causing isspace() to return false? Is there a better way to accomplish this?
Use the test:
# --- for s a str value, like paragraph.text ---
if s.strip() == "":
print("s is a blank line")
str.isspace() returns True if the string contains only whitespace. An empty str contains nothing, and so therefore does not contain whitespace.
I need a regular expression to select all the text between two outer brackets.
Example:
START_TEXT(text here(possible text)text(possible text(more text)))END_TXT
^ ^
Result:
(text here(possible text)text(possible text(more text)))
I want to add this answer for quickreference. Feel free to update.
.NET Regex using balancing groups:
\((?>\((?<c>)|[^()]+|\)(?<-c>))*(?(c)(?!))\)
Where c is used as the depth counter.
Demo at Regexstorm.com
Stack Overflow: Using RegEx to balance match parenthesis
Wes' Puzzling Blog: Matching Balanced Constructs with .NET Regular Expressions
Greg Reinacker's Weblog: Nested Constructs in Regular Expressions
PCRE using a recursive pattern:
\((?:[^)(]+|(?R))*+\)
Demo at regex101; Or without alternation:
\((?:[^)(]*(?R)?)*+\)
Demo at regex101; Or unrolled for performance:
\([^)(]*+(?:(?R)[^)(]*)*+\)
Demo at regex101; The pattern is pasted at (?R) which represents (?0).
Perl, PHP, Notepad++, R: perl=TRUE, Python: PyPI regex module with (?V1) for Perl behaviour.
(the new version of PyPI regex package already defaults to this → DEFAULT_VERSION = VERSION1)
Ruby using subexpression calls:
With Ruby 2.0 \g<0> can be used to call full pattern.
\((?>[^)(]+|\g<0>)*\)
Demo at Rubular; Ruby 1.9 only supports capturing group recursion:
(\((?>[^)(]+|\g<1>)*\))
Demo at Rubular (atomic grouping since Ruby 1.9.3)
JavaScript API :: XRegExp.matchRecursive
XRegExp.matchRecursive(str, '\\(', '\\)', 'g');
Java: An interesting idea using forward references by #jaytea.
Without recursion up to 3 levels of nesting:
(JS, Java and other regex flavors)
To prevent runaway if unbalanced, with * on innermost [)(] only.
\((?:[^)(]|\((?:[^)(]|\((?:[^)(]|\([^)(]*\))*\))*\))*\)
Demo at regex101; Or unrolled for better performance (preferred).
\([^)(]*(?:\([^)(]*(?:\([^)(]*(?:\([^)(]*\)[^)(]*)*\)[^)(]*)*\)[^)(]*)*\)
Demo at regex101; Deeper nesting needs to be added as required.
Reference - What does this regex mean?
RexEgg.com - Recursive Regular Expressions
Regular-Expressions.info - Regular Expression Recursion
Mastering Regular Expressions - Jeffrey E.F. Friedl 1 2 3 4
Regular expressions are the wrong tool for the job because you are dealing with nested structures, i.e. recursion.
But there is a simple algorithm to do this, which I described in more detail in this answer to a previous question. The gist is to write code which scans through the string keeping a counter of the open parentheses which have not yet been matched by a closing parenthesis. When that counter returns to zero, then you know you've reached the final closing parenthesis.
You can use regex recursion:
\(([^()]|(?R))*\)
[^\(]*(\(.*\))[^\)]*
[^\(]* matches everything that isn't an opening bracket at the beginning of the string, (\(.*\)) captures the required substring enclosed in brackets, and [^\)]* matches everything that isn't a closing bracket at the end of the string. Note that this expression does not attempt to match brackets; a simple parser (see dehmann's answer) would be more suitable for that.
This answer explains the theoretical limitation of why regular expressions are not the right tool for this task.
Regular expressions can not do this.
Regular expressions are based on a computing model known as Finite State Automata (FSA). As the name indicates, a FSA can remember only the current state, it has no information about the previous states.
In the above diagram, S1 and S2 are two states where S1 is the starting and final step. So if we try with the string 0110 , the transition goes as follows:
0 1 1 0
-> S1 -> S2 -> S2 -> S2 ->S1
In the above steps, when we are at second S2 i.e. after parsing 01 of 0110, the FSA has no information about the previous 0 in 01 as it can only remember the current state and the next input symbol.
In the above problem, we need to know the no of opening parenthesis; this means it has to be stored at some place. But since FSAs can not do that, a regular expression can not be written.
However, an algorithm can be written to do this task. Algorithms are generally falls under Pushdown Automata (PDA). PDA is one level above of FSA. PDA has an additional stack to store some additional information. PDAs can be used to solve the above problem, because we can 'push' the opening parenthesis in the stack and 'pop' them once we encounter a closing parenthesis. If at the end, stack is empty, then opening parenthesis and closing parenthesis matches. Otherwise not.
(?<=\().*(?=\))
If you want to select text between two matching parentheses, you are out of luck with regular expressions. This is impossible(*).
This regex just returns the text between the first opening and the last closing parentheses in your string.
(*) Unless your regex engine has features like balancing groups or recursion. The number of engines that support such features is slowly growing, but they are still not a commonly available.
It is actually possible to do it using .NET regular expressions, but it is not trivial, so read carefully.
You can read a nice article here. You also may need to read up on .NET regular expressions. You can start reading here.
Angle brackets <> were used because they do not require escaping.
The regular expression looks like this:
<
[^<>]*
(
(
(?<Open><)
[^<>]*
)+
(
(?<Close-Open>>)
[^<>]*
)+
)*
(?(Open)(?!))
>
I was also stuck in this situation when dealing with nested patterns and regular-expressions is the right tool to solve such problems.
/(\((?>[^()]+|(?1))*\))/
This is the definitive regex:
\(
(?<arguments>
(
([^\(\)']*) |
(\([^\(\)']*\)) |
'(.*?)'
)*
)
\)
Example:
input: ( arg1, arg2, arg3, (arg4), '(pip' )
output: arg1, arg2, arg3, (arg4), '(pip'
note that the '(pip' is correctly managed as string.
(tried in regulator: http://sourceforge.net/projects/regulator/)
I have written a little JavaScript library called balanced to help with this task. You can accomplish this by doing
balanced.matches({
source: source,
open: '(',
close: ')'
});
You can even do replacements:
balanced.replacements({
source: source,
open: '(',
close: ')',
replace: function (source, head, tail) {
return head + source + tail;
}
});
Here's a more complex and interactive example JSFiddle.
Adding to bobble bubble's answer, there are other regex flavors where recursive constructs are supported.
Lua
Use %b() (%b{} / %b[] for curly braces / square brackets):
for s in string.gmatch("Extract (a(b)c) and ((d)f(g))", "%b()") do print(s) end (see demo)
Raku (former Perl6):
Non-overlapping multiple balanced parentheses matches:
my regex paren_any { '(' ~ ')' [ <-[()]>+ || <&paren_any> ]* }
say "Extract (a(b)c) and ((d)f(g))" ~~ m:g/<&paren_any>/;
# => (｢(a(b)c)｣ ｢((d)f(g))｣)
Overlapping multiple balanced parentheses matches:
say "Extract (a(b)c) and ((d)f(g))" ~~ m:ov:g/<&paren_any>/;
# => (｢(a(b)c)｣ ｢(b)｣ ｢((d)f(g))｣ ｢(d)｣ ｢(g)｣)
See demo.
Python re non-regex solution
See poke's answer for How to get an expression between balanced parentheses.
Java customizable non-regex solution
Here is a customizable solution allowing single character literal delimiters in Java:
public static List<String> getBalancedSubstrings(String s, Character markStart,
Character markEnd, Boolean includeMarkers)
{
List<String> subTreeList = new ArrayList<String>();
int level = 0;
int lastOpenDelimiter = -1;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == markStart) {
level++;
if (level == 1) {
lastOpenDelimiter = (includeMarkers ? i : i + 1);
}
}
else if (c == markEnd) {
if (level == 1) {
subTreeList.add(s.substring(lastOpenDelimiter, (includeMarkers ? i + 1 : i)));
}
if (level > 0) level--;
}
}
return subTreeList;
}
}
Sample usage:
String s = "some text(text here(possible text)text(possible text(more text)))end text";
List<String> balanced = getBalancedSubstrings(s, '(', ')', true);
System.out.println("Balanced substrings:\n" + balanced);
// => [(text here(possible text)text(possible text(more text)))]
The regular expression using Ruby (version 1.9.3 or above):
/(?<match>\((?:\g<match>|[^()]++)*\))/
Demo on rubular
The answer depends on whether you need to match matching sets of brackets, or merely the first open to the last close in the input text.
If you need to match matching nested brackets, then you need something more than regular expressions. - see #dehmann
If it's just first open to last close see #Zach
Decide what you want to happen with:
abc ( 123 ( foobar ) def ) xyz ) ghij
You need to decide what your code needs to match in this case.
"""
Here is a simple python program showing how to use regular
expressions to write a paren-matching recursive parser.
This parser recognises items enclosed by parens, brackets,
braces and <> symbols, but is adaptable to any set of
open/close patterns. This is where the re package greatly
assists in parsing.
"""
import re
# The pattern below recognises a sequence consisting of:
# 1. Any characters not in the set of open/close strings.
# 2. One of the open/close strings.
# 3. The remainder of the string.
#
# There is no reason the opening pattern can't be the
# same as the closing pattern, so quoted strings can
# be included. However quotes are not ignored inside
# quotes. More logic is needed for that....
pat = re.compile("""
( .*? )
( \( | \) | \[ | \] | \{ | \} | \< | \> |
\' | \" | BEGIN | END | $ )
( .* )
""", re.X)
# The keys to the dictionary below are the opening strings,
# and the values are the corresponding closing strings.
# For example "(" is an opening string and ")" is its
# closing string.
matching = { "(" : ")",
"[" : "]",
"{" : "}",
"<" : ">",
'"' : '"',
"'" : "'",
"BEGIN" : "END" }
# The procedure below matches string s and returns a
# recursive list matching the nesting of the open/close
# patterns in s.
def matchnested(s, term=""):
lst = []
while True:
m = pat.match(s)
if m.group(1) != "":
lst.append(m.group(1))
if m.group(2) == term:
return lst, m.group(3)
if m.group(2) in matching:
item, s = matchnested(m.group(3), matching[m.group(2)])
lst.append(m.group(2))
lst.append(item)
lst.append(matching[m.group(2)])
else:
raise ValueError("After <<%s %s>> expected %s not %s" %
(lst, s, term, m.group(2)))
# Unit test.
if __name__ == "__main__":
for s in ("simple string",
""" "double quote" """,
""" 'single quote' """,
"one'two'three'four'five'six'seven",
"one(two(three(four)five)six)seven",
"one(two(three)four)five(six(seven)eight)nine",
"one(two)three[four]five{six}seven<eight>nine",
"one(two[three{four<five>six}seven]eight)nine",
"oneBEGINtwo(threeBEGINfourENDfive)sixENDseven",
"ERROR testing ((( mismatched ))] parens"):
print "\ninput", s
try:
lst, s = matchnested(s)
print "output", lst
except ValueError as e:
print str(e)
print "done"
You need the first and last parentheses. Use something like this:
str.indexOf('('); - it will give you first occurrence
str.lastIndexOf(')'); - last one
So you need a string between,
String searchedString = str.substring(str1.indexOf('('),str1.lastIndexOf(')');
because js regex doesn't support recursive match, i can't make balanced parentheses matching work.
so this is a simple javascript for loop version that make "method(arg)" string into array
push(number) map(test(a(a()))) bass(wow, abc)
$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)
const parser = str => {
let ops = []
let method, arg
let isMethod = true
let open = []
for (const char of str) {
// skip whitespace
if (char === ' ') continue
// append method or arg string
if (char !== '(' && char !== ')') {
if (isMethod) {
(method ? (method += char) : (method = char))
} else {
(arg ? (arg += char) : (arg = char))
}
}
if (char === '(') {
// nested parenthesis should be a part of arg
if (!isMethod) arg += char
isMethod = false
open.push(char)
} else if (char === ')') {
open.pop()
// check end of arg
if (open.length < 1) {
isMethod = true
ops.push({ method, arg })
method = arg = undefined
} else {
arg += char
}
}
}
return ops
}
// const test = parser(`$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)`)
const test = parser(`push(number) map(test(a(a()))) bass(wow, abc)`)
console.log(test)
the result is like
[ { method: 'push', arg: 'number' },
{ method: 'map', arg: 'test(a(a()))' },
{ method: 'bass', arg: 'wow,abc' } ]
[ { method: '$$', arg: 'groups' },
{ method: 'filter',
arg: '{type:\'ORGANIZATION\',isDisabled:{$ne:true}}' },
{ method: 'pickBy', arg: '_id,type' },
{ method: 'map', arg: 'test()' },
{ method: 'as', arg: 'groups' } ]
While so many answers mention this in some form by saying that regex does not support recursive matching and so on, the primary reason for this lies in the roots of the Theory of Computation.
Language of the form {a^nb^n | n>=0} is not regular. Regex can only match things that form part of the regular set of languages.
Read more # here
I didn't use regex since it is difficult to deal with nested code. So this snippet should be able to allow you to grab sections of code with balanced brackets:
def extract_code(data):
""" returns an array of code snippets from a string (data)"""
start_pos = None
end_pos = None
count_open = 0
count_close = 0
code_snippets = []
for i,v in enumerate(data):
if v =='{':
count_open+=1
if not start_pos:
start_pos= i
if v=='}':
count_close +=1
if count_open == count_close and not end_pos:
end_pos = i+1
if start_pos and end_pos:
code_snippets.append((start_pos,end_pos))
start_pos = None
end_pos = None
return code_snippets
I used this to extract code snippets from a text file.
This do not fully address the OP question but I though it may be useful to some coming here to search for nested structure regexp:
Parse parmeters from function string (with nested structures) in javascript
Match structures like:
matches brackets, square brackets, parentheses, single and double quotes
Here you can see generated regexp in action
/**
* get param content of function string.
* only params string should be provided without parentheses
* WORK even if some/all params are not set
* #return [param1, param2, param3]
*/
exports.getParamsSAFE = (str, nbParams = 3) => {
const nextParamReg = /^\s*((?:(?:['"([{](?:[^'"()[\]{}]*?|['"([{](?:[^'"()[\]{}]*?|['"([{][^'"()[\]{}]*?['")}\]])*?['")}\]])*?['")}\]])|[^,])*?)\s*(?:,|$)/;
const params = [];
while (str.length) { // this is to avoid a BIG performance issue in javascript regexp engine
str = str.replace(nextParamReg, (full, p1) => {
params.push(p1);
return '';
});
}
return params;
};
This might help to match balanced parenthesis.
\s*\w+[(][^+]*[)]\s*
This one also worked
re.findall(r'\(.+\)', s)
So, I parsed HTML code from FIFA worldcup website, and want to get all the matches:
wcup <- htmlTreeParse("http://www.fifa.com/worldcup/matches/", useInternalNodes=T)
However, the field for one country is 't-nText kern' and for the rest of countries is 't-nText '.
<span class="t-nText kern">Bosnia and Herzegovina</span>
Therefore, if I use this command, I will miss 'Bosnia and Herzegovina', like this command:
xpathSApply(wcup, "//span[#class='t-nText ']", xmlValue)
So, is there any way that I can search for both attributes 't-nText ' and 't-nText kern' at the same time? Or do you have any other solution? I want to keep the order of the matches as is.
xpath doesn't support logical OR:
xpathSApply(wcup, "//span[#class='t-nText ' || 't-nText kern']", xmlValue)
XPath error : Invalid expression
//span[#class='t-nText ' || 't-nText kern']
^
XPath error : Invalid expression
//span[#class='t-nText ' || 't-nText kern']
^
Error in xpathApply.XMLInternalDocument(doc, path, fun, ..., namespaces = namespaces, :
error evaluating xpath expression //span[#class='t-nText ' || 't-nText kern']
Use 'or' or perhaps 'starts-with()',
wcup["//span[#class='t-nText kern' or #class='t-nText ']"]
wcup["//span[starts-with(#class, 't-nText ')]"]
I originally posted this ,, then noticed order was needed, so I searched SO for "XPath OR"
Why not just append the results of the two searches together:
c( xpathSApply(wcup, "//span[#class='t-nText kern']", xmlValue),
xpathSApply(wcup, "//span[#class='t-nText ']", xmlValue)
)
Lo and behold I came up with:
xpathSApply(wcup, "//*[starts-with(#class,'t-nText')]", xmlValue)
Which appears mighty similar to Martin Morgan's solution. I had not realized that XPath was it's own language. Guess I'm at least 10 years behind the times.
Is it possible for MySQL database to generate a 5 or 6 digit code comprised of only numbers and letters when I insert a record? If so how?
Just like goo.gl, bit.ly and jsfiddle do it. For exaple:
http://bit.ly/3PKQcJ
http://jsfiddle.net/XzKvP
cZ6ahF, 3t5mM, xGNPN, xswUdS...
So UUID_SHORT() will not work because it returns a value like 23043966240817183
Requirements:
Must be unique (non-repeating)
Can be but not required to be based off of primary key integer value
Must scale (grow by one character when all possible combinations have been used)
Must look random. (item 1234 cannot be BCDE while item 1235 be BCDF)
Must be generated on insert.
Would greatly appreciate code examples.
Try this:
SELECT LEFT(UUID(), 6);
I recommend using Redis for this task, actually. It has all the features that make this task suitable for its use. Foremost, it is very good at searching a big list for a value.
We will create two lists, buffered_ids, and used_ids. A cronjob will run every 5 minutes (or whatever interval you like), which will check the length of buffered_ids and keep it above, say, 5000 in length. When you need to use an id, pop it from buffered_ids and add it to used_ids.
Redis has sets, which are unique items in a collection. Think of it as a hash where the keys are unique and all the values are "true".
Your cronjob, in bash:
log(){ local x=$1 n=2 l=-1;if [ "$2" != "" ];then n=$x;x=$2;fi;while((x));do let l+=1 x/=n;done;echo $l; }
scale=`redis-cli SCARD used_ids`
scale=`log 16 $scale`
scale=$[ scale + 6]
while [ `redis-cli SCARD buffered_ids` -lt 5000 ]; do
uuid=`cat /dev/urandom | tr -cd "[:alnum:]" | head -c ${1:-$scale}`
if [ `redis-cli SISMEMBER used_ids $uuid` == 1]; then
continue
fi
redis-cli SADD buffered_ids $uuid
done
To grab the next uid for use in your application (in pseudocode because you did not specify a language)
$uid = redis('SPOP buffered_ids');
redis('SADD used_ids ' . $uid);
edit actually there's a race condition there. To safely pop a value, add it to used_ids first, then remove it from buffered_ids.
$uid = redis('SRANDMEMBER buffered_ids');
redis('SADD used_ids ' . $uid);
redis('SREM buffered_ids ' . $uid);