For a fun exercise I wondered if I could tokenize simple arithmetic expressions (containing only positive integers and the four basic operations) using a regular expression, so I came up with the following:
But the test cases below do not behave as I expected due to the failures listed at the end (Go Playground):
func TestParseCalcExpression(t *testing.T) {
re := regexp.MustCompile(`^(\d+)(?:([*/+-])(\d+))*$`)
for _, eg := range []struct {
input string
expected [][]string
}{
{"1", [][]string{{"1", "1", "", ""}}},
{"1+1", [][]string{{"1+1", "1", "+", "1"}}},
{"22/7", [][]string{{"22/7", "22", "/", "7"}}},
{"1+2+3", [][]string{{"1+2+3", "1", "+", "2", "+", "3"}}},
{"2*3+5/6", [][]string{{"2*3+5/6", "2", "*", "3", "+", "5", "/", "6"}}},
} {
actual := re.FindAllStringSubmatch(eg.input, -1)
if !reflect.DeepEqual(actual, eg.expected) {
t.Errorf("expected parse(%q)=%#v, got %#v", eg.input, eg.expected, actual)
}
}
}
// === RUN TestParseCalcExpression
// prog.go:24: expected parse("1+2+3")=[][]string{[]string{"1+2+3", "1", "+", "2", "+", "3"}}, got [][]string{[]string{"1+2+3", "1", "+", "3"}}
// prog.go:24: expected parse("2*3+5/6")=[][]string{[]string{"2*3+5/6", "2", "*", "3", "+", "5", "/", "6"}}, got [][]string{[]string{"2*3+5/6", "2", "/", "6"}}
// --- FAIL: TestParseCalcExpression (0.00s)
// FAIL
I was hoping that the "zero or more repetition" of the non-matching subgroup ((?:...)*) which identifies and groups operators and numbers (([*/+-])(\d+)) would match all occurrences of that sub-expression but it only appears to match the last one.
On the one hand, this makes sense because the regex literally has only three matching groups, so it follows that any resulting match could only have three matches. However, the "zero or more repetition" makes it seem like it's missing all the "middle" repeated items in the failed tests (e.g. +2 in 1+2+3).
// expected parse("1+2+3")=
// [][]string{[]string{"1+2+3", "1", "+", "2", "+", "3"}},
// got [][]string{[]string{"1+2+3", "1", "+", "3"}}
Is there a way to parse these kinds of arithmetic expressions using go regular expressions or is this a fundamental limitation of regular expressions (or go/re2 regexps, or the general combination of non/capturing groups)?
(I realize I could just split by word boundaries and scan the tokens to validate the structure but I'm more interested in this limitation of non/capturing groups than the example problem.)
package main
import (
"reflect"
"regexp"
"testing"
)
func TestParseCalcExpression(t *testing.T) {
re := regexp.MustCompile(`(\d+)([*/+-]?)`)
for _, eg := range []struct {
input string
expected [][]string
}{
{"1", [][]string{{"1", "1", ""}}},
{"1+1", [][]string{{"1+", "1", "+"}, {"1", "1", ""}}},
{"22/7", [][]string{{"22/", "22", "/"}, {"7", "7", ""}}},
{"1+2+3", [][]string{{"1+", "1", "+"}, {"2+", "2", "+"}, {"3", "3", ""}}},
{"2*3+5/6", [][]string{{"2*", "2", "*"}, {"3+", "3", "+"}, {"5/", "5", "/"}, {"6", "6", ""}}},
} {
actual := re.FindAllStringSubmatch(eg.input, -1)
if !reflect.DeepEqual(actual, eg.expected) {
t.Errorf("expected parse(%q)=%#v, got %#v", eg.input, eg.expected, actual)
}
}
}
Playground link
As mentioned in this question about Swift (I'm not a Swift or regex expert so I'm just guessing this applies to Go as well), you can only return one match for each matching group in your regex. It seems to just identify the last match if the group is repeating.
From the Go standard library regexp package documentation:
If 'Submatch' is present, the return value is a slice identifying the successive submatches of the expression. Submatches are matches of parenthesized subexpressions (also known as capturing groups) within the regular expression, numbered from left to right in order of opening parenthesis. Submatch 0 is the match of the entire expression, submatch 1 the match of the first parenthesized subexpression, and so on.
Given this convention, returning multiple matches per match group would break the numbering and therefore you wouldn't know which items were associated with each matching group. It seems it's possible that a regex engine could return multiple matches per group, but this package couldn't do that without breaking this convention stated in the documentation.
My solution is to make your problem more regular. Instead of treating the entire expression as one match, which gave us the problem that we can only return finitely many strings per match, we treat the entire expression as simply a series of pairs.
Each pair is composed of a number (\d+), and an optional operator ([*/+-]?).
Then doing a FindAllStringSubmatch on the whole expression, we extract a series of these pairs and get the number and operator for each.
For example:
"1+2+3"
returns
[][]string{{"1+", "1", "+"}, {"2+", "2", "+"}, {"3", "3", ""}}}
This only tokenizes the expression; it doesn't validate it. If you need the expression to be validated, then you'll need another initial regex match to verify that the string is indeed an unbroken series of these pairs.
Related
I am checking in my validation form if there are repeated more than two times the same character.
I have tried this expression ([a-zA-Z0-9])\1{2,} but it doesn't work properly because if I add aaA it founds the string and it shouldn't because "aaA" is permitted. Also it doesn't check if it is repeated a special character.
Here is how I applied my code:
this.form = this.formBuilder.group(
{
newpassword: new FormControl(
'',
Validators.compose([
Validators.required,
CustomValidators.patternValidator(/[(\[a-zA-Z0-9\])\\1{2,}]/, {
hasRepeatedCharacters: true,
}),
])
),
},
{ validators: this.password }
);
Any idea?
If I understand correctly what you are considering to be invalid, you want this:
/(.)\1{2,}/
Use the following regex to detect any character repeated 2 or more times:
(.)\1{2,}
In order to capture aaA (repeated letters irrespective of their case) as well, you'll need to add the case-insensitive i flag.
You can use /(.)(?=\1.*\1)/, assuming you allow the repeated characters to be non-consecutive:
const pat = /(.)(?=.*\1.*\1)/;
[
"a",
"aa",
"aaa",
"zba1a1za",
"aaA",
"aaAA",
"aAaAa",
"aAbbAb",
].forEach(e => console.log(`'${e}' => ${pat.test(e)}`));
I have a code snippet like below
[ "sortBy", "String", "sort by method" ],
[ "sortOrder", "String", "sort order includes ascend and descend" ],
[ "count", "Int", "The number of results to return." ],
[ "names", "Array<String>", "array of strings represents name" ]
I want to use regular expression to match and replace and align so that the result would be look like this:
{ Name = "sortBy"; Ref = "String"; Description = Some "sort by method" }
{ Name = "sortOrder"; Ref = "String"; Description = Some "sort order includes ascend and descend" }
{ Name = "count"; Ref = "Int"; Description = Some "The number of results to return." }
{ Name = "names"; Ref = "Array<String>"; Description = Some "array of strings represents name" }
and each column should be aligned. I am stuck at the beginning how to group match it and align the result. My search is this
*\[ *"(.*)", *"(.*)", *"(.*)" *\],
in visual studio code but it only match the first row. Instead I want to to match all rows at once and replace it and then align it.
The point here is to match and capture only the parts you need to keep, and just match other parts.
You may use
^( *)\[( *)(".*?"),( *)(".*?"),( *)(".*?" *)\],?$
Replace with $1{$2Name = $3;$4Ref = $5;$6Description = Some $7}.
See the regex demo
Details
^ - start of line
( *) - Group 1 ($1): leading spaces
\[ - a [ char (will be replaced with {)
( *) - Group 2 ($2): spaces after [
(".*?") - Group 3 ($3): "..." substring
, - a comma (will be replaced with ;)
( *) - Group 4 ($4): spaces after the first ,
(".*?") - Group 5 ($5): "..." substring
, - a comma (will be replaced with ;)
( *) - Group 6 ($6): spaces after the second ,
(".*?" *) - Group 7 ($7): "..." substring and 0+ spaces after
\],?$ - ], an optional , and end of line.
Here is an answer using a macro extension. Because you need to run two separate regex's (although the second regex is very simple). First a demo with your original text first, some badly formatted text second and your desired results last:
Select your text first and then trigger the macro. I am using alt+r as the keybinding but you can choose whatever you want.
Using the macro extension multi-command put this into your settings.json:
"multiCommand.commands": [
{
"command": "multiCommand.insertAlignRows",
"sequence": [
"editor.action.insertCursorAtEndOfEachLineSelected",
"cursorHomeSelect",
{
"command": "editor.action.insertSnippet",
"args": {
"snippet": "${TM_SELECTED_TEXT/^(\\s*)\\[\\s*(.{12})\\s*(.{18})\\s*([^\\]]*)\\],?/$1{ Name = $2 Ref = $3Description = Some $4}/g}",
}
},
"cursorHomeSelect",
{
"command": "editor.action.insertSnippet",
"args": {
"snippet": "${TM_SELECTED_TEXT/,/;/g}",
}
},
]
}
]
In keybindings.json:
{
"key": "alt+r", // choose whatever keybinding you want
"command": "extension.multiCommand.execute",
"args": { "command": "multiCommand.insertAlignRows" },
"when": "editorTextFocus"
},
The regex that is doing almost all of the work is:
^(\s*)\[\s*(.{12})\s*(.{18})\s*([^\]]*)\],?
I removed the double escapes necessary in snippets but not in the find/replace widget so you could just use this regex in your find input (and not do the macro at all) and
$1{ Name = $2 Ref = $3Description = Some $4}
in the replace field. And then just replace , with ; after that.
Back to that regex: ^(\s*)\[\s*(.{12})\s*(.{18})\s*([^\]]*)\],? which looks brittle because of the "magic numbers" 12 and 18 derived from your sample text. But it isn't as bad as it first seems as the demo with the bad original formatting shows. They are just counting characters and as long as your input is reasonably close to what you presented it'll work.
The 12 can actually be from 12-16, with the 12 being the length of your longest first item (like "sortOrder",) and the 16 being the minimum number from the beginning of the first items to where the second items (like "String") begin.
Likewise the 18 could be 17-24 given your input and where you want the final column to start. Play with the numbers, it is pretty easy in regex101 demo.
I think the only restriction is that your input not look like this:
[ "names", "Array<String>", "array of strings represents name" ]
[ "sortOrder","String", "sort order includes ascend and descend" ],
where a later column starts before the end of the previous column - as in column 3 starts before all the column 2's end. Likewise for some column 2 item starting before all the column 1 items have ended like
[ "sortOrder", "String", "sort order includes ascend and descend" ],
[ "names", "Array<String>", "array of strings represents name" ]
If your input is that bad you could fix it first with some simple regex's.
Remember you can also adjust where the columns start in your replace by adding/subtracting spaces, as between the $2 Ref in my example above or $3Description - you can add space(s) after the $3 if you wish.
How to regexp_replace the phone_num & phone_ext with only numeric instead of characters.
[ {
"phone_type":"HOME",
"phone_num":"(+1)123-456-7890",
"phone_ext":"-85254-",
"phone_status":"Y",
},
{
"phone_type":"HOME",
"phone_num":"+001-123-456-7890",
"phone_ext":"85-254",
"phone_status":"N",
}
]
should be displayed as
[ {
"phone_type":"HOME",
"phone_num":"11234567890",
"phone_ext":"85254",
"phone_status":"Y",
},
{
"phone_type":"HOME",
"phone_num":"0011234567890",
"phone_ext":"85254",
"phone_status":"N",
}
]
Well, finding the text is fairly easy.
/phone_(num|ext)"\s*:\s*"([^"]*)",/gmi
Next part is finding the second grouping ([^"]*) within your match function and strip all none numeric characters. This will vary by application.
I have a problem with counting words
I want to count word in projects.log.subject.
ex) count [A],[B],[C]..
I searched how to use map reduce.. but I don't understand how to use it for result i want.
{
"_id": ObjectID("569f3a3e9d2540764d8bde59"),
"A": "book",
"server": "us",
"projects": [
{
"domainArray": [
{
~~~~
}
],
"log": [
{
~~~~~,
"subject": "[A][B]I WANT THIS"
}
],
"before": "234234234"
},
{
"domainArray": [
{
~~~~
}
],
"log": [
{
~~~~~,
"subject": "[B][C]I WANT THIS"
}
],
"before": "234234234"
},....
] //end of projects
}//end of document
This is a basic principle of using regular expressions and testing each string against the source string and emitting the found count for the result. In mapReduce terms, you want your "mapper" function to possibly emit multiple values for each "term" as a key, and for every array element present in each document.
So you basically want a source array of regular expressions to process ( likely just a word list ) to iterate and test and also iterate each array member.
Basically something like this:
db.collection.mapReduce(
function() {
var list = ["the", "quick", "brown" ]; // words you want to count
this.projects.forEach(function(project) {
project.log.forEach(function(log) {
list.forEach(function(word) {
var res = log.subject.match(new RegExp("\\b" + word + "\\b","ig"));
if ( res != null )
emit(word,res.length); // returns number of matches for word
});
});
});
},
function(key,values) {
return Array.sum(values);
},
{ "out": { "inline": 1 } }
)
So the loop processes the array elements in the document and then applies each word to look for with a regular expression to test. The .match() method will return an array of matches in the string or null if done was found. Note the i and g options for the regex in order to search case insensitive and beyond just the first match. You might need m for multi-line if your text includes line break characters as well.
If null is not returned, then we emit the current word as the "key" and the count as the length of the matched array.
The reducer then takes all output values from those emit calls in the mapper and simply adds up the emitted counts.
The result will be one document keyed by each "word/term" provided and the count of total occurances in the inspected field within the collection. For more fields, just add more logic to sum up the results, or similarly just keep "emitting" in the mapper and let the reducer do the work.
Note the "\\b" represents a word boundary expression to wrap each term escaped by` in order to construct the expression from strings. You need these to discriminate "the" from "then" for example, by specifying where the word/term ends.
Also that as regular expressions, characters like [] are reserved, so if you actually were looking for strings like that the you similarly escape, i.e:
"\[A\]"
But if you were actually doing that, then remove the word boundary characters:
new RegExp( "\[A\]", "ig" )
As that is enough of a complete match in itself.
I am extracting info from a string using regex.
auto version = { // comments shows the expected output
// version // output : (year, month, sp#, patch#)
"2012.12", // "2012", "12", "", ""
"2012.12-1", // "2012", "12", "", "1"
"2012.12-SP1", // "2012", "12", "SP1", ""
"2012.12-SP2-1", // "2012", "12", "SP2", "1"
"I-2013.12-2", // "2013", "12", "", "2"
"J-2014.09", // "2014", "09", "", ""
"J-2014.09-SP2-1", // "2014", "09", "SP2", "1"
};
The regex I have is the following:
// J - 2014 . 09 - SP2 - 1
std::regex regexExpr("[A-Z]?-?([0-9]{4})\\.([0-9]{2})-?(SP[1-9])?-?([1-9])?.*");
and this seems to work well. I am not very confident about this since I don't have much expertise in regex. Is the regex right and can this be improved?
You can just use \w{2,}|\d as your regex that match any combinations of word characters with length 2 or more (\w{2,})(to avoid of matching the j at the beginning of some strings) or a digit with length 1 (\d)(for match the 1 at the end of some strings)!
Demo
You can use sub_match class template for this aim:
The class template sub_match is used by the regular expression engine to denote sequences of characters matched by marked sub-expressions. A match is a [begin, end) pair within the target range matched by the regular expression, but with additional observer functions to enhance code clarity.