Trying to replace equal sign characters in a string with an escaped equal sign character in coffee script (for use with calling pip via Fabric).
== should become \=\=
coffee> "==".replace /\=\=/, "=="
'=='
coffee> "==".replace /\=\=/, "\=\="
'=='
coffee> "==".replace /\=\=/, "\\=\\="
'\\=\\='
How can I do this?
That last one is probably correct. It just prints funny. Try x.length to confirm (should be four).
coffee> "==".replace(/\=\=/, "\\=\\=").length
4
Related
Using q’s like function, how can we achieve the following match using a single regex string regstr?
q) ("foo7"; "foo8"; "foo9"; "foo10"; "foo11"; "foo12"; "foo13") like regstr
>>> 0111110b
That is, like regstr matches the foo-strings which end in the numbers 8,9,10,11,12.
Using regstr:"foo[8-12]" confuses the square brackets (how does it interpret this?) since 12 is not a single digit, while regstr:"foo[1[0-2]|[1-9]]" returns a type error, even without the foo-string complication.
As the other comments and answers mentioned, this can't be done using a single regex. Another alternative method is to construct the list of strings that you want to compare against:
q)str:("foo7";"foo8";"foo9";"foo10";"foo11";"foo12";"foo13")
q)match:{x in y,/:string z[0]+til 1+neg(-/)z}
q)match[str;"foo";8 12]
0111110b
If your eventual goal is to filter on the matching entries, you can replace in with inter:
q)match:{x inter y,/:string z[0]+til 1+neg(-/)z}
q)match[str;"foo";8 12]
"foo8"
"foo9"
"foo10"
"foo11"
"foo12"
A variation on Cillian’s method: test the prefix and numbers separately.
q)range:{x+til 1+y-x}.
q)s:"foo",/:string 82,range 7 13 / include "foo82" in tests
q)match:{min(x~/:;in[;string range y]')#'flip count[x]cut'z}
q)match["foo";8 12;] s
00111110b
Note how unary derived functions x~/: and in[;string range y]' are paired by #' to the split strings, then min used to AND the result:
q)flip 3 cut's
"foo" "foo" "foo" "foo" "foo" "foo" "foo" "foo"
"82" ,"7" ,"8" ,"9" "10" "11" "12" "13"
q)("foo"~/:;in[;string range 8 12]')#'flip 3 cut's
11111111b
00111110b
Compositions rock.
As the comments state, regex in kdb+ is extremely limited. If the number of trailing digits is known like in the example above then the following can be used to check multiple patterns
q)str:("foo7"; "foo8"; "foo9"; "foo10"; "foo11"; "foo12"; "foo13"; "foo3x"; "foo123")
q)any str like/:("foo[0-9]";"foo[0-9][0-9]")
111111100b
Checking for a range like 8-12 is not currently possible within kdb+ regex. One possible workaround is to write a function to implement this logic. The function range checks a list of strings start with a passed string and end with a number within the range specified.
range:{
/ checking for strings starting with string y
s:((c:count y)#'x)like y;
/ convert remainder of string to long, check if within range
d:("J"$c _'x)within z;
/ find strings satisfying both conditions
s&d
}
Example use:
q)range[str;"foo";8 12]
011111000b
q)str where range[str;"foo";8 12]
"foo8"
"foo9"
"foo10"
"foo11"
"foo12"
This could be made more efficient by checking the trailing digits only on the subset of strings starting with "foo".
For your example you can pad, fill with a char, and then simple regex works fine:
("."^5$("foo7";"foo8";"foo9";"foo10";"foo11";"foo12";"foo13")) like "foo[1|8-9][.|0-2]"
Lets examine the following perl code
if ($a lt 0.00 or $a gt 100.000)
{
print "a must be between 0 and 100 \n";
exit 1
}
exit 0
Lets say a equals 5. The above code will exit with failure status because a isn't between 0 and 100.
Simply replacing the lt and gt with the actual operators they represent, < and > respectively, yields the expected results. Replacing the 100 with a number starting with a 9 will also yield the expected result.
Why are Perl's comparison operators telling me 5 is not between 0 and 100?
lt and gt are string operators, with numbers you want to use plain old < and >. Perl is polymorphic on values, so it's monomorphic on operators (unlike for example python which is the other way around).
In perl the lt and gt operators are not the same as the < and >. the perl documentation details this here perlop under rational operators, below is extracted from the documentation:
Binary "<" returns true if the left argument is numerically less than the right argument.
Binary ">" returns true if the left argument is numerically greater than the right argument.
Binary "<=" returns true if the left argument is numerically less than or equal to the right argument.
Binary ">=" returns true if the left argument is numerically greater than or equal to the right argument.
Binary "lt" returns true if the left argument is stringwise less than the right argument.
Binary "gt" returns true if the left argument is stringwise greater than the right argument.
Binary "le" returns true if the left argument is stringwise less than or equal to the right argument.
Binary "ge" returns true if the left argument is stringwise greater than or equal to the right argument.
Since perl does not have a string object and integer object perl has to make a guess at the context of the object. The only way perl can know if you you are comparing a string or an integer is by ensuring that the rational operators for lt and gt force the context for comparisons as a sting and that < and > operators for the context for comparisons as integers
I want to implement a Scala-style string interpolation in Scala. Here is an example,
val str = "hello ${var1} world ${var2}"
At runtime I want to replace "${var1}" and "${var2}" with some runtime strings. However, when trying to use Regex.replaceAllIn(target: CharSequence, replacer: (Match) ⇒ String), I ran into the following problem:
import scala.util.matching.Regex
val placeholder = new Regex("""(\$\{\w+\})""")
placeholder.replaceAllIn(str, m => s"A${m.matched}B")
java.lang.IllegalArgumentException: No group with name {var1}
at java.util.regex.Matcher.appendReplacement(Matcher.java:800)
at scala.util.matching.Regex$Replacement$class.replace(Regex.scala:722)
at scala.util.matching.Regex$MatchIterator$$anon$1.replace(Regex.scala:700)
at scala.util.matching.Regex$$anonfun$replaceAllIn$1.apply(Regex.scala:410)
at scala.util.matching.Regex$$anonfun$replaceAllIn$1.apply(Regex.scala:410)
at scala.collection.Iterator$class.foreach(Iterator.scala:743)
at scala.collection.AbstractIterator.foreach(Iterator.scala:1174)
at scala.util.matching.Regex.replaceAllIn(Regex.scala:410)
... 32 elided
However, when I removed '$' from the regular expression, it worked:
val placeholder = new Regex("""(\{\w+\})""")
placeholder.replaceAllIn(str, m => s"A${m.matched}B")
res2: String = hello $A{var1}B world $A{var2}B
So my question is that whether this is a bug in Scala Regex. And if so, are there other elegant ways to achieve the same goal (other than brutal force replaceAllLiterally on all placeholders)?
$ is a treated specially in the replacement string. This is described in the documentation of replaceAllIn:
In the replacement String, a dollar sign ($) followed by a number will be interpreted as a reference to a group in the matched pattern, with numbers 1 through 9 corresponding to the first nine groups, and 0 standing for the whole match. Any other character is an error. The backslash (\) character will be interpreted as an escape character and can be used to escape the dollar sign. Use Regex.quoteReplacement to escape these characters.
(Actually, that doesn't mention named group references, so I guess it's only sort of documented.)
Anyway, the takeaway here is that you need to escape the $ characters in the replacement string if you don't want them to be treated as references.
new scala.util.matching.Regex("""(\$\{\w+\})""")
.replaceAllIn("hello ${var1} world ${var2}", m => s"A\\${m.matched}B")
// "hello A${var1}B world A${var2}B"
It's hard to tell what you're expecting the behavior to do. The issue is that s"${m.matched}" is turning into "${var1}" (and "${var2}"). The '$' is special character to say "place the group with name {var1} here instead".
For example:
scala> placeholder.replaceAllIn(str, m => "$1")
res0: String = hello ${var1} world ${var2}
It replaces the match with the first capturing group (which is m itself).
It's hard to tell exactly what you're doing, but you could escape any $ like so:
scala> placeholder.replaceAllIn(str, m => s"${m.matched.replace("$","\\$")}")
res1: String = hello ${var1} world ${var2}
If what you really want to do is evaluate var1/var2 for some variables in the local scope of the method; that's not possible. In fact, the s"Hello, $name" pattern is actually converted into new StringContext("Hello, ", "").s(name) at compile time.
I am trying to compare characters to see if they match. I can't figure out why it doesn't work. I'm expecting true on the output, but I'm getting false.
character: "a"
word: "aardvark"
(first word) = character ; expecting true, getting false
So "a" in Rebol is not a character, it is actually a string.
A single unicode character is its own independent type, with its own literal syntax, e.g. #"a". For example, it can be converted back and forth from INTEGER! to get a code point, which the single-letter string "a" cannot:
>> to integer! #"a"
== 97
>> to integer! "a"
** Script error: cannot MAKE/TO integer! from: "a"
** Where: to
** Near: to integer! "a"
A string is not a series of one-character STRING!s, it's a series of CHAR!. So what you want is therefore:
character: #"a"
word: "aardvark"
(first word) = character ;-- true!
(Note: Interestingly, binary conversions of both a single character string and that character will be equivalent:
>> to binary! "μ"
== #{CEBC}
>> to binary! #"μ"
== #{CEBC}
...those are UTF-8 byte representations.)
I recommend for cases like this, when things start to behave in a different way than you expected, to use things like probe and type?. This will help you get a sense of what's going on, and you can use the interactive Rebol console on small pieces of code.
For instance:
>> character: "a"
>> word: "aardvark"
>> type? first word
== char!
>> type? character
== string!
So you can indeed see that the first element of word is a character #"a", while your character is the string! "a". (Although I agree with #HostileFork that comparing a string of length 1 and a character is for a human the same.)
Other places you can test things are http://tryrebol.esperconsultancy.nl or in the chat room with RebolBot
Just puzzling to me.
Related, but different question:
What does “0 but true” mean in Perl?
Perl doesn't distinguish kinds of numbers. Looking at all of those with a non-CS/programmer eye, they all mean the same thing to me as well: zero. (This is one of the foundations of Perl: it tries to work like people, not like computers. "If it looks like a duck....")
So, if you use them as numbers, they're all the same thing. If you use them as strings, they differ. This does lead to situations where you may need to force one interpretation ("0 but true"; see also "nancy typing"). but by and large it "does the right thing" automatically.
I don't understand, what else should they mean?
You give integer, scientific, floating point, signed integers and octal notations of zero. Why should they differ?
0==0 as everyone, including Larry Wall, knows.
Perl interprets every scalar value as both a string and (potentially) a number. All of those string representations of zero can convert to the integer value 0 , according to perl's conversion rules:
"0", "0.0", "-0", "+0", "000" => Simplest case of straight string to numeric conversion.
"0e0" => In a numeric context, only the leading valid numeric characters are converted, so only the leading "0" is used. For example, "1984abcdef2112" would be interpreted numerically as 1984.
"0 but true" in perl means that a string like "0e0" will evalutate numerically to 0, but in a boolean context will be "true" because the conversion to boolean follows different rules than the strict numeric conversion.
Perl works in contexts. In string context, they are all different. In numeric context, they are all zero.
print "same string\n" if '0' eq '0.0';
print "same number\n" if 0 == 0.0;
'0 but true' in boolean context is true:
print "boolean context\n" if '0 but true';
print "string context\n" if '0 but true' eq '0';
print "numeric context\n" if '0 but true' == 0;