I apply a function, but looks so bad.
function find_without_pattern(s1,s2)
for i =1,#s1-#s2+1 do
local t = string.sub(s1,i,#s2+i-1)
if t == s2 then
return i,i+#s2-1
end
end
end
The string.find method provides an optional 4th parameter to enforce a plaintext search by itself.
For example:
string.find("he#.*o", "e#.*o", 1, true)
will give you the correct results.
Quoting the Lua manual pages:
string.find (s, pattern [, init [, plain]])
A value of true as a fourth, optional argument plain turns off the pattern matching facilities, so the function does a plain "find substring" operation, with no characters in pattern being considered magic. Note that if plain is given, then init must be given as well.
Related
I can use occursin function, but its haystack argument cannot be a regular expression, which means I have to pass the entire alphanumeric string to it. Is there a neat way of doing this in Julia?
I'm not sure your assumption about occursin is correct:
julia> occursin(r"[a-zA-z]", "ABC123")
true
julia> occursin(r"[a-zA-z]", "123")
false
but its haystack argument cannot be a regular expression, which means I have to pass the entire alphanumeric string to it.
If you mean its needle argument, it can be a Regex, for eg.:
julia> occursin(r"^[[:alnum:]]*$", "adf24asg24y")
true
julia> occursin(r"^[[:alnum:]]*$", "adf24asg2_4y")
false
This checks that the given haystack string is alphanumeric using Unicode-aware character class
[[:alnum:]] which you can think of as equivalent to [a-zA-Z\d], extended to non-English characters too. (As always with Unicode, a "perfect" solution involves more work and complication, but this takes you most of the way.)
If you do mean you want the haystack argument to be a Regex, it's not clear why you'd want that here, and also why "I have to pass the entire alphanumeric string to it" is a bad thing.
As has been noted, you can indeed use regexes with occursin, and it works well. But you can also roll your own version, quite simply:
isalphanumeric(c::AbstractChar) = isletter(c) || ('0' <= c <= '9')
isalphanumeric(str::AbstractString) = all(isalphanumeric, str)
I have a string S = '02143' and a list A = ['a','b','c','d','e']. I want to replace all those digits in 'S' with their corresponding element in list A.
For example, replace 0 with A[0], 2 with A[2] and so on. Final output should be S = 'acbed'.
I tried:
S = re.sub(r'([0-9])', A[int(r'\g<1>')], S)
However this gives an error ValueError: invalid literal for int() with base 10: '\\g<1>'. I guess it is considering backreference '\g<1>' as a string. How can I solve this especially using re.sub and capture-groups, else alternatively?
The reason the re.sub(r'([0-9])',A[int(r'\g<1>')],S) does not work is that \g<1> (which is an unambiguous representation of the first backreference otherwise written as \1) backreference only works when used in the string replacement pattern. If you pass it to another method, it will "see" just \g<1> literal string, since the re module won't have any chance of evaluating it at that time. re engine only evaluates it during a match, but the A[int(r'\g<1>')] part is evaluated before the re engine attempts to find a match.
That is why it is made possible to use callback methods inside re.sub as the replacement argument: you may pass the matched group values to any external methods for advanced manipulation.
See the re documentation:
re.sub(pattern, repl, string, count=0, flags=0)
If repl is a function, it is called for every non-overlapping
occurrence of pattern. The function takes a single match object
argument, and returns the replacement string.
Use
import re
S = '02143'
A = ['a','b','c','d','e']
print(re.sub(r'[0-9]',lambda x: A[int(x.group())],S))
See the Python demo
Note you do not need to capture the whole pattern with parentheses, you can access the whole match with x.group().
Need to find all non alpha numeric characters in string (called lorem_ipsum) and assign the outcome to a variable, results.
So would I put:
results = lorem_ipsum.findall()
?
And if so, what goes in the parentheses?
There are two different ways to use findall with regular expressions.
If you want to use it as a method, the leading expression has to be a compiled regular expression object, not a string. The syntax is:
regex.findall(string[, pos[, endpos ]])
If you want to use it as a function, the syntax is:
re.findall(pattern, string, flags=0)
I'm trying to find exact matches of strings in Lua including, special characters. I want the example below to return that it is an exact match, but because of the - character it returns nil
index = string.find("test-string", "test-string")
returns nil
index = string.find("test-string", "test-")
returns 1
index = string.find("test-string", "test")
also returns 1
How can I get it to do full matching?
- is a pattern operator in a Lua string pattern, so when you say test-string, you're telling find() to match the string test as few times as possible. So what happens is it looks at test-string, sees test in there, and since - isn't an actual minus sign in this case, it's really looking for teststring.
Do as Mike has said and escape it with the % character.
I found this helpful for better understanding patterns.
You can also ask for a plain substring match that ignores magic characters:
string.find("test-string", "test-string",1,true)
you need to escape special characters in the pattern with the % character.
so in this case you are looking for
local index = string.find('test-string', 'test%-string')
With Lua's string.find function, there is an optional fourth argument you can pass to enable plain searching. From the Lua wiki:
The pattern argument also allows more complex searches. See the
PatternsTutorial for more information. We can turn off the pattern
matching feature by using the optional fourth argument plain. plain
takes a boolean value and must be preceeded by index. E.g.,
= string.find("Hello Lua user", "%su") -- find a space character followed by "u"
10 11
= string.find("Hello Lua user", "%su", 1, true) -- turn on plain searches, now not found
nil
Basically, I was wondering how I can accomplish the same plain searching using Lua's string.gsub function.
I expected there to be something in the standard library for this, but there isn't. The solution, then, is to escape the special characters in the pattern so they don't perform their usual functions.
Here's the general idea:
obtain the pattern string
replace any special characters with % followed by it (for example, % becomes %%, [ becomes %[
use this as your search pattern for replacing the text
Here is a simple library function for text replacement:
function string.replace(text, old, new)
local b,e = text:find(old,1,true)
if b==nil then
return text
else
return text:sub(1,b-1) .. new .. text:sub(e+1)
end
end
This function can be called as newtext = text:replace(old,new).
Note that this only replaces the first occurrence of old in text.
Use this function to escape all magic characters (and only those) in your search string.
function escape_magic(s)
return (s:gsub('[%^%$%(%)%%%.%[%]%*%+%-%?]','%%%1'))
end