I need to match string and return (\aHEX) and (text).
And use it like
for k, v in string.gmatch("Default \aFFFFFFFFWhite text \aFF0000FFJust red text", "<pattern>") do
-- k = "" / "\aFFFFFFFF" / "\aFF0000FF"
-- v = "Default" / "White text " / "Just red text"
end
tried this
local text = "Default \aFF00FFFFRed text\a00FF00FFGreen";
for k, v in text:gmatch("(\a%x%x%x%x%x%x%x%x)(%w+)") do
print(string.format("%s: %s", k, v));
-- \aFF00FFFF: Red
-- \a00FF00FF: Green
end
Missing "Default" and " text"
The problem is that you cannot make a sequence of patterns optional with Lua patterns.
You can use
local text = "Default \aFF00FFFFRed text\a00FF00FFGreen";
for word in text:gmatch("\a?[^\a]+") do
k, v = word:match("^(\a%x%x%x%x%x%x%x%x)(%w+)$")
if k then
print(string.format("%s: %s", k, v));
else
print(word)
end
end
Output:
Default
FF00FFFFRed text
00FF00FF: Green
Details:
\a?[^\a]+ matches all occurrences of sequences that start with an optional \a, and then contain one or more chars other than \a
then ^(\a%x%x%x%x%x%x%x%x)(%w+)$ is applied to each found substring, and if there is a match, the key:value pair is produced, else the whole match is the required value.
Related
I have this string text:
text = "hotkey=F4,value=,autoSend=false, hotkey=Shift+F9,value=,autoSend=false, hotkey=F5,value=,autoSend=false"
and I would like to convert it to a table like this one:
local table = {
{hotkey='F4', value=nil, autoSend=false};
{hotkey='Shift+F9', value=nil, autoSend=false};
{hotkey='F5', value=nil, autoSend=false}
}
This solution is limited in scope and will not cover all complexities in the input string. A simple pattern matching could generate tables you are looking for, but use this code to build a better/robust regex for the diversity of your strings
s = "hotkey=F4,value=,autoSend=false, hotkey=Shift+F9,value=,autoSend=false, hotkey=F5,value=,autoSend=false"
local words = {}
for w in s:gmatch("(hotkey=%g-,value=%g-,autoSend=%w*)") do
-- Split string in more managebale parts
-- i-g w = 'hotkey=F4,value=,autoSend=false, hotkey=Shift+F9'
-- Extract indivisual k,v pairs and insert into table as desired
local _hotkey = string.match(w,"hotkey=(%g-),")
local _value = string.match(w,"value=(%g-),")
local _autoSend = string.match(w,"autoSend=(%w+)")
table.insert(words,{hotkey=_hotkey, value=_value, autoSend=_autoSend})
end
for _, w in ipairs(words) do
for k, v in pairs(w) do
print(k .. ':' .. v)
end
end
Regex Explanation
(): Capture string
%g: printable characters except for spaces
%w: alphanumeric characters
* : 0 or more repetitions
- : 0 or more lazy repetitions
s = "That that occurs sometimes. It sometimes means that which, and sometimes just that"
target = "that"
words = s.split()
b = []
for i,w in enumerate(words):
if w == target:
if i > 0:
b = words[i-1]
print([b].sep="",end",")
"I used, end=",sep=",but nothing worked.I need the output horizontally, with square brackets, commas and quotations marks. the brackets appear in the middle, and a comma at the end."
"Current output"
['That'],['means'],['just'],
"I need this output"
['That','means','just']
try this code it will work fine
s = "That that occurs sometimes. It sometimes means that which, and sometimes just that"
target = "that"
words = s.split()
b = []
for i,w in enumerate(words):
if w == target:
if i > 0:
b.append(words[i-1])
print(b,sep="")
I'm trying to write a conditional statement where I can skip a specific space then start reading all the characters after it.
I was thinking to use substring but that wouldn't help because substring will only work if I know the exact number of characters I want to skip but in this case, I want to skip a specific space to read characters afterward.
For example:
String text = "ABC DEF W YZ" //number of characters before the spaces are unknown
String test = "A"
if ( test == "A") {
return text (/*escape the first two space and return anything after that*/)
}
You can split your string on " " with tokenize, remove the first N elements from the returned array (where N is the number of spaces you want to ignore) and join what's left with " ".
Supposing your N is 2:
String text = "ABC DEF W YZ" //number of characters before the spaces are unknown
String test = "A"
if ( test == "A") {
return text.tokenize(" ").drop(2).join(" ")
}
I'm trying to find the difference in text between two string values in Lua, and I'm just not quite sure how to do this effectively. I'm not very experienced in working with string patterns, and I'm sure that's my downfall on this one. Here's an example:
-- Original text
local text1 = "hello there"
-- Changed text
local text2 = "hello.there"
-- Finding the alteration of original text with some "pattern"
print(text2:match("pattern"))
In the example above, I'd want to output the text ".", since that's the difference between the two texts. Same goes for cases where the difference could be sensitive to a string pattern, like this:
local text1 = "hello there"
local text2 = "hello()there"
print(text2:match("pattern"))
In this example, I'd want to print "(" since at that point the new string is no longer consistent with the old one.
If anyone has any insight on this, I'd really appreciate it. Sorry I couldn't give more to work with code-wise, I'm just not sure where to begin.
Just iterate over the strings and find when they don't match.
function StringDifference(str1,str2)
for i = 1,#str1 do --Loop over strings
if str1:sub(i,i) ~= str2:sub(i,i) then --If that character is not equal to it's counterpart
return i --Return that index
end
end
return #str1+1 --Return the index after where the shorter one ends as fallback.
end
print(StringDifference("hello there", "hello.there"))
local function get_inserted_text(old, new)
local prv = {}
for o = 0, #old do
prv[o] = ""
end
for n = 1, #new do
local nxt = {[0] = new:sub(1, n)}
local nn = new:sub(n, n)
for o = 1, #old do
local result
if nn == old:sub(o, o) then
result = prv[o-1]
else
result = prv[o]..nn
if #nxt[o-1] <= #result then
result = nxt[o-1]
end
end
nxt[o] = result
end
prv = nxt
end
return prv[#old]
end
Usage:
print(get_inserted_text("hello there", "hello.there")) --> .
print(get_inserted_text("hello there", "hello()there")) --> ()
print(get_inserted_text("hello there", "hello htere")) --> h
print(get_inserted_text("hello there", "heLlloU theAre")) --> LUA
I have a list of words in Pandas (DF)
Words
Shirt
Blouse
Sweater
What I'm trying to do is swap out certain letters in those words with letters from my dictionary one letter at a time.
so for example:
mydict = {"e":"q,w",
"a":"z"}
would create a new list that first replaces all the "e" in a list one at a time, and then iterates through again replacing all the "a" one at a time:
Words
Shirt
Blouse
Sweater
Blousq
Blousw
Swqater
Swwater
Sweatqr
Sweatwr
Swezter
I've been looking around at solutions here: Mass string replace in python?
and have tried the following code but it changes all instances "e" instead of doing so one at a time -- any help?:
mydict = {"e":"q,w"}
s = DF
for k, v in mydict.items():
for j in v:
s['Words'] = s["Words"].str.replace(k, j)
DF["Words"] = s
this doesn't seem to work either:
s = DF.replace({"Words": {"e": "q","w"}})
This answer is very similar to Brian's answer, but a little bit sanitized and the output has no duplicates:
words = ["Words", "Shirt", "Blouse", "Sweater"]
md = {"e": "q,w", "a": "z"}
md = {k: v.split(',') for k, v in md.items()}
newwords = []
for word in words:
newwords.append(word)
for c in md:
occ = word.count(c)
pos = 0
for _ in range(occ):
pos = word.find(c, pos)
for r in md[c]:
tmp = word[:pos] + r + word[pos+1:]
newwords.append(tmp)
pos += 1
Content of newwords:
['Words', 'Shirt', 'Blouse', 'Blousq', 'Blousw', 'Sweater', 'Swqater', 'Swwater', 'Sweatqr', 'Sweatwr', 'Swezter']
Prettyprint:
Words
Shirt
Blouse
Blousq
Blousw
Sweater
Swqater
Swwater
Sweatqr
Sweatwr
Swezter
Any errors are a result of the current time. ;)
Update (explanation)
tl;dr
The main idea is to find the occurences of the character in the word one after another. For each occurence we are then replacing it with the replacing-char (again one after another). The replaced word get's added to the output-list.
I will try to explain everything step by step:
words = ["Words", "Shirt", "Blouse", "Sweater"]
md = {"e": "q,w", "a": "z"}
Well. Your basic input. :)
md = {k: v.split(',') for k, v in md.items()}
A simpler way to deal with replacing-dictionary. md now looks like {"e": ["q", "w"], "a": ["z"]}. Now we don't have to handle "q,w" and "z" differently but the step for replacing is just the same and ignores the fact, that "a" only got one replace-char.
newwords = []
The new list to store the output in.
for word in words:
newwords.append(word)
We have to do those actions for each word (I assume, the reason is clear). We also append the world directly to our just created output-list (newwords).
for c in md:
c as short for character. So for each character we want to replace (all keys of md), we do the following stuff.
occ = word.count(c)
occ for occurrences (yeah. count would fit as well :P). word.count(c) returns the number of occurences of the character/string c in word. So "Sweater".count("o") => 0 and "Sweater".count("e") => 2.
We use this here to know, how often we have to take a look at word to get all those occurences of c.
pos = 0
Our startposition to look for c in word. Comes into use in the next loop.
for _ in range(occ):
For each occurence. As a continual number has no value for us here, we "discard" it by naming it _. At this point where c is in word. Yet.
pos = word.find(c, pos)
Oh. Look. We found c. :) word.find(c, pos) returns the index of the first occurence of c in word, starting at pos. At the beginning, this means from the start of the string => the first occurence of c. But with this call we already update pos. This plus the last line (pos += 1) moves our search-window for the next round to start just behind the previous occurence of c.
for r in md[c]:
Now you see, why we updated mc previously: we can easily iterate over it now (a md[c].split(',') on the old md would do the job as well). So we are doing the replacement now for each of the replacement-characters.
tmp = word[:pos] + r + word[pos+1:]
The actual replacement. We store it in tmp (for debug-reasons). word[:pos] gives us word up to the (current) occurence of c (exclusive c). r is the replacement. word[pos+1:] adds the remaining word (again without c).
newwords.append(tmp)
Our so created new word tmp now goes into our output-list (newwords).
pos += 1
The already mentioned adjustment of pos to "jump over c".
Additional question from OP: Is there an easy way to dictate how many letters in the string I want to replace [(meaning e.g. multiple at a time)]?
Surely. But I have currently only a vague idea on how to achieve this. I am going to look at it, when I got my sleep. ;)
words = ["Words", "Shirt", "Blouse", "Sweater", "multipleeee"]
md = {"e": "q,w", "a": "z"}
md = {k: v.split(',') for k, v in md.items()}
num = 2 # this is the number of replaces at a time.
newwords = []
for word in words:
newwords.append(word)
for char in md:
for r in md[char]:
pos = multiples = 0
current_word = word
while current_word.find(char, pos) != -1:
pos = current_word.find(char, pos)
current_word = current_word[:pos] + r + current_word[pos+1:]
pos += 1
multiples += 1
if multiples == num:
newwords.append(current_word)
multiples = 0
current_word = word
Content of newwords:
['Words', 'Shirt', 'Blouse', 'Sweater', 'Swqatqr', 'Swwatwr', 'multipleeee', 'multiplqqee', 'multipleeqq', 'multiplwwee', 'multipleeww']
Prettyprint:
Words
Shirt
Blouse
Sweater
Swqatqr
Swwatwr
multipleeee
multiplqqee
multipleeqq
multiplwwee
multipleeww
I added multipleeee to demonstrate, how the replacement works: For num = 2 it means the first two occurences are replaced, after them, the next two. So there is no intersection of the replaced parts. If you would want to have something like ['multiplqqee', 'multipleqqe', 'multipleeqq'], you would have to store the position of the "first" occurence of char. You can then restore pos to that position in the if multiples == num:-block.
If you got further questions, feel free to ask. :)
Because you need to replace letters one at a time, this doesn't sound like a good problem to solve with pandas, since pandas is about doing everything at once (vectorized operations). I would dump out your DataFrame into a plain old list and use list operations:
words = DF.to_dict()["Words"].values()
for find, replace in reversed(sorted(mydict.items())):
for word in words:
occurences = word.count(find)
if not occurences:
print word
continue
start_index = 0
for i in range(occurences):
for replace_char in replace.split(","):
modified_word = list(word)
index = modified_word.index(find, start_index)
modified_word[index] = replace_char
modified_word = "".join(modified_word)
print modified_word
start_index = index + 1
Which gives:
Words
Shirt
Blousq
Blousw
Swqater
Swwater
Sweatqr
Sweatwr
Words
Shirt
Blouse
Swezter
Instead of printing the words, you can append them to a list and re-create a DataFrame if that's what you want to end up with.
If you are looping, you need to update s at each cycle of the loop. You also need to loop over v.
mydict = {"e":"q,w"}
s=deduped
for k, v in mydict.items():
for j in v:
s = s.replace(k, j)
Then reassign it to your dataframe:
df["Words"] = s
If you can write this as a function that takes in a 1d array (list, numpy array etc...), you can use df.apply to apply it to any column, using df.apply().