Python regular expressions with Foreign characters in python PyQT5 - python-3.x

This problem might be very simple but I find it a bit confusing & that is why I need help.
With relevance to this question I posted that got solved, I got a new issue that I just noticed.
Source code:
from PyQt5 import QtCore,QtWidgets
app=QtWidgets.QApplication([])
def scroll():
#QtCore.QRegularExpression(r'\b'+'cat'+'\b')
item = listWidget.findItems(r'\bcat\b', QtCore.Qt.MatchRegularExpression)
for d in item:
print(d.text())
window = QtWidgets.QDialog()
window.setLayout(QtWidgets.QVBoxLayout())
listWidget = QtWidgets.QListWidget()
window.layout().addWidget(listWidget)
cats = ["love my cat","catirization","cat in the clouds","catść"]
for i,cat in enumerate(cats):
QtWidgets.QListWidgetItem(f"{i} {cat}", listWidget)
btn = QtWidgets.QPushButton('Scroll')
btn.clicked.connect(scroll)
window.layout().addWidget(btn)
window.show()
app.exec_()
Output GUI:
Now as you can see I am just trying to print out the text data based on the regex r"\bcat\b" when I press the "Scroll" button and it works fine!
Output:
0 love my cat
2 cat in the clouds
3 catść
However... as you can see on the #3, it should not be printed out cause it obviously does not match with the mentioned regular expression which is r"\bcat\b". However it does & I am thinking it has something to do with that special foreign character ść that makes it a match & prints it out (which it shouldn't right?).
I'm expecting an output like:
0 love my cat
2 cat in the clouds
Researches I have tried
I found this question and it says something about this \p{L} & based on the answer it means:
If all you want to match is letters (including "international"
letters) you can use \p{L}.
To be honest I'm not so sure how to apply that with PyQT5 also still I've made some tries & and I tried changing the regex to like this r'\b'+r'\p{cat}'+r'\b'. However I got this error.
QString::contains: invalid QRegularExpression object
QString::contains: invalid QRegularExpression object
QString::contains: invalid QRegularExpression object
QString::contains: invalid QRegularExpression object
Obviously the error says it's not a valid regex. Can someone educate me on how to solve this issue? Thank you!

In general, when you need to make your shorthand character classes and word boundaries Unicode-aware, you need to pass the QRegularExpression.UseUnicodePropertiesOption option to the regex compiler. See the QRegularExpression.UseUnicodePropertiesOption reference:
The meaning of the \w, \d, etc., character classes, as well as the meaning of their counterparts (\W, \D, etc.), is changed from matching ASCII characters only to matching any character with the corresponding Unicode property. For instance, \d is changed to match any character with the Unicode Nd (decimal digit) property; \w to match any character with either the Unicode L (letter) or N (digit) property, plus underscore, and so on. This option corresponds to the /u modifier in Perl regular expressions.
In Python, you could declare it as
rx = QtCore.QRegularExpression(r'\bcat\b', QtCore.QRegularExpression.UseUnicodePropertiesOption)
However, since the QListWidget.findItems does not support a QRegularExpression as argument and only allows the regex as a string object, you can only use the (*UCP) PCRE
verb as an alternative:
r'(*UCP)\bcat\b'
Make sure you define it at the regex beginning.

Related

How can I remove all characters inside angle brackets python?

How can I remove all characters inside angle brackets including the brackets in a string? How can I also remove all the text between ("\r\n") and ("."+"any 3 characters") Is this possible? I am currently using the solution by #xkcdjerry
e.g
body = """Dear Students roads etc. you place a tree take a snapshot, then when you place a\r\nbuilding, take a snapshot. Place at least 5-6 objects and then have 5-6\r\nsnapshots. Please keep these snapshots with you as everyone will be asked\r\nto share them during the class.\r\n\r\nI am attaching one PowerPoint containing instructions and one video of\r\nexplanation for your reference.\r\n\r\nKind regards,\r\nTeacher Name\r\n zoom_0.mp4\r\n<https://drive.google.com/file/d/1UX-klOfVhbefvbhZvIWijaBdQuLgh_-Uru4_1QTkth/view?usp=drive_web>"""
d = re.compile("\r\n.+?\\....")
body = d.sub('', body)
a = re.compile("<.*?>")
body = a.sub('', body)
print(body)```
For some reason the output is fine except that it has:
```gle.com/file/d/1UX-klOfVhbefvbhZvIWijaBdQuLgh_-Uru4_1QTkth/view?usp=drive_web>
randomly attached to the end How can I fix it.
Answer
Your problem can be solved by a regex:
Put this into the shell:
import re
a=re.compile("<.*?>")
a.sub('',"Keep this part of the string< Remove this part>Keep This part as well")
Output:
'Keep this part of the stringKeep This part as well'
Second question:
import re
re.compile("\r\n.*?\\..{3}")
a.sub('',"Hello\r\nFilename.png")
Output:
'Hello'
Breakdown
Regex is a robust way of finding, replacing, and mutating small strings inside bigger ones, for further reading,consult https://docs.python.org/3/library/re.html. Meanwhile, here are the breakdowns of the regex information used in this answer:
. means any char.
*? means as many of the before as needed but as little as possible(non-greedy match)
So .*? means any number of characters but as little as possible.
Note: The reason there is a \\. in the second regex is that a . in the match needs to be escaped by a \, which in its turn needs to be escaped as \\
The methods:
re.compile(patten:str) compiles a regex for farther use.
regex.sub(repl:str,string:str) replaces every match of regex in string with repl.
Hope it helps.

Python3 strip() get unexpect result

It's a weird problem
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120"
And two strings below:
s1="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\[Content_Types].xml"
s2="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\_rels\.rels"
When I use the command below:
s1.strip(to_be_stripped)
s2.strip(to_be_stripped)
I get these outputs:
'[Content_Types].x'
'_rels\\.'
If I use lstrip(), they will be:
'[Content_Types].xml'
'_rels\\.rels'
Which is the right outputs.
However, if we replace all Project Known with zeus_pipeline:
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120"
And:
s2="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120\\_rels\.rels"
s2.lstrip(to_be_stripped)will be '.rels'
If I use / instead of \\, nothing goes wrong. I am wondering why this problem happens.
strip isn't meant to remove full strings exactly. Rather, you give it a string, and every character in that string is removed from the start and of the string to be stripped.
In your case, the variable to_be_stripped contains the characters m and l, so those are stripped from the end of s1. However, it doesn't contain the character x, so the stripping stops there and no characters beyond that are removed.
Check out this question. The accepted answer is probably more extensive than you need - I like another user's suggestion of using replace instead of strip. This would look like:
s1.replace(to_be_stripped, "")

How to change a string into a variable

I want to write out some data into a file. I saved the filename as a variable. I wan to use % mode to substitude the variable to the text, but it gives an error:
IndentationError: unindent does not match any outer indentation level
writeafile = open('N:\myfile\%s.txt' , "a") % (variable)
Assuming we are talking about Python here, you should move variable next to the
'N:\\myfile\\%s.txt' string for correct syntax, like so:
writeafile = open("N:\\myfile\\%s.txt" % variable, "a")
However, using this style of formatting is not recommended by Pydocs:
The formatting operations described here exhibit a variety of quirks that lead to a number of common errors (such as failing to display tuples and dictionaries correctly). Using the newer formatted string literals, the str.format() interface, or template strings may help avoid these errors. Each of these alternatives provides their own trade-offs and benefits of simplicity, flexibility, and/or extensibility.
Source
So, I'd suggest using f-strings, which have been available in Python since 3.6. The double \\ is intentional here, otherwise Python will treat it as an escape character and you'll get undesired results.
writeafile = open(f"N:\\myfile\\{variable}.txt", "a")
Alternatively, you could also use str.format():
writeafile = open("N:\\myfile\\{name}.txt".format(name=variable), "a")

substitue string by index without using regular expressions

It should be very easy, but I am looking for an efficient way to perform it.
I know that I could split the string into two parts and insert the new value, but I have tried to substitute each line between the indexes 22-26 as follows:
line.replace(line[22:26],new_value)
The Problem
However, that function substitutes everything in the line that is similar to the pattern in line[22:26].
In the example below, I want to replace the marked number 1 with number 17:
Here are the results. Note the replacement of 1 with 17 in several places:
Thus I don't understand the behavior of replace command. Is there a simple explanation of what I'm doing wrong?
Why I don't want RE
The values between index 22-26 are not unified in form.
Note: I am using python 3.5 on Unix/Linux machines.
str.replace replaces 1 sub-string pattern with another everywhere in the string.
e.g.
'ab cd ab ab'.replace('ab', 'xy')
# produces output 'xy cd xy xy'
similarly,
mystr = 'ab cd ab ab'
mystr.replace(mystr[0:2], 'xy')
# also produces output 'xy cd xy xy'
what you could do instead, to replace just the characters in position 22-26
line = line[0:22] + new_value + line[26:]
Also, looking at your data, it seems to me to be a fixed-width text file. While my suggestion will work, a more robust way to process this data would be to read it & separate the different fields in the record first, before processing the data.
If you have access to the pandas library, it provides a useful function just for reading fixed-width files

What's the point of nesting brackets in Lua?

I'm currently teaching myself Lua for iOS game development, since I've heard lots of very good things about it. I'm really impressed by the level of documentation there is for the language, which makes learning it that much easier.
My problem is that I've found a Lua concept that nobody seems to have a "beginner's" explanation for: nested brackets for quotes. For example, I was taught that long strings with escaped single and double quotes like the following:
string_1 = "This is an \"escaped\" word and \"here\'s\" another."
could also be written without the overall surrounding quotes. Instead one would simply replace them with double brackets, like the following:
string_2 = [[This is an "escaped" word and "here's" another.]]
Those both make complete sense to me. But I can also write the string_2 line with "nested brackets," which include equal signs between both sets of the double brackets, as follows:
string_3 = [===[This is an "escaped" word and "here's" another.]===]
My question is simple. What is the point of the syntax used in string_3? It gives the same result as string_1 and string_2 when given as an an input for print(), so I don't understand why nested brackets even exist. Can somebody please help a noob (me) gain some perspective?
It would be used if your string contains a substring that is equal to the delimiter. For example, the following would be invalid:
string_2 = [[This is an "escaped" word, the characters ]].]]
Therefore, in order for it to work as expected, you would need to use a different string delimiter, like in the following:
string_3 = [===[This is an "escaped" word, the characters ]].]===]
I think it's safe to say that not a lot of string literals contain the substring ]], in which case there may never be a reason to use the above syntax.
It helps to, well, nest them:
print [==[malucart[[bbbb]]]bbbb]==]
Will print:
malucart[[bbbb]]]bbbb
But if that's not useful enough, you can use them to put whole programs in a string:
loadstring([===[print "o m g"]===])()
Will print:
o m g
I personally use them for my static/dynamic library implementation. In the case you don't know if the program has a closing bracket with the same amount of =s, you should determine it with something like this:
local c = 0
while contains(prog, "]" .. string.rep("=", c) .. "]") do
c = c + 1
end
-- do stuff

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