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, "")
Related
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.
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.
In TI-BASIC, the + operation is overloaded for string concatenation (in this, if nothing else, TI-BASIC joins the rest of the world).
However, any attempt to concatenate involving an empty string raises a Dimension Mismatch error:
"Fizz"+"Buzz"
FizzBuzz
"Fizz"+""
Error
""+"Buzz"
Error
""+""
Error
Why does this occur, and is there an elegant workaround? I've been using a starting space and truncating the string when necessary (doesn't always work well) or using a loop to add characters one at a time (slow).
The best way depends on what you are doing.
If you have a string (in this case, Str1) that you need to concatenate with another (Str2), and you don't know if it is empty, then this is a good general-case solution:
Str2
If length(Str1
Str1+Str2
If you need to loop and add a stuff to the string each time, then this is your best solution:
Before the loop:
" →Str1
In the loop:
Str1+<stuff_that_isn't_an_empty_string>→Str1
After the loop:
sub(Str1,2,length(Str1)-1→Str1
There are other situations, too, and if you have a specific situation, then you should post a simplified version of the relevant code.
Hope this helps!
It is very unfortunate that TI-Basic doesn't support empty strings. If you are starting with an empty string and adding chars, you have to do something like this:
"?
For(I,1,3
Prompt Str1
Ans+Str1
End
sub(Ans,2,length(Ans)-1
Another useful trick is that if you have a string that you are eventually going to evaluate using expr(, you can do "("+Str1+")"→Str1 and then freely do search and replace on the string. This is a necessary workaround since you can't search and replace any text involving the first or last character in a string.
I want to remove all the characters from a string expect whatever character is between a certain set of characters. So for example I have the input of Grade:2/2014-2015 and I want the output of just the grade, 2.
I'm thinking that I need to use the FIND function to grab whatever is between the : and the / , this also needs to work with double characters such 10 however I believe that it would work so long as the defining values with the FIND function are correct.
Unfortunately I am totally lost on this when using the FIND function however if there is another function that would work better I could probably figure it out myself if I knew what function.
It's not particularly elegant but =MID(A1,FIND(":",A1)+1,FIND("/",A1) - FIND(":",A1) - 1) would work.
MID takes start and length,FIND returns the index of a given character.
Edit:
As pointed out, "Grade:" is fixed length so the following would work just as well:
=MID(A1,7,FIND("/",A1) - 7)
You could use LEFT() to remove "Grade:"
And then use and then use LEFTB() to remove the year.
Look at this link here. This is the way I would go about it.
=SUBSTITUTE(SUBSTITUTE(C4, "Grade:", ""), "/2014-2015", "")
where C4 is the name of your cell.
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