Develop Reference

Escaping missing parenthesis using pandas str.match

I'm having trouble with regex. I'm trying to check if my database fully matches with the item name I'm working. The problem is that sometimes the data is incomplete and I'll get errors. I would like to ignore regex completely as it is not necessary at this point.
For example the code below returns re.error: missing ), unterminated subpattern at position 10 as the last item on the list is missing a parenthesis. I've tried using if database['Item Name'].str.match(item, regex=False).any(): but it's not enough as the items can be named quite similarly and I would need perfect match. I've also tried to read re module documentation but I do not understand it well enough to get rid of the problem.
Any ideas how could I bypass the issue?
database = pd.read_csv("database.csv", sep=";")
list = ["Test Name !", "Test Name (2020)", "Test name ("]
for item in list:
if database['Item Name'].str.match(item).any():
# do something
pass
else:
#do something else
pass

If I understand your post correctly, you are trying to use the data read to create a regex. Since you don't want these treated as regexes, you might simply use string comparisons.
However, if your application requires the use of regex, you can use re.escape() render the string as literal so the paren won’t be magic.
For example:
import re
string1 = 'this is a magic ( that will break your regex'
string2 = re.escape(string1) # escapes your string
re.match(string2, "this won't cause issues")
#re.match(string1, "this will cause issues")

Search a delimited string in a file - Python

I have the following read.json file
{:{"JOL":"EuXaqHIbfEDyvph%2BMHPdCOJWMDPD%2BGG2xf0u0mP9Vb4YMFr6v5TJzWlSqq6VL0hXy07VDkWHHcq3At0SKVUrRA7shgTvmKVbjhEazRqHpvs%3D-%1E2D%TL/xs23EWsc40fWD.tr","LAPTOP":"error"}
and python script :
import re
shakes = open("read.json", "r")
needed = open("needed.txt", "w")
for text in shakes:
if re.search('JOL":"(.+?).tr', text):
print >> needed, text,
I want it to find what's between two words (JOL":" and .tr) and then print it. But all it does is printing all the text set in "read.json".

You're calling re.search, but you're not doing anything with the returned match, except to check that there is one. Instead, you're just printing out the original text. So of course you get the whole line.
The solution is simple: just store the result of re.search in a variable, so you can use it. For example:
for text in shakes:
match = re.search('JOL":"(.+?).tr', text)
if match:
print >> needed, match.group(1)
In your example, the match is JOL":"EuXaqHIbfEDyvph%2BMHPdCOJWMDPD%2BGG2xf0u0mP9Vb4YMFr6v5TJzWlSqq6VL0hXy07VDkWHHcq3At0SKVUrRA7shgTvmKVbjhEazRqHpvs%3D-%1E2D%TL/xs23EWsc40fWD.tr, and the first (and only) group in it is EuXaqHIbfEDyvph%2BMHPdCOJWMDPD%2BGG2xf0u0mP9Vb4YMFr6v5TJzWlSqq6VL0hXy07VDkWHHcq3At0SKVUrRA7shgTvmKVbjhEazRqHpvs%3D-%1E2D%TL/xs23EWsc40fWD, which is (I think) what you're looking for.
However, a couple of side notes:
First, . is a special pattern in a regex, so you're actually matching anything up to any character followed by tr, not .tr. For that, escape the . with a \. (And, once you start putting backslashes into a regex, use a raw string literal.) So: r'JOL":"(.+?)\.tr'.
Second, this is making a lot of assumptions about the data that probably aren't warranted. What you really want here is not "everything between JOL":" and .tr", it's "the value associated with key 'JOL' in the JSON object". The only problem is that this isn't quite a JSON object, because of that prefixed :. Hopefully you know where you got the data from, and therefore what format it's actually in. For example, if you know it's actually a sequence of colon-prefixed JSON objects, the right way to parse it is:
d = json.loads(text[1:])
if 'JOL' in d:
print >> needed, d['JOL']
Finally, you don't actually have anything named needed in your code; you opened a file named 'needed.txt', but you called the file object love. If your real code has a similar bug, it's possible that you're overwriting some completely different file over and over, and then looking in needed.txt and seeing nothing changed each time…

If you know that your starting and ending matching strings only appear once, you can ignore that it's JSON. If that's OK, then you can split on the starting characters (JOL":"), take the 2nd element of the split array [1], then split again on the ending characters (.tr) and take the 1st element of the split array [0].
>>> text = '{:{"JOL":"EuXaqHIbfEDyvph%2BMHPdCOJWMDPD%2BGG2xf0u0mP9Vb4YMFr6v5TJzWlSqq6VL0hXy07VDkWHHcq3At0SKVUrRA7shgTvmKVbjhEazRqHpvs%3D-%1E2D%TL/xs23EWsc40fWD.tr","LAPTOP":"error"}'
>>> text.split('JOL":"')[1].split('.tr')[0]
'EuXaqHIbfEDyvph%2BMHPdCOJWMDPD%2BGG2xf0u0mP9Vb4YMFr6v5TJzWlSqq6VL0hXy07VDkWHHcq3At0SKVUrRA7shgTvmKVbjhEazRqHpvs%3D-%1E2D%TL/xs23EWsc40fWD'

regex search&replace a variable string including a regex statement

I want to use re.sub to replace a part of a string I know exactly what looks like. relevant part of code:
print "Regex statement: ", foundStatements[iterator]
print "string to replace with : \n", latexPreparedString
print "string to search&replace in: \n", fileAsString
processedString = re.sub(foundStatements[iterator], latexPreparedString, fileAsString)
print "processed string: \n", processedString
In my testing case, foundStatements[iterator] is "%#import script_example.py ( *out =(.|\n)*?return out)" But even though processedString contains foundStatements[iterator], processedString looks exactly like fileAsString, so it hasn't accomplished the re.sub task. What am I doing wrong?
EDIT: Ok, it definitely has something to do with the string I'm searching to replace containing regex code. Is there a way to make it just interpret it foundStatements[iterator] as a raw string to search for? The only solution I can think of is to create a function that replaces any regex symbols in a string with \regexsymbol (e.g. * -> \*), but it'd make sense for there to be a way to solve this with inbuilt functions. It'd also be a bit overkill since I'd have to make sure it works with every single regex symbol, of which there are quite a few :/
EDIT2: Well, just changing it to re.sub(re.escape(foundStatements[iterator]), latexPreparedString, fileAsString) seems to work. except when the regex statement doesn't hit anything in the original file. To explain, latexPreparedString is generated by using the regex-part of the foundStatements[iterator]. While it's logical that it shouldn't be able to set latexPreparedString to anything when the regex statement doesn't hit anything, I set latexPreparedString = "" by default, so in that case it should re.sub replace it with a blank string if it doesn't hit anything. Here's how to code looks at the moment: pastebin.com/wUedK3LN

First, for replacing an exact match in a string, you should use [string.replace()][1]:
processedString = fileAsString(foundStatements[iterator], latexPreparedString)
However, this will still fail in your case, because foundStatements[iterator] has a newline character in it. To escape it, you need to use the r prefix when declaring foundStatements[iterator].
If you still want to use re.sub, you have to both prefix the string with r and use re.escape(foundStatements[iterator]) instead of foundStatements[iterator]. You can read more about re.escape here.

in python find index in list if combination of strings exist

I'm writing my first script and trying to learn python.
But I'm stuck and can't get out of this one.
I'm writing a script to change file names.
Lets say I have a string = "this.is.tEst3.E00.erfeh.ervwer.vwtrt.rvwrv"
I want the result to be string = "This Is Test3 E00"
this is what I have so far:
l = list(string)
//Transform the string into list
for i in l:
if "E" in l:
p = l.index("E")
if isinstance((p+1), int () is True:
if isinstance((p+2), int () is True:
delp = p+3
a = p-3
del l[delp:]
new = "".join(l)
new = new.replace("."," ")
print (new)
get in index where "E" and check if after "E" there are 2 integers.
Then delete everything after the second integer.
However this will not work if there is an "E" anyplace else.
at the moment the result I get is:
this is tEst
because it is finding index for the first "E" on the list and deleting everything after index+3
I guess my question is how do I get the index in the list if a combination of strings exists.
but I can't seem to find how.
thanks for everyone answers.
I was going in other direction but it is also not working.
if someone could see why it would be awesome. It is much better to learn by doing then just coping what others write :)
this is what I came up with:
for i in l:
if i=="E" and isinstance((i+1), int ) is True:
p = l.index(i)
print (p)
anyone can tell me why this isn't working. I get an error.
Thank you so much

Have you ever heard of a Regular Expression?
Check out python's re module. Link to the Docs.
Basically, you can define a "regex" that would match "E and then two integers" and give you the index of it.
After that, I'd just use python's "Slice Notation" to choose the piece of the string that you want to keep.
Then, check out the string methods for str.replace to swap the periods for spaces, and str.title to put them in Title Case

An easy way is to use a regex to find up until the E followed by 2 digits criteria, with s as your string:
import re
up_until = re.match('(.*?E\d{2})', s).group(1)
# this.is.tEst3.E00
Then, we replace the . with a space and then title case it:
output = up_until.replace('.', ' ').title()
# This Is Test3 E00

The technique to consider using is Regular Expressions. They allow you to search for a pattern of text in a string, rather than a specific character or substring. Regular Expressions have a bit of a tough learning curve, but are invaluable to learn and you can use them in many languages, not just in Python. Here is the Python resource for how Regular Expressions are implemented:
http://docs.python.org/2/library/re.html
The pattern you are looking to match in your case is an "E" followed by two digits. In Regular Expressions (usually shortened to "regex" or "regexp"), that pattern looks like this:
E\d\d # ('\d' is the specifier for any digit 0-9)
In Python, you create a string of the regex pattern you want to match, and pass that and your file name string into the search() method of the the re module. Regex patterns tend to use a lot of special characters, so it's common in Python to prepend the regex pattern string with 'r', which tells the Python interpreter not to interpret the special characters as escape characters. All of this together looks like this:
import re
filename = 'this.is.tEst3.E00.erfeh.ervwer.vwtrt.rvwrv'
match_object = re.search(r'E\d\d', filename)
if match_object:
# The '0' means we want the first match found
index_of_Exx = match_object.end(0)
truncated_filename = filename[:index_of_Exx]
# Now take care of any more processing
Regular expressions can get very detailed (and complex). In fact, you can probably accomplish your entire task of fully changing the file name using a single regex that's correctly put together. But since I don't know the full details about what sorts of weird file names might come into your program, I can't go any further than this. I will add one more piece of information: if the 'E' could possibly be lower-case, then you want to add a flag as a third argument to your pattern search which indicates case-insensitive matching. That flag is 're.I' and your search() method would look like this:
match_object = re.search(r'E\d\d', filename, re.I)
Read the documentation on Python's 're' module for more information, and you can find many great tutorials online, such as this one:
http://www.zytrax.com/tech/web/regex.htm
And before you know it you'll be a superhero. :-)

The reason why this isn't working:
for i in l:
if i=="E" and isinstance((i+1), int ) is True:
p = l.index(i)
print (p)
...is because 'i' contains a character from the string 'l', not an integer. You compare it with 'E' (which works), but then try to add 1 to it, which errors out.

Python: pattern matching for a string

Im trying to check a file line by line for any_string=any_string. It must be that format, no spaces or anything else. The line must contain a string then a "=" and then another string and nothing else. Could someone help me with the syntax in python to find this please? =]
pattern='*\S\=\S*'
I have this, but im pretty sure its wrong haha.

Don't know if you are looking for lines with the same value on both = sides. If so then use:
the_same_re = re.compile(r'^(\S+)=(\1)$')
if values can differ then use
the_same_re = re.compile(r'^(\S+)=(\S+)$')
In this regexpes:
^ is the beginning of line
$ is the end of line
\S+ is one or more non space character
\1 is first group
r before regex string means "raw" string so you need not escape backslashes in string.

pattern = r'\S+=\S+'
If you want to be able to grab the left and right-hand sides, you could add capture groups:
pattern = r'(\S+)=(\S+)'
If you don't want to allow multiple equals signs in the line (which would do weird things), you could use this:
pattern = r'[^\s=]+=[^\s=]+'

I don't know what the tasks you want make use this pattern. Maybe you want parse configuration file.
If it is true you may use module ConfigParser.

Ok, so you want to find anystring=anystring and nothing else. Then no need regex.
>>> s="anystring=anystring"
>>> sp=s.split("=")
>>> if len(sp)==2:
... print "ok"
...
ok

Since Python 2.5 I prefer this to split. If you don't like spaces, just check.
left, _, right = any_string.partition("=")
if right and " " not in any_string:
# proceed
Also it never hurts to learn regular expressions.