I'm trying to check if a subString exists in a string using regular expression.
RE : re_string_literal = '^"[a-zA-Z0-9_ ]+"$'
The thing is, I don't want to match any substring. I'm reading data from a file:
Now one of the lines have this text:
cout<<"Hello"<<endl;
I just want to check if there's a string inside the line and if yes, store it in a list.
I have tried the re.match method but it only works if we have to match a pattern, but in this case, I just want to check if a string exists or not, if yes, store it somewhere.
re_string_lit = '^"[a-zA-Z0-9_ ]+"$'
text = 'cout<<"Hello World!"<<endl;'
re.match(re_string_lit,text)
It doesn't output anything.
In simple words,
I just want to extract everything inside ""
If you just want to extract everything inside "" then string splitting would be much simpler way of doing things.
>>> a = 'something<<"actualString">>something,else'
>>> b = a.split('"')[1]
>>> b
'actualString'
The above example would only work for not more than 2 instances of double quotes ("), but you could make it work by iterating over every substring extracted using split method and applying a much simpler Regular Expression.
This worked for me:
re.search('"(.+?)"', 'cout<<"Hello"<<endl')
Related
I have a text file formatted like a JSON file however everything is on a single line (could be a MongoDB File). Could someone please point me in the direction of how I could extract values using a Python regex method please?
The text shows up like this:
{"d":{"__type":"WikiFileNodeContent:http:\/\/samplesite.com.au\/ns\/business\/wiki","author":null,"description":null,"fileAssetId":"034b9317-60d9-45c2-b6d6-0f24b59e1991","filename":"Reports.pdf"},"createdBy":1531,"createdByUsername":"John Cash","icon":"\/Assets10.37.5.0\/pix\/16x16\/page_white_acrobat.png","id":3041,"inheritedPermissions":false,"name":"map","permissions":[23,87,35,49,65],"type":3,"viewLevel":2},{"__type":"WikiNode:http:\/\/samplesite.com.au\/ns\/business\/wiki","children":[],"content":
I am wanting to get the "fileAssetId" and filename". Ive tried to load the like with Pythons JSON module but I get an error
For the FileAssetid I tried this regex:
regex = re.compile(r"([0-9a-f]{8})\S*-\S*([0-9a-f]{4})\S*-\S*([0-9a-f]{4})\S*-\S*([0-9a-f]{4})\S*-\S*([0-9a-f]{12})")
But i get the following 034b9317, 60d9, 45c2, b6d6, 0f24b59e1991
Im not to sure how to get the data as its displayed.
How about using positive lookahead and lookbehind:
(?<=\"fileAssetId\":\")[a-fA-F0-9-]+?(?=\")
captures the fileAssetId and
(?<=\"filename\":\").+?(?=\")
matches the filename.
For a detailed explanation of the regex have a look at the Regex101-Example. (Note: I combined both in the example with an OR-Operator | to show both matches at once)
To get a list of all matches use re.findall or re.finditer instead of re.match.
re.findall(pattern, string) returns a list of matching strings.
re.finditer(pattern, string) returns an iterator with the objects.
You can use python's walk method and check each entry with re.match.
In case that the string you got is not convertable to a python dict, you can use just regex:
print re.match(r'.*fileAssetId\":\"([^\"]+)\".*', your_pattern).group(1)
Solution for your example:
import re
example_string = '{"d":{"__type":"WikiFileNodeContent:http:\/\/samplesite.com.u\/ns\/business\/wiki","author":null,"description":null,"fileAssetId":"034b9317-60d9-45c2-b6d6-0f24b59e1991","filename":"Reports.pdf"},"createdBy":1531,"createdByUsername":"John Cash","icon":"\/Assets10.37.5.0\/pix\/16x16\/page_white_acrobat.png","id":3041,"inheritedPermissions":false,"name":"map","permissions":[23,87,35,49,65],"type":3,"viewLevel":2},{"__type":"WikiNode:http:\/\/samplesite.com.au\/ns\/business\/wiki","children":[],"content"'
regex_pattern = r'.*fileAssetId\":\"([^\"]+)\".*'
match = re.match(regex_pattern, example_string)
fileAssetId = match.group(1)
print('fileAssetId: {}'.format(fileAssetId))
executing this yields:
34b9317-60d9-45c2-b6d6-0f24b59e1991
Try adding \n to the string that you are entering in to the file (\n means new line)
Based on the idea given here https://stackoverflow.com/a/3845829 and by following the JSON standard https://www.json.org/json-en.html, we can use Python + regex https://pypi.org/project/regex/ and do the following:
json_pattern = (
r'(?(DEFINE)'
r'(?P<whitespace>( |\n|\r|\t)*)'
r'(?P<boolean>true|false)'
r'(?P<number>-?(0|([1-9]\d*))(\.\d*[1-9])?([eE][+-]?\d+)?)'
r'(?P<string>"([^"\\]|\\("|\\|/|b|f|n|r|t|u[0-9a-fA-F]{4}))*")'
r'(?P<array>\[((?&whitespace)|(?&value)(,(?&value))*)\])'
r'(?P<key>(?&whitespace)(?&string)(?&whitespace))'
r'(?P<value>(?&whitespace)((?&boolean)|(?&number)|(?&string)|(?&array)|(? &object)|null)(?&whitespace))'
r'(?P<object>\{((?&whitespace)|(?&key):(?&value)(,(?&key):(?&value))*)\})'
r'(?P<document>(?&object)|(?&array))'
r')'
r'(?&document)'
)
json_regex = regex.compile(json_pattern)
match = json_regex.match(json_document_text)
You can change last line in json_pattern to match not document but individual objects replacing (?&document) by (?&object). I think the regex is easier than I expected, but I did not run extensive tests on this. It works fine for me and I have tested hundreds of files. I wil try to improve my answer in case I find any issue when running it.
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'
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.
In Python, I am extracting emails from a string like so:
split = re.split(" ", string)
emails = []
pattern = re.compile("^[a-zA-Z0-9_\.-]+#[a-zA-Z0-9-]+.[a-zA-Z0-9-\.]+$");
for bit in split:
result = pattern.match(bit)
if(result != None):
emails.append(bit)
And this works, as long as there is a space in between the emails. But this might not always be the case. For example:
Hello, foo#foo.com
would return:
foo#foo.com
but, take the following string:
I know my best friend mailto:foo#foo.com!
This would return null. So the question is: how can I make it so that a regex is the delimiter to split? I would want to get
foo#foo.com
in all cases, regardless of punctuation next to it. Is this possible in Python?
By "splitting by regex" I mean that if the program encounters the pattern in a string, it will extract that part and put it into a list.
I'd say you're looking for re.findall:
>>> email_reg = re.compile(r'[a-zA-Z0-9_.-]+#[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+')
>>> email_reg.findall('I know my best friend mailto:foo#foo.com!')
['foo#foo.com']
Notice that findall can handle more than one email address:
>>> email_reg.findall('Text text foo#foo.com, text text, baz#baz.com!')
['foo#foo.com', 'baz#baz.com']
Use re.search or re.findall.
You also need to escape your expression properly (. needs to be escaped outside of character classes, not inside) and remove/replace the anchors ^ and $ (for example with \b), eg:
r"\b[a-zA-Z0-9_.+-]+#[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+\b"
The problem I see in your regex is your use of ^ which matches the start of a string and $ which matches the end of your string. If you remove it and then run it with your sample test case it will work
>>> re.findall("[A-Za-z0-9\._-]+#[A-Za-z0-9-]+.[A-Za-z0-9-\.]+","I know my best friend mailto:foo#foo.com!")
['foo#foo.com']
>>> re.findall("[A-Za-z0-9\._-]+#[A-Za-z0-9-]+.[A-Za-z0-9-\.]+","Hello, foo#foo.com")
['foo#foo.com']
>>>
i got an string that might look like this
"myFunc('element','node','elementVersion','ext',12,0,0)"
i'm currently checking for validity using, which works fine
myFunc\((.+?)\,(.+?)\,(.+?)\,(.+?)\,(.+?)\,(.+?)\,(.+?)\)
now i'd like to replace whatever string is at the 3rd parameter.
unfortunately i cant just use a stringreplace on whatever sub-string on the 3rd position since the same 'sub-string' could be anywhere else in that string.
with this and a re.findall,
myFunc\(.+?\,.+?\,(.+?)\,.+?\,.+?\,.+?\,.+?\)
i was able to get the contents of the substring on the 3rd position, but re.sub does not replace the string it just returns me the string i want to replace with :/
here's my code
myRe = re.compile(r"myFunc\(.+?\,.+?\,(.+?)\,.+?\,.+?\,.+?\,.+?\)")
val = "myFunc('element','node','elementVersion','ext',12,0,0)"
print myRe.findall(val)
print myRe.sub("noVersion",val)
any idea what i've missed ?
thanks!
Seb
In re.sub, you need to specify a substitution for the whole matching string. That means that you need to repeat the parts that you don't want to replace. This works:
myRe = re.compile(r"(myFunc\(.+?\,.+?\,)(.+?)(\,.+?\,.+?\,.+?\,.+?\))")
print myRe.sub(r'\1"noversion"\3', val)
If your only tool is a hammer, all problems look like nails. A regular expression is a powerfull hammer but is not the best tool for every task.
Some tasks are better handled by a parser. In this case the argument list in the string is just like a Python tuple, sou you can cheat: use the Python builtin parser:
>>> strdata = "myFunc('element','node','elementVersion','ext',12,0,0)"
>>> args = re.search(r'\(([^\)]+)\)', strdata).group(1)
>>> eval(args)
('element', 'node', 'elementVersion', 'ext', 12, 0, 0)
If you can't trust the input ast.literal_eval is safer than eval for this. Once you have the argument list in the string decontructed I think you can figure out how to manipulate and reassemble it again, if needed.
Read the documentation: re.sub returns a copy of the string where every occurrence of the entire pattern is replaced with the replacement. It cannot in any case modify the original string, because Python strings are immutable.
Try using look-ahead and look-behind assertions to construct a regex that only matches the element itself:
myRe = re.compile(r"(?<=myFunc\(.+?\,.+?\,)(.+?)(?=\,.+?\,.+?\,.+?\,.+?\))")
Have you tried using named groups? http://docs.python.org/howto/regex.html#search-and-replace
Hopefully that will let you just target the 3rd match.
If you want to do this without using regex:
>>> s = "myFunc('element','node','elementVersion','ext',12,0,0)"
>>> l = s.split(",")
>>> l[2]="'noVersion'"
>>> s = ",".join(l)
>>> s
"myFunc('element','node','noVersion','ext',12,0,0)"