This is in reference to a question I asked before here
I received a solution to the problem in that question but ended up needing to go with regex for this particular part.
I need a regular expression to search and replace a string for instances of two vowels in a row that are the same, so the "oo" in "took", or the "ee" in "bees" and replace it with the one of the letters that was replaced and a :.
Some examples of expected behavior:
"took" should become "to:k"
"waaeek" should become "wa:e:k"
"raaag" should become "ra:ag"
Thank you for the help.
Try this:
re.sub(r'([aeiou])\1', r'\1:', str)
Search for ([aeiou])\1 and replace it with \1:
I don't know about python, but you should be able to make the regex case insensitive and global with something like /([aeiou])\1/gi
What NOT to do:
As noted, this will match any two vowels together. Leaving this answer as an example of what NOT to do. The correct answer (in this case) is to use backreferences as mentioned in numerous other answers.
import re
data = ["took","waaeek","raaag"]
for s in data:
print re.sub(r'([aeiou]){2}',r'\1:',s)
This matches exactly two occurrences {2} of any member of the set [aeiou]. and replaces it with the vowel, captured with the parens () and placed in the sub string by the \1 followed by a ':'
Output:
to:k
wa:e:k
ra:ag
You'll need to use a back reference in your search expression. Try something like: ([a-z])+\1 (or ([a-z])\1 for just a double).
Related
Sorry if this question seems too similar to other's I have found. This is a variation of using re.sub to replace exact characters in a string.
I have a string that looks like:
C1([*:5])C([*:6])C2=NC1=C([*:1])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5
I would like to only replace, for example, the '*:1' with 'Ar'. My current attempt looks like this:
smiles_all='C1([*:5])C([*:6])C2=NC1=C([*:1])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5'
print(smiles_all)
new_smiles=re.sub('[*:]1','Ar',smiles_all)
print(new_smiles)
C1([*:5])C([*:6])C2=NC1=C([*Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*Ar0])C(=N4)C([*:3])=C5C([*Ar1])=C([*Ar2])C(=C2([*:4]))N5
As you can see, this is still changing the values that were previously 10,11, etc. I've tried different variations where I select [*:1], but that is also incorrect. Any help here would be greatly appreciated. In my current output, the * also remains. That needs to be swapped so that *:1 becomes Ar
Here is an example of what the output should be
C1([*:5])C([*:6])C2=NC1=C([Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5
*Edit:
At one point this question was flagged as answered by this question:
Escaping regex string
When I implement re.escape as suggested, I still get an error:
new_smiles=re.sub(re.escape('*:1'),'Ar',smiles_all)
C1([*:5])C([*:6])C2=NC1=C([*:1])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5
C1([*:5])C([*:6])C2=NC1=C([Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([Ar0])C(=N4)C([*:3])=C5C([Ar1])=C([Ar2])C(=C2([*:4]))N5
Given:
smiles_all='C1([*:5])C([*:6])C2=NC1=C([*:1])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5'
desired='C1([*:5])C([*:6])C2=NC1=C([Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5'
You are trying to replace the literal string [*:1] with [Ar]. In a regex, the expression [*:1] is a character class that matches a single one of the characters inside the class with one match. If you add any regex repetition to a character class, it will match those characters in any order up to the repetition limit.
The easiest way to to replace the literal [*:1] with [Ar] is to use Python's string methods:
>>> smiles_all.replace('[*:1]','[Ar]')==desired
True
If you want to use a regex, you need to escape those metacharaters to get a literal string:
>>> re.sub(r'\[\*:1\]', "[Ar]", smiles_all)==desired
True
Or let Python do the escaping for you:
>>> re.sub(re.escape(r'[*:1]'), "[Ar]", smiles_all)==desired
True
You can try:
re.sub(r"[*:]+1(?=])", "Ar", smiles_all)
Difference from yours is to allow 1+ repetitions of literal * and : followed by 1 which is also ensured to be followed by a ] via the ?=, i.e., positive lookahead.
to get
"C1([*:5])C([*:6])C2=NC1=C([Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5"
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.
Is there any way to directly replace all groups using regex syntax?
The normal way:
re.match(r"(?:aaa)(_bbb)", string1).group(1)
But I want to achieve something like this:
re.match(r"(\d.*?)\s(\d.*?)", "(CALL_GROUP_1) (CALL_GROUP_2)")
I want to build the new string instantaneously from the groups the Regex just captured.
Have a look at re.sub:
result = re.sub(r"(\d.*?)\s(\d.*?)", r"\1 \2", string1)
This is Python's regex substitution (replace) function. The replacement string can be filled with so-called backreferences (backslash, group number) which are replaced with what was matched by the groups. Groups are counted the same as by the group(...) function, i.e. starting from 1, from left to right, by opening parentheses.
The accepted answer is perfect. I would add that group reference is probably better achieved by using this syntax:
r"\g<1> \g<2>"
for the replacement string. This way, you work around syntax limitations where a group may be followed by a digit. Again, this is all present in the doc, nothing new, just sometimes difficult to spot at first sight.
i am using python with re module for regular expressions
i want to remove all the characters from the string except numbers and characters.
To achieve this i am using sub function
Code Snippet:-
>>> text="foo.bar"
>>> re.sub("[^A-Z][^a-z]","",text)
'fobar'
I wanted to know why above expression removes the "o."?
I am not able to understand why it removes the "o"
Can someone please explain me what is going on behind this?
I know to correct solution of this problem is
>>> re.sub("[^A-Z ^a-z]","",text)
'foobar'
Thanks in advance
A very important aspect to realize is that [^A-Z][^a-z] represents two characters (one for each character group), while [^A-Za-z] represents only one.
Explained in detail:
The [^A-Z] means all characters except uppercase A to Z, the second o in foo.bar is not uppercase so it matches as a matter of fact everything in foo.bar is matched at this point.
Then you add [^a-z] so you look for a character that is not lowercase, only the dot matches.
Combine both and you look for a non-uppercase character followed by a non-lowercase character so this matches o.
The solution is the one proposed by Ignacio.
Because o matches [^A-Z] and . matches [^a-z].
And the correct solution is [^A-Za-z0-9].
One rule that I need is that if the last vowel (aeiou) of a string is before a character from the set ('t','k','s','tk'), then a : needs to be added right after the vowel.
So, in Python if I have the string "orchestras" I need a rule that will turn it into "orchestra:s"
edit: The (t, k, s, tk) would be the final character(s) in the string
re.sub(r"([aeiou])(t|k|s|tk)([^aeiou]*)$", r"\1:\2\3", "orchestras")
re.sub(r"([aeiou])(t|k|s|tk)$", r"\1:\2", "orchestras")
You don't say if there can be other consonants after the t/k/s/tk. The first regex allows for this as long as there aren't any more vowels, so it'll change "fist" to "fi:st" for instance. If the word must end with the t/k/s/tk then use the second regex, which will do nothing for "fist".
If you have not figured it out yet, I recommend trying [python_root]/tools/scripts/redemo.py It is a nice testing area.
Another take on the replacement regex:
re.sub("(?<=[aeiou])(?=(?:t|k|s|tk)$)", ":", "orchestras")
This one does not need to replace using remembered groups.