This is how the string splitting works for me right now:
output = string.encode('UTF8').split('}/n}')[0]
output += '}\n}'
But I am wondering if there is a more pythonic way to do it.
The goal is to get everything before this '}/n}' including '}/n}'.
This might be a good use of str.partition.
string = '012za}/n}ddfsdfk'
parts = string.partition('}/n}')
# ('012za', '}/n}', 'ddfsdfk')
''.join(parts[:-1])
# 012za}/n}
Or, you can find it explicitly with str.index.
repl = '}/n}'
string[:string.index(repl) + len(repl)]
# 012za}/n}
This is probably better than using str.find since an exception will be raised if the substring isn't found, rather than producing nonsensical results.
It seems like anything "more elegant" would require regular expressions.
import re
re.search('(.*?}/n})', string).group(0)
# 012za}/n}
It can be done with with re.split() -- the key is putting parens around the split pattern to preserve what you split on:
import re
output = "".join(re.split(r'(}/n})', string.encode('UTF8'))[:2])
However, I doubt that this is either the most efficient nor most Pythonic way to achieve what you want. I.e. I don't think this is naturally a split sort of problem. For example:
tag = '}/n}'
encoded = string.encode('UTF8')
output = encoded[:encoded.index(tag)] + tag
or if you insist on a one-liner:
output = (lambda string, tag: string[:string.index(tag)] + tag)(string.encode('UTF8'), '}/n}')
or returning to regex:
output = re.match(r".*}/n}", string.encode('UTF8')).group(0)
>>> string_to_split = 'first item{\n{second item'
>>> sep = '{\n{'
>>> output = [item + sep for item in string_to_split.split(sep)]
NOTE: output = ['first item{\n{', 'second item{\n{']
then you can use the result:
for item_with_delimiter in output:
...
It might be useful to look up os.linesep if you're not sure what the line ending will be. os.linesep is whatever the line ending is under your current OS, so '\r\n' under Windows or '\n' under Linux or Mac. It depends where input data is from, and how flexible your code needs to be across environments.
Adapted from Slice a string after a certain phrase?, you can combine find and slice to get the first part of the string and retain }/n}.
str = "012za}/n}ddfsdfk"
str[:str.find("}/n}")+4]
Will result in 012za}/n}
Related
What are the most efficient ways to extract text from a string? Are there some available functions or regex expressions, or some other way?
For example, my string is below and I want to extract the IDs as well
as the ScreenNames, separately.
[User(ID=1234567890, ScreenName=RandomNameHere), User(ID=233323490, ScreenName=AnotherRandomName), User(ID=4459284, ScreenName=YetAnotherName)]
Thank you!
Edit: These are the text strings that I want to pull. I want them to be in a list.
Target_IDs = 1234567890, 233323490, 4459284
Target_ScreenNames = RandomNameHere, AnotherRandomName, YetAnotherName
import re
str = '[User(ID=1234567890, ScreenName=RandomNameHere), User(ID=233323490, ScreenName=AnotherRandomName), User(ID=4459284, ScreenName=YetAnotherName)]'
print 'Target IDs = ' + ','.join( re.findall(r'ID=(\d+)', str) )
print 'Target ScreenNames = ' + ','.join( re.findall(r' ScreenName=(\w+)', str) )
Output :
Target IDs = 1234567890,233323490,4459284
Target ScreenNames = RandomNameHere,AnotherRandomName,YetAnotherName
It depends. Assuming that all your text comes in the form of
TagName = TagValue1, TagValue2, ...
You need just two calls to split.
tag, value_string = string.split('=')
values = value_string.split(',')
Remove the excess space (probably a couple of rstrip()/lstrip() calls will suffice) and you are done. Or you can take regex. They are slightly more powerful, but in this case I think it's a matter of personal taste.
If you want more complex syntax with nonterminals, terminals and all that, you'll need lex/yacc, which will require some background in parsers. A rather interesting thing to play with, but not something you'll want to use for storing program options and such.
The regex I'd use would be:
(?:ID=|ScreenName=)+(\d+|[\w\d]+)
However, this assumes that ID is only digits (\d) and usernames are only letters or numbers ([\w\d]).
This regex (when combined with re.findall) would return a list of matches that could be iterated through and sorted in some fashion like so:
import re
s = "[User(ID=1234567890, ScreenName=RandomNameHere), User(ID=233323490, ScreenName=AnotherRandomName), User(ID=4459284, ScreenName=YetAnotherName)]"
pattern = re.compile(r'(?:ID=|ScreenName=)+(\d+|[\w\d]+)');
ids = []
names = []
for p in re.findall(pattern, s):
if p.isnumeric():
ids.append(p)
else:
names.append(p)
print(ids, names)
Let's say I have a string that looks like this:
myStr = '(Txt_l1 (Txt_l2)) or (Txt2_l1 (Txt2_l2))'
What I would like to obtain in the end would be:
myStr_l1 = '(Txt_l1) or (Txt2_l1)'
and
myStr_l2 = '(Txt_l2) or (Txt2_l2)'
Some properties:
all "Txt_"-elements of the string start with an uppercase letter
the string can contain much more elements (so there could also be Txt3, Txt4,...)
the suffixes '_l1' and '_l2' look different in reality; they cannot be used for matching (I chose them for demonstration purposes)
I found a way to get the first part done by using:
myStr_l1 = re.sub('\(\w+\)','',myStr)
which gives me
'(Txt_l1 ) or (Txt2_l1 )'
However, I don't know how to obtain myStr_l2. My idea was to remove everything between two open parentheses. But when I do something like this:
re.sub('\(w+\(', '', myStr)
the entire string is returned.
re.sub('\(.*\(', '', myStr)
removes - of course - far too much and gives me
'Txt2_l2))'
Does anyone have an idea how to get myStr_l2?
When there is an "and" instead of an "or", the strings look slightly different:
myStr2 = '(Txt_l1 (Txt_l2) and Txt2_l1 (Txt2_l2))'
Then I can still use the command from above:
re.sub('\(\w+\)','',myStr2)
which gives:
'(Txt_l1 and Txt2_l1 )'
but I again fail to get myStr2_l2. How would I do this for these kind of strings?
And how would one then do this for mixed expressions with "and" and "or" e.g. like this:
myStr3 = '(Txt_l1 (Txt_l2) and Txt2_l1 (Txt2_l2)) or (Txt3_l1 (Txt3_l2) and Txt4_l1 (Txt2_l2))'
re.sub('\(\w+\)','',myStr3)
gives me
'(Txt_l1 and Txt2_l1 ) or (Txt3_l1 and Txt4_l1 )'
but again: How would I obtain myStr3_l2?
Regexp is not powerful enough for nested expressions (in your case: nested elements in parentheses). You will have to write a parser. Look at https://pyparsing.wikispaces.com/
I'm not entirely sure what you want but I wrote this to strip everything between the parenthesis.
import re
mystr = '(Txt_l1 (Txt_l2)) or (Txt2_l1 (Txt2_l2))'
sets = mystr.split(' or ')
noParens = []
for line in sets:
mat = re.match(r'\((.* )\((.*\)\))', line, re.M)
if mat:
noParens.append(mat.group(1))
noParens.append(mat.group(2).replace(')',''))
print(noParens)
This takes all the parenthesis away and puts your elements in a list. Here's an alternate way of doing it without using Regular Expressions.
mystr = '(Txt_l1 (Txt_l2)) or (Txt2_l1 (Txt2_l2))'
noParens = []
mystr = mystr.replace(' or ', ' ')
mystr = mystr.replace(')','')
mystr = mystr.replace('(','')
noParens = mystr.split()
print(noParens)
I have a string in python as below:
"\\B1\\B1xxA1xxMdl1zzInoAEROzzMofIN"
I want to get the string as
"B1xxA1xxMdl1zzInoAEROzzMofIN"
I think this can be done using regex but could not achieve it yet. Please give me an idea.
st = "\B1\B1xxA1xxMdl1zzInoAEROzzMofIN"
s = re.sub(r"\\","",st)
idx = s.rindex("B1")
print s[idx:]
output = 'B1xxA1xxMdl1zzInoAEROzzMofIN'
OR
st = "\B1\B1xxA1xxMdl1zzInoAEROzzMofIN"
idx = st.rindex("\\")
print st[idx+1:]
output = 'B1xxA1xxMdl1zzInoAEROzzMofIN'
Here is a try:
import re
s = "\\B1\\B1xxA1xxMdl1zzInoAEROzzMofIN"
s = re.sub(r"\\[^\\]+\\","", s)
print s
Tested on http://py-ide-online.appspot.com (couldn't find a way to share though)
[EDIT] For some explanation, have a look at the Python regex documentation page and the first comment of this SO question:
How to remove symbols from a string with Python?
because using brackets [] can be tricky (IMHO)
In this case, [^\\] means anything but two backslashes \\.
So [^\\]+ means one or more character that matches anything but two backslashes \\.
If the desired section of the string is always on the RHS of a \ char then you could use:
string = "\\B1\\B1xxA1xxMdl1zzInoAEROzzMofIN"
string.rpartition("\\")[2]
output = 'B1xxA1xxMdl1zzInoAEROzzMofIN'
How to find values in string, add specific value to each of them and replace output with fixed string.
import re
def _replace(content):
#x = float(content.group(4))+20
#y = float(content.group(6))+20
return content.group(6)
print re.sub('<g(\s)transform="matrix\((.*)(\s)(.*)(\s)(.*)\)\"', _replace, '<g transform="matrix(0.412445 -0.0982513 0.0982513 0.412445 -5.77618 67.0025)">')
First off, I should repeat the usual warning about not parsing XML with regexes. It's a bad idea, and it will never work for all cases. If you're actually trying to parse the full xml document, use an XML parser.
That having been said, I'm guilty of doing quick and dirty stuff like this all the time. If you really just need a one-off solution, a simple regex can often get the job done. Just be aware that it will come back to haunt you as soon as you run into something more complex!
Next, I confess to not being much of a regex wiz, but here's how I'd modify your code snippet:
import re
def _replace(content):
values = [float(val) for val in content.group(2).split()]
values[3] += 20
values[5] += 100
values = ['{0}'.format(val) for val in values]
return content.group(1) + ' '.join(values) + content.group(3)
test_string = '<g transform="matrix(0.412445 -0.0982513 0.0982513 0.412445 -5.77618 67.0025)">'
pattern = r'(transform=\"matrix\()(.*?)(\))'
print test_string
print re.sub(pattern, _replace, test_string)
I am wanting to verify and then parse this string (in quotes):
string = "start: c12354, c3456, 34526; other stuff that I don't care about"
//Note that some codes begin with 'c'
I would like to verify that the string starts with 'start:' and ends with ';'
Afterward, I would like to have a regex parse out the strings. I tried the following python re code:
regx = r"start: (c?[0-9]+,?)+;"
reg = re.compile(regx)
matched = reg.search(string)
print ' matched.groups()', matched.groups()
I have tried different variations but I can either get the first or the last code but not a list of all three.
Or should I abandon using a regex?
EDIT: updated to reflect part of the problem space I neglected and fixed string difference.
Thanks for all the suggestions - in such a short time.
In Python, this isn’t possible with a single regular expression: each capture of a group overrides the last capture of that same group (in .NET, this would actually be possible since the engine distinguishes between captures and groups).
Your easiest solution is to first extract the part between start: and ; and then using a regular expression to return all matches, not just a single match, using re.findall('c?[0-9]+', text).
You could use the standard string tools, which are pretty much always more readable.
s = "start: c12354, c3456, 34526;"
s.startswith("start:") # returns a boolean if it starts with this string
s.endswith(";") # returns a boolean if it ends with this string
s[6:-1].split(', ') # will give you a list of tokens separated by the string ", "
This can be done (pretty elegantly) with a tool like Pyparsing:
from pyparsing import Group, Literal, Optional, Word
import string
code = Group(Optional(Literal("c"), default='') + Word(string.digits) + Optional(Literal(","), default=''))
parser = Literal("start:") + OneOrMore(code) + Literal(";")
# Read lines from file:
with open('lines.txt', 'r') as f:
for line in f:
try:
result = parser.parseString(line)
codes = [c[1] for c in result[1:-1]]
# Do something with teh codez...
except ParseException exc:
# Oh noes: string doesn't match!
continue
Cleaner than a regular expression, returns a list of codes (no need to string.split), and ignores any extra characters in the line, just like your example.
import re
sstr = re.compile(r'start:([^;]*);')
slst = re.compile(r'(?:c?)(\d+)')
mystr = "start: c12354, c3456, 34526; other stuff that I don't care about"
match = re.match(sstr, mystr)
if match:
res = re.findall(slst, match.group(0))
results in
['12354', '3456', '34526']