>>>user_sentence = "hello \t how are you?"
>>>import re
>>>user_sentenceSplit = re.findall(r"([\s]|[\w']+|[.,!?;])",user_sentence)
>>>print user_sentenceSplit
I get ['hello', '\t', 'how', 'are', 'you', '?']
I don't know how to create any code that will replace the '\t' with 'tab'.
I do not believe that replacing \t in the original string will ever work, you have two issues:
Your code also outputs spaces as tokens, but you do not want to have them
The \t in between letters will become a part of a word token.
So, you need to replace [\s] with [^\S ] pattern that matches any whitespace but a regular space (add more excluded whitespace symbols if necessary into the negated character class) and you need to iterate through all the tokens and check if a token is equal to a tab, and then replace it with tab value. So, the best is to use re.finditer and push the found values into a list variable, see sample code below:
import re
user_sentence = "hello \t how are you?"
user_sentenceSplit = []
for x in re.finditer(r"[^\S ]|[\w']+|[.,!?;]",user_sentence):
if x.group() == "\t": # if it is a tab, replace the value
user_sentenceSplit.append("tab")
else: # else, push the match value
user_sentenceSplit.append(x.group())
print(user_sentenceSplit)
See the Python demo
I think str.replace would do the job.
user_sentence.replace('\t', 'tab')
Do this before splitting the string.
It is behavior of Python's compiler. You should not be worrying about it. Pyhton's Compiler store tab as \t. You need not to do anything on it as it will treat it as tab while performing any action over it. For example:
>>> my_string = 'Yes Hello So?' # <- String with tab
>>> my_string
'Yes\tHello\tSo?' # <- Stored tab as '\t'
>>> print my_string
Yes Hello So? # While printing, again tab
However you exact requirement is not clear to me. In case you want to replace the value of \t with tab string, you may do:
>>> my_string = my_string.replace('\t', 'tab')
>>> my_string
'YestabHellotabSo?'
where my_string is holding the value I mentioned in previous example.
Related
I have a String that looks like
test = '20170125NBCNightlyNews'
I am trying to split it into two parts, the digits, and the name. The format will always be [date][show] the date is stripped of format and is digit only in the direction of YYYYMMDD (dont think that matters)
I am trying to use re. I have a working version by writing.
re.split('(\d+)',test)
Simple enough, this gives me the values I need in a list.
['', '20170125', 'NBCNightlyNews']
However, as you will note, there is an empty string in the first position. I could theoretically just ignore it, but I want to learn why it is there in the first place, and if/how I can avoid it.
I also tried telling it to match the begininning of the string as well, and got the same results.
>>> re.split('(^\d+)',test)
['', '20170125', 'NBCNightlyNews']
>>> re.split('^(\d+)',test)
['', '20170125', 'NBCNightlyNews']
>>>
Does anyone have any input as to why this is there / how I can avoid the empty string?
Other answers have explained why what you're doing does what it does, but if you have a constant format for the date, there is no reason to abuse a re.split to parse this data:
test[:8], test[8:]
Will split your strings just fine.
What you are actually doing by entering re.split('(^\d+)', test) is, that your test string is splitted on any occurence of a number with at least one character.
So, if you have
test = '20170125NBCNightlyNews'
This is happening:
20170125 NBCNightlyNews
^^^^^^^^
The string is split into three parts, everything before the number, the number itself and everything after the number.
Maybe it is easier to understand if you have a sentence of words, separated by a whitespace character.
re.split(' ', 'this is a house')
=> ['this', 'is', 'a', 'house']
re.split(' ', ' is a house')
=> ['', 'is', 'a', 'house']
You're getting an empty result in the beginning because your input string starts with digits and you're splitting it by digits only. Hence you get an empty string which is before first set of digits.
To avoid that you can use filter:
>>> print filter(None, re.split('(\d+)',test))
['20170125', 'NBCNightlyNews']
Why re.split when you can just match and get the groups?...
import re
test = '20170125NBCNightlyNews'
pattern = re.compile('(\d+)(\w+)')
result = re.match(pattern, test)
result.groups()[0] # for the date part
result.groups()[1] # for the show name
I realize now the intention was to parse the text, not fix the regex usage. I'm with the others, you shouldn't use regex for this simple task when you already know the format won't change and the date is fixed size and will always be first. Just use string indexing.
From the documentation:
If there are capturing groups in the separator and it matches at the start of the string, the result will start with an empty string. The same holds for the end of the string. That way, separator components are always found at the same relative indices within the result list.
So if you have:
test = 'test20170125NBCNightlyNews'
The indexes would remain unaffected:
>>>re.split('(\d+)',test)
['test', '20170125', 'NBCNightlyNews']
If the date is always 8 digits long, I would access the substrings directly (without using regex):
>>> [test[:8], test[8:]]
['20170125', 'NBCNightlyNews']
If the length of the date might vary, I would use:
>>> s = re.search('^(\d*)(.*)$', test)
>>> [s.group(1), s.group(2)]
['20170125', 'NBCNightlyNews']
I have a string like:
text1 = 'python...is...fun...'
I want to replace the multiple '.'s to one '.' only when they are at the end of the string, i want the output to be:
python...is...fun.
So when there is only one '.' at the end of the string, then it won't be replaced
text2 = 'python...is...fun.'
and the output is just the same as text2
My regex is like this:
text = re.sub(r'(.*)\.{2,}$', r'\1.', text)
which i want to match any string then {2 to n} of '.' at the end of the string, but the output is:
python...is...fun..
any ideas how to do this?
Thx in advance!
You are making it a bit complex, you can easily do it by using regex as \.+$ and replace the regex pattern with single . character.
>>> text1 = 'python...is...fun...'
>>> new_text = re.sub(r"\.+$", ".", text1)
>>> 'python...is...fun.'
You may extend this regex further to handle the cases with input such as ... only, etc but the main concept was that there is no need to counting the number of ., as you have done in your answer.
Just look for the string ending with three periods, and replace them with a single one.
import re
x = "foo...bar...quux..."
print(re.sub('\.{2,}$', '.', x))
// foo...bar...quux.
import re
print(re.sub(r'\.{2,}$', '.', 'I...love...python...'))
As simple as that. Note that you need to escape the . because otherwise, it means whichever char
except \n.
I want to replace the multiple '.'s to one '.' only when they are at
the end of the string
For such simple case it's easier to substitute without importing re module, checking the value of the last 3 characters:
text1 = 'python...is...fun...'
text1 = text1[:-2] if text1[-3:] == '...' else text1
print(text1)
The output:
python...is...fun.
Basically, I'm asking the user to input a string of text into the console, but the string is very long and includes many line breaks. How would I take the user's string and delete all line breaks to make it a single line of text. My method for acquiring the string is very simple.
string = raw_input("Please enter string: ")
Is there a different way I should be grabbing the string from the user? I'm running Python 2.7.4 on a Mac.
P.S. Clearly I'm a noob, so even if a solution isn't the most efficient, the one that uses the most simple syntax would be appreciated.
How do you enter line breaks with raw_input? But, once you have a string with some characters in it you want to get rid of, just replace them.
>>> mystr = raw_input('please enter string: ')
please enter string: hello world, how do i enter line breaks?
>>> # pressing enter didn't work...
...
>>> mystr
'hello world, how do i enter line breaks?'
>>> mystr.replace(' ', '')
'helloworld,howdoienterlinebreaks?'
>>>
In the example above, I replaced all spaces. The string '\n' represents newlines. And \r represents carriage returns (if you're on windows, you might be getting these and a second replace will handle them for you!).
basically:
# you probably want to use a space ' ' to replace `\n`
mystring = mystring.replace('\n', ' ').replace('\r', '')
Note also, that it is a bad idea to call your variable string, as this shadows the module string. Another name I'd avoid but would love to use sometimes: file. For the same reason.
You can try using string replace:
string = string.replace('\r', '').replace('\n', '')
You can split the string with no separator arg, which will treat consecutive whitespace as a single separator (including newlines and tabs). Then join using a space:
In : " ".join("\n\nsome text \r\n with multiple whitespace".split())
Out: 'some text with multiple whitespace'
https://docs.python.org/2/library/stdtypes.html#str.split
The canonic answer, in Python, would be :
s = ''.join(s.splitlines())
It splits the string into lines (letting Python doing it according to its own best practices). Then you merge it. Two possibilities here:
replace the newline by a whitespace (' '.join())
or without a whitespace (''.join())
updated based on Xbello comment:
string = my_string.rstrip('\r\n')
read more here
Another option is regex:
>>> import re
>>> re.sub("\n|\r", "", "Foo\n\rbar\n\rbaz\n\r")
'Foobarbaz'
If anybody decides to use replace, you should try r'\n' instead '\n'
mystring = mystring.replace(r'\n', ' ').replace(r'\r', '')
A method taking into consideration
additional white characters at the beginning/end of string
additional white characters at the beginning/end of every line
various end-line characters
it takes such a multi-line string which may be messy e.g.
test_str = '\nhej ho \n aaa\r\n a\n '
and produces nice one-line string
>>> ' '.join([line.strip() for line in test_str.strip().splitlines()])
'hej ho aaa a'
UPDATE:
To fix multiple new-line character producing redundant spaces:
' '.join([line.strip() for line in test_str.strip().splitlines() if line.strip()])
This works for the following too
test_str = '\nhej ho \n aaa\r\n\n\n\n\n a\n '
Regular expressions is the fastest way to do this
s='''some kind of
string with a bunch\r of
extra spaces in it'''
re.sub(r'\s(?=\s)','',re.sub(r'\s',' ',s))
result:
'some kind of string with a bunch of extra spaces in it'
The problem with rstrip() is that it does not work in all cases (as I myself have seen few). Instead you can use
text = text.replace("\n"," ")
This will remove all new line '\n' with a space.
You really don't need to remove ALL the signs: lf cr crlf.
# Pythonic:
r'\n', r'\r', r'\r\n'
Some texts must have breaks, but you probably need to join broken lines to keep particular sentences together.
Therefore it is natural that line breaking happens after priod, semicolon, colon, but not after comma.
My code considers above conditions. Works well with texts copied from pdfs.
Enjoy!:
def unbreak_pdf_text(raw_text):
""" the newline careful sign removal tool
Args:
raw_text (str): string containing unwanted newline signs: \\n or \\r or \\r\\n
e.g. imported from OCR or copied from a pdf document.
Returns:
_type_: _description_
"""
pat = re.compile((r"[, \w]\n|[, \w]\r|[, \w]\r\n"))
breaks = re.finditer(pat, raw_text)
processed_text = raw_text
raw_text = None
for i in breaks:
processed_text = processed_text.replace(i.group(), i.group()[0]+" ")
return processed_text
After replacing all word characters in a string with the character '^', using re.sub("\w", "^" , stringorphrase) I'm left with :
>>> '^^^ ^^ ^^^^'
Is there any way to remove the single quotes so it looks cleaner?
>>> ^^^ ^^ ^^^^
Are you sure it's just not how it's displayed in the interactive prompt or something (and there aren't actually apost's in your string)?
If the ' is actually part of the string, and is first/last then either:
string = string.strip("'")
or:
string = string[1:-1] # lop ending characters off
Use the print statement. The quotes aren't actually part of the string.
To remove all occurrences of single quotes:
mystr = some_string_with_single_quotes
answer = mystr.replace("'", '')
To remove single quotes ONLY at the ends of the string:
mystr = some_string_with_single_quotes
answer = mystr.strip("'")
Hope this helps
I am close but I am not sure what to do with the restuling match object. If I do
p = re.search('[/#.* /]', str)
I'll get any words that start with # and end up with a space. This is what I want. However this returns a Match object that I dont' know what to do with. What's the most computationally efficient way of finding and returning a string which is prefixed with a #?
For example,
"Hi there #guy"
After doing the proper calculations, I would be returned
guy
The following regular expression do what you need:
import re
s = "Hi there #guy"
p = re.search(r'#(\w+)', s)
print p.group(1)
It will also work for the following string formats:
s = "Hi there #guy " # notice the trailing space
s = "Hi there #guy," # notice the trailing comma
s = "Hi there #guy and" # notice the next word
s = "Hi there #guy22" # notice the trailing numbers
s = "Hi there #22guy" # notice the leading numbers
That regex does not do what you think it does.
s = "Hi there #guy"
p = re.search(r'#([^ ]+)', s) # this is the regex you described
print p.group(1) # first thing matched inside of ( .. )
But as usually with regex, there are tons of examples that break this, for example if the text is s = "Hi there #guy, what's with the comma?" the result would be guy,.
So you really need to think about every possible thing you want and don't want to match. r'#([a-zA-Z]+)' might be a good starting point, it literally only matches letters (a .. z, no unicode etc).
p.group(0) should return guy. If you want to find out what function an object has, you can use the dir(p) method to find out. This will return a list of attributes and methods that are available for that object instance.
As it's evident from the answers so far regex is the most efficient solution for your problem. Answers differ slightly regarding what you allow to be followed by the #:
[^ ] anything but space
\w in python-2.x is equivalent to [A-Za-z0-9_], in py3k is locale dependent
If you have better idea what characters might be included in the user name you might adjust your regex to reflect that, e.g., only lower case ascii letters, would be:
[a-z]
NB: I skipped quantifiers for simplicity.
(?<=#)\w+
will match a word if it's preceded by a # (without adding it to the match, a so-called positive lookbehind). This will match "words" that are composed of letters, numbers, and/or underscore; if you don't want those, use (?<=#)[^\W\d_]+
In Python:
>>> strg = "Hi there #guy!"
>>> p = re.search(r'(?<=#)\w+', strg)
>>> p.group()
'guy'
You say: """If I do p = re.search('[/#.* /]', str) I'll get any words that start with # and end up with a space."" But this is incorrect -- that pattern is a character class which will match ONE character in the set #/.* and space. Note: there's a redundant second / in the pattern.
For example:
>>> re.findall('[/#.* /]', 'xxx#foo x/x.x*x xxxx')
['#', ' ', '/', '.', '*', ' ']
>>>
You say that you want "guy" returned from "Hi there #guy" but that conflicts with "and end up with a space".
Please edit your question to include what you really want/need to match.