I'm not sure how to exactly convey what I'm trying to do, but I'm trying to create a function to split off a part of my string (the leading whitespace) so that I can edit it with different parts of my script, then add it again to my string after it has been altered.
So lets say I have the string:
" That's four spaces"
I want to split it so I end up with:
" " and "That's four spaces"
You can use re.match:
>>> import re
>>> re.match('(\s*)(.*)', " That's four spaces").groups()
(' ', "That's four spaces")
>>>
(\s*) captures zero or more whitespace characters at the start of the string and (.*) gets everything else.
Remember though that strings are immutable in Python. Technically, you cannot edit their contents; you can only create new string objects.
For a non-Regex solution, you could try something like this:
>>> mystr = " That's four spaces"
>>> n = next(i for i, c in enumerate(mystr) if c != ' ') # Count spaces at start
>>> (' ' * n, mystr[n:])
(' ', "That's four spaces")
>>>
The main tools here are next, enumerate, and a generator expression. This solution is probably faster than the Regex one, but I personally think that the first is more elegant.
Why don't you try matching instead of splitting?
>>> import re
>>> s = " That's four spaces"
>>> re.findall(r'^\s+|.+', s)
[' ', "That's four spaces"]
Explanation:
^\s+ Matches one or more spaces at the start of a line.
| OR
.+ Matches all the remaining characters.
One solution is to lstrip the string, then figure out how many characters you've removed. You can then 'modify' the string as desired and finish by adding the whitespace back to your string. I don't think this would work properly with tab characters, but for spaces only it seems to get the job done:
my_string = " That's four spaces"
no_left_whitespace = my_string.lstrip()
modified_string = no_left_whitespace + '!'
index = my_string.index(no_left_whitespace)
final_string = (' ' * index) + modified_string
print(final_string) # That's four spaces!
And a simple test to ensure that we've done it right, which passes:
assert final_string == my_string + '!'
One thing you can do it make a list out of string.that is
x=" That's four spaces"
y=list(x)
z="".join(y[0:4]) #if this is variable you can apply a loop over here to detect spaces from start
k="".join(y[4:])
s=[]
s.append(z)
s.append(k)
print s
This is a non regex solution which will not require any imports
Related
I want to eliminate all the whitespace from a string, on both ends, and in between words.
I have this Python code:
def my_handle(self):
sentence = ' hello apple '
sentence.strip()
But that only eliminates the whitespace on both sides of the string. How do I remove all whitespace?
If you want to remove leading and ending spaces, use str.strip():
>>> " hello apple ".strip()
'hello apple'
If you want to remove all space characters, use str.replace() (NB this only removes the “normal” ASCII space character ' ' U+0020 but not any other whitespace):
>>> " hello apple ".replace(" ", "")
'helloapple'
If you want to remove duplicated spaces, use str.split() followed by str.join():
>>> " ".join(" hello apple ".split())
'hello apple'
To remove only spaces use str.replace:
sentence = sentence.replace(' ', '')
To remove all whitespace characters (space, tab, newline, and so on) you can use split then join:
sentence = ''.join(sentence.split())
or a regular expression:
import re
pattern = re.compile(r'\s+')
sentence = re.sub(pattern, '', sentence)
If you want to only remove whitespace from the beginning and end you can use strip:
sentence = sentence.strip()
You can also use lstrip to remove whitespace only from the beginning of the string, and rstrip to remove whitespace from the end of the string.
An alternative is to use regular expressions and match these strange white-space characters too. Here are some examples:
Remove ALL spaces in a string, even between words:
import re
sentence = re.sub(r"\s+", "", sentence, flags=re.UNICODE)
Remove spaces in the BEGINNING of a string:
import re
sentence = re.sub(r"^\s+", "", sentence, flags=re.UNICODE)
Remove spaces in the END of a string:
import re
sentence = re.sub(r"\s+$", "", sentence, flags=re.UNICODE)
Remove spaces both in the BEGINNING and in the END of a string:
import re
sentence = re.sub("^\s+|\s+$", "", sentence, flags=re.UNICODE)
Remove ONLY DUPLICATE spaces:
import re
sentence = " ".join(re.split("\s+", sentence, flags=re.UNICODE))
(All examples work in both Python 2 and Python 3)
"Whitespace" includes space, tabs, and CRLF. So an elegant and one-liner string function we can use is str.translate:
Python 3
' hello apple '.translate(str.maketrans('', '', ' \n\t\r'))
OR if you want to be thorough:
import string
' hello apple'.translate(str.maketrans('', '', string.whitespace))
Python 2
' hello apple'.translate(None, ' \n\t\r')
OR if you want to be thorough:
import string
' hello apple'.translate(None, string.whitespace)
For removing whitespace from beginning and end, use strip.
>> " foo bar ".strip()
"foo bar"
' hello \n\tapple'.translate({ord(c):None for c in ' \n\t\r'})
MaK already pointed out the "translate" method above. And this variation works with Python 3 (see this Q&A).
In addition, strip has some variations:
Remove spaces in the BEGINNING and END of a string:
sentence= sentence.strip()
Remove spaces in the BEGINNING of a string:
sentence = sentence.lstrip()
Remove spaces in the END of a string:
sentence= sentence.rstrip()
All three string functions strip lstrip, and rstrip can take parameters of the string to strip, with the default being all white space. This can be helpful when you are working with something particular, for example, you could remove only spaces but not newlines:
" 1. Step 1\n".strip(" ")
Or you could remove extra commas when reading in a string list:
"1,2,3,".strip(",")
Be careful:
strip does a rstrip and lstrip (removes leading and trailing spaces, tabs, returns and form feeds, but it does not remove them in the middle of the string).
If you only replace spaces and tabs you can end up with hidden CRLFs that appear to match what you are looking for, but are not the same.
eliminate all the whitespace from a string, on both ends, and in between words.
>>> import re
>>> re.sub("\s+", # one or more repetition of whitespace
'', # replace with empty string (->remove)
''' hello
... apple
... ''')
'helloapple'
https://en.wikipedia.org/wiki/Whitespace_character
Python docs:
https://docs.python.org/library/stdtypes.html#textseq
https://docs.python.org/library/stdtypes.html#str.replace
https://docs.python.org/library/string.html#string.replace
https://docs.python.org/library/re.html#re.sub
https://docs.python.org/library/re.html#regular-expression-syntax
I use split() to ignore all whitespaces and use join() to concatenate
strings.
sentence = ''.join(' hello apple '.split())
print(sentence) #=> 'helloapple'
I prefer this approach because it is only a expression (not a statement).
It is easy to use and it can use without binding to a variable.
print(''.join(' hello apple '.split())) # no need to binding to a variable
import re
sentence = ' hello apple'
re.sub(' ','',sentence) #helloworld (remove all spaces)
re.sub(' ',' ',sentence) #hello world (remove double spaces)
In the following script we import the regular expression module which we use to substitute one space or more with a single space. This ensures that the inner extra spaces are removed. Then we use strip() function to remove leading and trailing spaces.
# Import regular expression module
import re
# Initialize string
a = " foo bar "
# First replace any number of spaces with a single space
a = re.sub(' +', ' ', a)
# Then strip any leading and trailing spaces.
a = a.strip()
# Show results
print(a)
I found that this works the best for me:
test_string = ' test a s test '
string_list = [s.strip() for s in str(test_string).split()]
final_string = ' '.join(string_array)
# final_string: 'test a s test'
It removes any whitespaces, tabs, etc.
try this.. instead of using re i think using split with strip is much better
def my_handle(self):
sentence = ' hello apple '
' '.join(x.strip() for x in sentence.split())
#hello apple
''.join(x.strip() for x in sentence.split())
#helloapple
I want to be able to remove all punctuation and single quotes ' from a string, unless the single quote ' is in the middle of a word.
At this point I have the following code:
with open('test.txt','r') as f:
for line in f:
line = line.lower()
line = re.sub('[^a-z\ \']+', " ", line)
print line
if there happens to be a line in test.txt like:
Here is some stuff. 'Now there are quotes.' Now there's not.
The result I want is:
here is some stuff now there are quotes now there's not
But the result I get is:
here is some stuff 'now there are quotes' now there's not
How can I remove the single quotes ' from a string if they're at the beginning or end of the word but not in the middle? Thanks for the help!
Split the string, use strip() on each word to remove leading and trailing characters on it, then join it all back together.
>>> s = "'here is some stuff 'now there are quotes' now there's not'"
>>> print(' '.join(w.strip("'") for w in s.split()).lower())
here is some stuff now there are quotes now there's not
Using regular expressions, you could first remove 's that don't follow a letter, then remove 's that don't precede a letter (thus only keeping ones that both follow and precede a letter):
line = "Here is some stuff. 'Now there are quotes.' Now there's not."
print re.sub(r"'([^A-Za-z])", r"\1", re.sub(r"([^A-Za-z])'", r"\1", line))
# Here is some stuff. Now there are quotes. Now there's not.
Probably more efficient to do it #TigerhawkT3's way. Though they produce different results if you have something like 'this'. If you want to remove that second ' too, then the regular expressions method is probably the simplest you can do.
Here's another solution using regular expressions with lookarounds.
This method will preserve any whitespace your string may have.
import re
rgx = re.compile(r"(?<!\w)\'|\'(?!\w)")
# Regex explanation:
# (?<!\w)\' match any quote not preceded by a word
# | or
# \'(?!\w) match any quote not followed by a word
s = "'here is some stuff 'now there are quotes' now there's not'"
print rgx.sub('', s) # here is some stuff now there are quotes now there's not
If a word is a sequence of 1+ letters, digits and underscores that can be matched with \w+ you may use
re.sub(r"(?!\b'\b)'", "", text)
See the regex demo. Here, ' is matched when it is not preceded nor followed with letters/digits/_.
Or, if words are strictly linguistic words that only consist of letters, use
re.sub(r"'(?!(?<=[a-zA-Z]')[a-zA-Z])", "", text) # ASCII only
re.sub(r"'(?!(?<=[^\W\d_]')[^\W\d_])", "", text) # any Unicode letter support
See Demo #2 (ASCII only letters) and Demo #3 (see last line in the demo text). Here, ' is only matched if it is not preceded nor followed with a letter (ASCII or any).
Python demo:
import re
text = "'text... 'some quotes', there's none'. three 'four' can't, '2'4', '_'_', 'l'école'"
print( re.sub(r"(?!\b'\b)'", "", text) )
# => text... some quotes, there's none. three four can't, 2'4, _'_, l'école
print( re.sub(r"'(?!(?<=[a-zA-Z]')[a-zA-Z])", "", text) )
# => text... some quotes, there's none. three four can't, 24, __, lécole
print( re.sub(r"'(?!(?<=[^\W\d_]')[^\W\d_])", "", text) )
# => text... some quotes, there's none. three four can't, 24, __, l'école
Here is complete solution to remove whatever you don't want in a string:
def istext (text):
ok = 0
for x in text: ok += x.isalnum()
return ok>0
def stripit (text, ofwhat):
for x in ofwhat: text = text.strip(x)
return text
def purge (text, notwanted="'\"!#$%&/()=?*+-.,;:_<>|\\[]{}"):
text = text.splitlines()
text = [" ".join([stripit(word, notwanted) for word in line.split() if istext(word)]) for line in text]
return "\n".join(text)
>>> print purge("'Nice, .to, see! you. Isn't it?'")
Nice to see you Isn't it
Note: this will kill all whitespaces too and transform them to space or remove them completely.
I want to replace more than one white spaces from string with "#".
If one white space is there it should be intact but if there is more than one consecutive whitespace then it will keep one and append #. For example
s = "Hello how are you."
would become
"Hello how #are ##you"
Python 2.7:
import re
s = "Hello how are you"
s = re.sub("(?<= ) ", "#" ,s)
print s
Or, if you want to have only one # signifying "multiple spaces", change to ("(?<= ) +", "#" ,s)
Explanation: The regex contains a positive lookbehind (?<= ) : it only finds spaces that are preceded by another space, but does not include the first space in the results. Because of that, when the results are replaced by an #, the first space remains intact (it is not preceded by another one), all the others are replaced in a one-by-one fashion by #.
Adding + to the main expression means that it will collect all multiple spaces except for the first one (due to positive lookbehind) and replace them with a single #.
This pattern will only get the " ", if you want to cover tabs, too, you'd need to change to \s
Edit: Having looked at the source of your question, my answer didn't fix it. I couldn't see the extra spaces in your source string. A regular expression as sg.sysel suggests will do the job nicely.
In case you did want to do it yourself with a loop:
def addats(s):
i = 0
r = ''
for c in s:
if c == ' ':
if i > 0:
r += '#'
else:
r += ' '
i += 1
else:
r += c
i = 0
return r
Note that for a real application, you should use something mutable like a list instead of a string for r there, but this should solve your immediate problem.
I use part of code to read a website and scrap some information and place it into Google and print some directions.
I'm having an issue as some of the information. the site i use sometimes adds a # followed by 3 random numbers then a / and another 3 numbers e.g #037/100
how can i use python to ignore this "#037/100" string?
I currently use
for i, part in enumerate(list(addr_p)):
if '#' in part:
del addr_p[i]
break
to remove the # if found but I'm not sure how to do it for the random numbers
Any ideas ?
If you find yourself wanting to remove "three digits followed by a forward slash followed by three digits" from a string s, you could do
import re
s = "this is a string #123/234 with other stuff"
t = re.sub('#\d{3}\/\d{3}', '', s)
print t
Result:
'this is a string with other stuff'
Explanation:
# - literal character '#'
\d{3} - exactly three digits
\/ - forward slash (escaped since it can have special meaning)
\d{3} - exactly three digits
And the whole thing that matches the above (if it's present) is replaced with '' - i.e. "removed".
import re
re.sub('#[0-9]+\/[0-9]+$', '', addr_p[i])
I'm no wizzard with regular expressions but i'd imagine you could so something like this.
You could even handle '#' in the regexp as well.
If the format is always the same, then you could check if the line starts with a #, then set the string to itself without the first 8 characters.
if part[0:1] == '#':
part = part[8:]
if the first letter is a #, it sets the string to itself, from the 8th character to the end.
I'd double your problems and match against a regular expression for this.
import re
regex = re.compile(r'([\w\s]+)#\d+\/\d+([\w\s]+)')
m = regex.match('This is a string with a #123/987 in it')
if m:
s = m.group(1) + m.group(2)
print(s)
A more concise way:
import re
s = "this is a string #123/234 with other stuff"
t = re.sub(r'#\S+', '', s)
print(t)
I would like to replace strings like 'HDMWhoSomeThing' to 'HDM Who Some Thing' with regex.
So I would like to extract words which starts with an upper-case letter or consist of upper-case letters only. Notice that in the string 'HDMWho' the last upper-case letter is in the fact the first letter of the word Who - and should not be included in the word HDM.
What is the correct regex to achieve this goal? I have tried many regex' similar to [A-Z][a-z]+ but without success. The [A-Z][a-z]+ gives me 'Who Some Thing' - without 'HDM' of course.
Any ideas?
Thanks,
Rukki
#! /usr/bin/env python
import re
from collections import deque
pattern = r'([A-Z]{2,}(?=[A-Z]|$)|[A-Z](?=[a-z]|$))'
chunks = deque(re.split(pattern, 'HDMWhoSomeMONKEYThingXYZ'))
result = []
while len(chunks):
buf = chunks.popleft()
if len(buf) == 0:
continue
if re.match(r'^[A-Z]$', buf) and len(chunks):
buf += chunks.popleft()
result.append(buf)
print ' '.join(result)
Output:
HDM Who Some MONKEY Thing XYZ
Judging by lines of code, this task is a much more natural fit with re.findall:
pattern = r'([A-Z]{2,}(?=[A-Z]|$)|[A-Z][a-z]*)'
print ' '.join(re.findall(pattern, 'HDMWhoSomeMONKEYThingX'))
Output:
HDM Who Some MONKEY Thing X
Try to split with this regular expression:
/(?=[A-Z][a-z])/
And if your regular expression engine does not support splitting empty matches, try this regular expression to put spaces between the words:
/([A-Z])(?![A-Z])/
Replace it with " $1" (space plus match of the first group). Then you can split at the space.
one liner :
' '.join(a or b for a,b in re.findall('([A-Z][a-z]+)|(?:([A-Z]*)(?=[A-Z]))',s))
using regexp
([A-Z][a-z]+)|(?:([A-Z]*)(?=[A-Z]))
So 'words' in this case are:
Any number of uppercase letters - unless the last uppercase letter is followed by a lowercase letter.
One uppercase letter followed by any number of lowercase letters.
so try:
([A-Z]+(?![a-z])|[A-Z][a-z]*)
The first alternation includes a negative lookahead (?![a-z]), which handles the boundary between an all-caps word and an initial caps word.
May be '[A-Z]*?[A-Z][a-z]+'?
Edit: This seems to work: [A-Z]{2,}(?![a-z])|[A-Z][a-z]+
import re
def find_stuff(str):
p = re.compile(r'[A-Z]{2,}(?![a-z])|[A-Z][a-z]+')
m = p.findall(str)
result = ''
for x in m:
result += x + ' '
print result
find_stuff('HDMWhoSomeThing')
find_stuff('SomeHDMWhoThing')
Prints out:
HDM Who Some Thing
Some HDM Who Thing