I haven't read a lot into the other examples, but I have just created this method for consolidating multiple consecutive space characters.
It does not use any libraries, and whilst it is relatively long in terms of script length, it is not a complex implementation:
def spaceMatcher(command):
"""
Function defined to consolidate multiple whitespace characters in
strings to a single space
"""
# Initiate index to flag if more than one consecutive character
iteration
space_match = 0
space_char = ""
for char in command:
if char == " ":
space_match += 1
space_char += " "
elif (char != " ") & (space_match > 1):
new_command = command.replace(space_char, " ")
space_match = 0
space_char = ""
elif char != " ":
space_match = 0
space_char = ""
return new_command
command = None
command = str(input("Please enter a command ->"))
print(spaceMatcher(command))
print(list(spaceMatcher(command)))
Solution for Python developers:
import re
text1 = 'Python Exercises Are Challenging Exercises'
print("Original string: ", text1)
print("Without extra spaces: ", re.sub(' +', ' ', text1))
Output:
Original string: Python Exercises Are Challenging Exercises
Without extra spaces: Python Exercises Are Challenging Exercises
This does and will do: :)
# python... 3.x
import operator
...
# line: line of text
return " ".join(filter(lambda a: operator.is_not(a, ""), line.strip().split(" ")))
import re
Text = " You can select below trims for removing white space!! BR Aliakbar "
# trims all white spaces
print('Remove all space:',re.sub(r"\s+", "", Text), sep='')
# trims left space
print('Remove leading space:', re.sub(r"^\s+", "", Text), sep='')
# trims right space
print('Remove trailing spaces:', re.sub(r"\s+$", "", Text), sep='')
# trims both
print('Remove leading and trailing spaces:', re.sub(r"^\s+|\s+$", "", Text), sep='')
# replace more than one white space in the string with one white space
print('Remove more than one space:',re.sub(' +', ' ',Text), sep='')
Result:
Remove all space:Youcanselectbelowtrimsforremovingwhitespace!!BRAliakbar
Remove leading space:You can select below trims for removing white space!! BR Aliakbar
Remove trailing spaces: You can select below trims for removing white space!! BR Aliakbar
Remove leading and trailing spaces:You can select below trims for removing white space!! BR Aliakbar
Remove more than one space: You can select below trims for removing white space!! BR Aliakbar
" ".join(foo.split()) is not quite correct with respect to the question asked because it also entirely removes single leading and/or trailing white spaces. So, if they shall also be replaced by 1 blank, you should do something like the following:
" ".join(('*' + foo + '*').split()) [1:-1]
Of course, it's less elegant.
In some cases it's desirable to replace consecutive occurrences of every whitespace character with a single instance of that character. You'd use a regular expression with backreferences to do that.
(\s)\1{1,} matches any whitespace character, followed by one or more occurrences of that character. Now, all you need to do is specify the first group (\1) as the replacement for the match.
Wrapping this in a function:
import re
def normalize_whitespace(string):
return re.sub(r'(\s)\1{1,}', r'\1', string)
>>> normalize_whitespace('The fox jumped over the log.')
'The fox jumped over the log.'
>>> normalize_whitespace('First line\t\t\t \n\n\nSecond line')
'First line\t \nSecond line'
One line of code to remove all extra spaces before, after, and within a sentence:
sentence = " The fox jumped over the log. "
sentence = ' '.join(filter(None,sentence.split(' ')))
Explanation:
*The remaining elements should be words or words with punctuations, etc. I did not test this extensively, but this should be a good starting point. All the best!
To remove white space, considering leading, trailing and extra white space in between words, use:
(?<=\s) +|^ +(?=\s)| (?= +[\n\0])
The first or deals with leading white space, the second or deals with start of string leading white space, and the last one deals with trailing white space.
For proof of use, this link will provide you with a test.
https://regex101.com/r/meBYli/4
This is to be used with the re.split function.
I have tried the following method and it even works with the extreme case like:
str1=' I live on earth '
' '.join(str1.split())
But if you prefer a regular expression it can be done as:
re.sub('\s+', ' ', str1)
Although some preprocessing has to be done in order to remove the trailing and ending space.
Using regexes with "\s" and doing simple string.split()'s will also remove other whitespace - like newlines, carriage returns, tabs. Unless this is desired, to only do multiple spaces, I present these examples.
I used 11 paragraphs, 1000 words, 6665 bytes of Lorem Ipsum to get realistic time tests and used random-length extra spaces throughout:
original_string = ''.join(word + (' ' * random.randint(1, 10)) for word in lorem_ipsum.split(' '))
The one-liner will essentially do a strip of any leading/trailing spaces, and it preserves a leading/trailing space (but only ONE ;-).
# setup = '''
import re
def while_replace(string):
while ' ' in string:
string = string.replace(' ', ' ')
return string
def re_replace(string):
return re.sub(r' {2,}' , ' ', string)
def proper_join(string):
split_string = string.split(' ')
# To account for leading/trailing spaces that would simply be removed
beg = ' ' if not split_string[ 0] else ''
end = ' ' if not split_string[-1] else ''
# versus simply ' '.join(item for item in string.split(' ') if item)
return beg + ' '.join(item for item in split_string if item) + end
original_string = """Lorem ipsum ... no, really, it kept going... malesuada enim feugiat. Integer imperdiet erat."""
assert while_replace(original_string) == re_replace(original_string) == proper_join(original_string)
#'''
# while_replace_test
new_string = original_string[:]
new_string = while_replace(new_string)
assert new_string != original_string
# re_replace_test
new_string = original_string[:]
new_string = re_replace(new_string)
assert new_string != original_string
# proper_join_test
new_string = original_string[:]
new_string = proper_join(new_string)
assert new_string != original_string
NOTE: The " Keep in mind that the main while version" made a copy of the original_string, as I believe once modified on the first run, successive runs would be faster (if only by a bit). As this adds time, I added this string copy to the other two so that the times showed the difference only in the logic.stmt on timeit instances will only be executed once; the original way I did this, the while loop worked on the same label, original_string, thus the second run, there would be nothing to do. The way it's set up now, calling a function, using two different labels, that isn't a problem. I've added assert statements to all the workers to verify we change something every iteration (for those who may be dubious). E.g., change to this and it breaks:
# while_replace_test
new_string = original_string[:]
new_string = while_replace(new_string)
assert new_string != original_string # will break the 2nd iteration
while ' ' in original_string:
original_string = original_string.replace(' ', ' ')
Tests run on a laptop with an i5 processor running Windows 7 (64-bit).
timeit.Timer(stmt = test, setup = setup).repeat(7, 1000)
test_string = 'The fox jumped over\n\t the log.' # trivial
Python 2.7.3, 32-bit, Windows
test | minum | maximum | average | median
---------------------+------------+------------+------------+-----------
while_replace_test | 0.001066 | 0.001260 | 0.001128 | 0.001092
re_replace_test | 0.003074 | 0.003941 | 0.003357 | 0.003349
proper_join_test | 0.002783 | 0.004829 | 0.003554 | 0.003035
Python 2.7.3, 64-bit, Windows
test | minum | maximum | average | median
---------------------+------------+------------+------------+-----------
while_replace_test | 0.001025 | 0.001079 | 0.001052 | 0.001051
re_replace_test | 0.003213 | 0.004512 | 0.003656 | 0.003504
proper_join_test | 0.002760 | 0.006361 | 0.004626 | 0.004600
Python 3.2.3, 32-bit, Windows
test | minum | maximum | average | median
---------------------+------------+------------+------------+-----------
while_replace_test | 0.001350 | 0.002302 | 0.001639 | 0.001357
re_replace_test | 0.006797 | 0.008107 | 0.007319 | 0.007440
proper_join_test | 0.002863 | 0.003356 | 0.003026 | 0.002975
Python 3.3.3, 64-bit, Windows
test | minum | maximum | average | median
---------------------+------------+------------+------------+-----------
while_replace_test | 0.001444 | 0.001490 | 0.001460 | 0.001459
re_replace_test | 0.011771 | 0.012598 | 0.012082 | 0.011910
proper_join_test | 0.003741 | 0.005933 | 0.004341 | 0.004009
test_string = lorem_ipsum
# Thanks to http://www.lipsum.com/
# "Generated 11 paragraphs, 1000 words, 6665 bytes of Lorem Ipsum"
Python 2.7.3, 32-bit
test | minum | maximum | average | median
---------------------+------------+------------+------------+-----------
while_replace_test | 0.342602 | 0.387803 | 0.359319 | 0.356284
re_replace_test | 0.337571 | 0.359821 | 0.348876 | 0.348006
proper_join_test | 0.381654 | 0.395349 | 0.388304 | 0.388193
Python 2.7.3, 64-bit
test | minum | maximum | average | median
---------------------+------------+------------+------------+-----------
while_replace_test | 0.227471 | 0.268340 | 0.240884 | 0.236776
re_replace_test | 0.301516 | 0.325730 | 0.308626 | 0.307852
proper_join_test | 0.358766 | 0.383736 | 0.370958 | 0.371866
Python 3.2.3, 32-bit
test | minum | maximum | average | median
---------------------+------------+------------+------------+-----------
while_replace_test | 0.438480 | 0.463380 | 0.447953 | 0.446646
re_replace_test | 0.463729 | 0.490947 | 0.472496 | 0.468778
proper_join_test | 0.397022 | 0.427817 | 0.406612 | 0.402053
Python 3.3.3, 64-bit
test | minum | maximum | average | median
---------------------+------------+------------+------------+-----------
while_replace_test | 0.284495 | 0.294025 | 0.288735 | 0.289153
re_replace_test | 0.501351 | 0.525673 | 0.511347 | 0.508467
proper_join_test | 0.422011 | 0.448736 | 0.436196 | 0.440318
For the trivial string, it would seem that a while-loop is the fastest, followed by the Pythonic string-split/join, and regex pulling up the rear.
For non-trivial strings, seems there's a bit more to consider. 32-bit 2.7? It's regex to the rescue! 2.7 64-bit? A while loop is best, by a decent margin. 32-bit 3.2, go with the "proper" join. 64-bit 3.3, go for a while loop. Again.
In the end, one can improve performance if/where/when needed, but it's always best to remember the mantra:
IANAL, YMMV, Caveat Emptor!
Another alternative:
>>> import re
>>> str = 'this is a string with multiple spaces and tabs'
>>> str = re.sub('[ \t]+' , ' ', str)
>>> print str
this is a string with multiple spaces and tabs
This also seems to work:
while " " in s:
s = s.replace(" ", " ")
Where the variable s represents your string.
I have to agree with Paul McGuire's comment. To me,
' '.join(the_string.split())
is vastly preferable to whipping out a regex.
My measurements (Linux and Python 2.5) show the split-then-join to be almost five times faster than doing the "re.sub(...)", and still three times faster if you precompile the regex once and do the operation multiple times. And it is by any measure easier to understand -- much more Pythonic.
Because @pythonlarry asked here are the missing generator based versions
The groupby join is easy. Groupby will group elements consecutive with same key. And return pairs of keys and list of elements for each group. So when the key is an space an space is returne else the entire group.
from itertools import groupby
def group_join(string):
return ''.join(' ' if chr==' ' else ''.join(times) for chr,times in groupby(string))
The group by variant is simple but very slow. So now for the generator variant. Here we consume an iterator, the string, and yield all chars except chars that follow an char.
def generator_join_generator(string):
last=False
for c in string:
if c==' ':
if not last:
last=True
yield ' '
else:
last=False
yield c
def generator_join(string):
return ''.join(generator_join_generator(string))
So i meassured the timings with some other lorem ipsum.
With Hello and World separated by 64KB of spaces
Not forget the original sentence
Interesting here for nearly space only strings group join is not that worse Timing showing always median from seven runs of a thousand times each.
I've got a simple method without splitting:
a = "Lorem Ipsum Darum Diesrum!"
while True:
count = a.find(" ")
if count > 0:
a = a.replace(" ", " ")
count = a.find(" ")
continue
else:
break
print(a)
foo is your string:
" ".join(foo.split())
Be warned though this removes "all whitespace characters (space, tab, newline, return, formfeed)" (thanks to hhsaffar, see comments). I.e., "this is \t a test\n" will effectively end up as "this is a test".
Similar to the previous solutions, but more specific: replace two or more spaces with one:
>>> import re
>>> s = "The fox jumped over the log."
>>> re.sub('\s{2,}', ' ', s)
'The fox jumped over the log.'
You can also use the string splitting technique in a Pandas DataFrame without needing to use .apply(..), which is useful if you need to perform the operation quickly on a large number of strings. Here it is on one line:
df['message'] = (df['message'].str.split()).str.join(' ')
A simple soultion
>>> import re
>>> s="The fox jumped over the log."
>>> print re.sub('\s+',' ', s)
The fox jumped over the log.
import re
s = "The fox jumped over the log."
re.sub("\s\s+" , " ", s)
or
re.sub("\s\s+", " ", s)
since the space before comma is listed as a pet peeve in PEP 8, as mentioned by user Martin Thoma in the comments.
def unPretty(S):
# Given a dictionary, JSON, list, float, int, or even a string...
# return a string stripped of CR, LF replaced by space, with multiple spaces reduced to one.
return ' '.join(str(S).replace('\n', ' ').replace('\r', '').split())
The fastest you can get for user-generated strings is:
if ' ' in text:
while ' ' in text:
text = text.replace(' ', ' ')
The short circuiting makes it slightly faster than pythonlarry's comprehensive answer. Go for this if you're after efficiency and are strictly looking to weed out extra whitespaces of the single space variety.
I have my simple method which I have used in college.
line = "I have a nice day."
end = 1000
while end != 0:
line.replace(" ", " ")
end -= 1
This will replace every double space with a single space and will do it 1000 times. It means you can have 2000 extra spaces and will still work. :)
import re
string = re.sub('[ \t\n]+', ' ', 'The quick brown \n\n \t fox')
This will remove all the tabs, new lines and multiple white spaces with single white space.
sentence = "The fox jumped over the log."
word = sentence.split()
result = ""
for string in word:
result += string+" "
print(result)
Quite surprising - no one posted simple function which will be much faster than ALL other posted solutions. Here it goes:
def compactSpaces(s):
os = ""
for c in s:
if c != " " or (os and os[-1] != " "):
os += c
return os
string = 'This is a string full of spaces and taps'
string = string.split(' ')
while '' in string:
string.remove('')
string = ' '.join(string)
print(string)
Results:
This is a string full of spaces and taps
If it's whitespace you're dealing with, splitting on None will not include an empty string in the returned value.
Source: Stackoverflow.com