I have a string with multiple successive instances of , (comma+space) that I want to replace with a single instance. Is there a clean way to do so? I suppose RegEx can be helpful.
A naive example:
s = 'a, b, , c, , , d, , , e, , , , , , , f
The desired output:
'a, b, c, d, e, f
Naturally, the text can change, so the search should be for successive instances of ,.
So the regular expression searches for two or more instances of , (comma + space) and then in sub function you replace it with only a single ,.
import re
pattern = re.compile(r'(,\s){2,}')
test_string = 'a, b, , c, , , d, , , e, , , , , , , f'
print re.sub(pattern, ', ', test_string)
>>> a, b, c, d, e, f
and without a regular expression (as @Casimir et Hippolyte suggested in comment)
test_string = 'a, b, , c, , , d, , , e, , , , , , , f'
test_string_parts = test_string.split(',')
test_string_parts = [part.strip() for part in test_string_parts if part != ' ']
print ', '.join(test_string_parts)
>>> a, b, c, d, e, f
re.sub(r'(,\s){2,}', ', ', test_string) ? Or, is there some notable difference or corner case?
compile here: stackoverflow.com/questions/452104/…', ' I have '\n' how would I adapt this find & replace? (\\n){2,} seems to work in regex101.com, but not in Python. Any suggestions?r'\n{2,}'?You can use reduce:
>>> from functools import reduce
>>> reduce( (lambda x, y: x+', '+y if y else x), s.split(', '))
(Where x is the carry and y the item)
the simplest way for your problem would be:
>>> s = 'a, b, , c, , , d, , , e, , , , , , , f'
>>> s = [x for x in s if x.isalpha()]
>>> print(s)
['a', 'b', 'c', 'd', 'e', 'f']
then, use join()
>>> ', '.join(s)
'a, b, c, d, e, f'
do it in one line:
>>> s = ', '.join([x for x in s if x.isalpha()])
>>> s
'a, b, c, d, e, f'
Just figure other way:
>>> s = 'a, b, , c, , , d, , , e, , , , , , , f'
>>> s = s.split() #split all ' '(<- space)
>>> s
['a,', 'b,', ',', 'c,', ',', ',', 'd,', ',', ',', 'e,', ',', ',', ',', ',', ',', ',', 'f']
>>> while ',' in s:
... s.remove(',')
>>> s
['a,', 'b,', 'c,', 'd,', 'e,', 'f']
>>> ''.join(s)
'a,b,c,d,e,f'
s = ", , a, b, , c, , , d, , , e, , , , , , , f,,,,"
s = [o for o in s.replace(' ', '').split(',') if len(o)]
print (s)
One more solution: go through the combination of the list and the same list shifted by one (in other words, by the pairs of consecutive items) and select the second item from each pair where the first (previous) item differs from the second (next) item:
s = 'a. b. . c. . . d. . . e. . . . . . . f'
test = []
for i in s:
if i != ' ':
test.append(i)
res = [test[0]] + [y for x,y in zip(test, test[1:]) if x!=y]
for x in res:
print(x, end='')
yields
a.b.c.d.e.f
[Program finished]
