Python: Removing spaces from list objects
Question:
I have a list of objects appended from a mysql database and contain spaces. I wish to remove the spaces such as below, but the code im using doesnt work?
hello = ['999 ',' 666 ']
k = []
for i in hello:
str(i).replace(' ','')
k.append(i)
print k
Answers:
result = map(str.strip, hello)
Strings in Python are immutable (meaning that their data cannot be modified) so the replace method doesn’t modify the string – it returns a new string. You could fix your code as follows:
for i in hello:
j = i.replace(' ','')
k.append(j)
However a better way to achieve your aim is to use a list comprehension. For example the following code removes leading and trailing spaces from every string in the list using strip:
hello = [x.strip(' ') for x in hello]
String methods return the modified string.
k = [x.replace(' ', '') for x in hello]
replace() does not operate in-place, you need to assign its result to something. Also, for a more concise syntax, you could supplant your for loop with a one-liner: hello_no_spaces = map(lambda x: x.replace(' ', ''), hello)
Presuming that you don’t want to remove internal spaces:
def normalize_space(s):
"""Return s stripped of leading/trailing whitespace
and with internal runs of whitespace replaced by a single SPACE"""
# This should be a str method :-(
return ' '.join(s.split())
replacement = [normalize_space(i) for i in hello]
List comprehension [num.strip() for num in hello] is the fastest.
>>> import timeit
>>> hello = ['999 ',' 666 ']
>>> t1 = lambda: map(str.strip, hello)
>>> timeit.timeit(t1)
1.825870468015296
>>> t2 = lambda: list(map(str.strip, hello))
>>> timeit.timeit(t2)
2.2825958750515269
>>> t3 = lambda: [num.strip() for num in hello]
>>> timeit.timeit(t3)
1.4320335103944899
>>> t4 = lambda: [num.replace(' ', '') for num in hello]
>>> timeit.timeit(t4)
1.7670568718943969
for element in range(0,len(hello)):
d[element] = hello[element].strip()
I have a list of objects appended from a mysql database and contain spaces. I wish to remove the spaces such as below, but the code im using doesnt work?
hello = ['999 ',' 666 ']
k = []
for i in hello:
str(i).replace(' ','')
k.append(i)
print k
result = map(str.strip, hello)
Strings in Python are immutable (meaning that their data cannot be modified) so the replace method doesn’t modify the string – it returns a new string. You could fix your code as follows:
for i in hello:
j = i.replace(' ','')
k.append(j)
However a better way to achieve your aim is to use a list comprehension. For example the following code removes leading and trailing spaces from every string in the list using strip:
hello = [x.strip(' ') for x in hello]
String methods return the modified string.
k = [x.replace(' ', '') for x in hello]
replace() does not operate in-place, you need to assign its result to something. Also, for a more concise syntax, you could supplant your for loop with a one-liner: hello_no_spaces = map(lambda x: x.replace(' ', ''), hello)
Presuming that you don’t want to remove internal spaces:
def normalize_space(s):
"""Return s stripped of leading/trailing whitespace
and with internal runs of whitespace replaced by a single SPACE"""
# This should be a str method :-(
return ' '.join(s.split())
replacement = [normalize_space(i) for i in hello]
List comprehension [num.strip() for num in hello] is the fastest.
>>> import timeit
>>> hello = ['999 ',' 666 ']
>>> t1 = lambda: map(str.strip, hello)
>>> timeit.timeit(t1)
1.825870468015296
>>> t2 = lambda: list(map(str.strip, hello))
>>> timeit.timeit(t2)
2.2825958750515269
>>> t3 = lambda: [num.strip() for num in hello]
>>> timeit.timeit(t3)
1.4320335103944899
>>> t4 = lambda: [num.replace(' ', '') for num in hello]
>>> timeit.timeit(t4)
1.7670568718943969
for element in range(0,len(hello)):
d[element] = hello[element].strip()