How to find the last non-digit in a string using python3
0
I use:
a_string = '00:00:02'
for char in reversed(a_string):
if not char.isdigit():
break
and "char" will be the last non-digit, but it feels hackish. Is there a more elegant way?
string python-3.x
asked Nov 27 '18 at 14:26
user3765030user3765030
133
add a comment |
0
I use:
a_string = '00:00:02'
for char in reversed(a_string):
if not char.isdigit():
break
and "char" will be the last non-digit, but it feels hackish. Is there a more elegant way?
string python-3.x
asked Nov 27 '18 at 14:26
user3765030user3765030
133
Yes, the list comprehension is my winner. Thank you!
– user3765030
Dec 4 '18 at 10:04
add a comment |
0
0
0
I use:
a_string = '00:00:02'
for char in reversed(a_string):
if not char.isdigit():
break
and "char" will be the last non-digit, but it feels hackish. Is there a more elegant way?
string python-3.x
asked Nov 27 '18 at 14:26
user3765030user3765030
133
I use:
a_string = '00:00:02'
for char in reversed(a_string):
if not char.isdigit():
break
and "char" will be the last non-digit, but it feels hackish. Is there a more elegant way?
string python-3.x
string python-3.x
asked Nov 27 '18 at 14:26
user3765030user3765030
133
asked Nov 27 '18 at 14:26
user3765030user3765030
133
asked Nov 27 '18 at 14:26
user3765030user3765030
133
asked Nov 27 '18 at 14:26
user3765030user3765030
133
asked Nov 27 '18 at 14:26
user3765030user3765030
133
133
Yes, the list comprehension is my winner. Thank you!
– user3765030
Dec 4 '18 at 10:04
add a comment |
Yes, the list comprehension is my winner. Thank you!
– user3765030
Dec 4 '18 at 10:04
Yes, the list comprehension is my winner. Thank you!
– user3765030
Dec 4 '18 at 10:04
Yes, the list comprehension is my winner. Thank you!
– user3765030
Dec 4 '18 at 10:04
add a comment |
1 Answer
1
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0
Apart from a dedicated function? Pick any:
# regex
print (re.findall(r'(D)d*$', a_string)[0])
# list comprehension
print ([x for x in a_string if not x.isdigit()][-1])
# filter
print ([x for x in filter(lambda x: not x.isdigit(), a_string)][-1])
answered Nov 27 '18 at 14:43
usr2564301usr2564301
17.8k73370
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
Apart from a dedicated function? Pick any:
# regex
print (re.findall(r'(D)d*$', a_string)[0])
# list comprehension
print ([x for x in a_string if not x.isdigit()][-1])
# filter
print ([x for x in filter(lambda x: not x.isdigit(), a_string)][-1])
answered Nov 27 '18 at 14:43
usr2564301usr2564301
17.8k73370
add a comment |
0
Apart from a dedicated function? Pick any:
# regex
print (re.findall(r'(D)d*$', a_string)[0])
# list comprehension
print ([x for x in a_string if not x.isdigit()][-1])
# filter
print ([x for x in filter(lambda x: not x.isdigit(), a_string)][-1])
answered Nov 27 '18 at 14:43
usr2564301usr2564301
17.8k73370
add a comment |
0
0
0
Apart from a dedicated function? Pick any:
# regex
print (re.findall(r'(D)d*$', a_string)[0])
# list comprehension
print ([x for x in a_string if not x.isdigit()][-1])
# filter
print ([x for x in filter(lambda x: not x.isdigit(), a_string)][-1])
answered Nov 27 '18 at 14:43
usr2564301usr2564301
17.8k73370
Apart from a dedicated function? Pick any:
# regex
print (re.findall(r'(D)d*$', a_string)[0])
# list comprehension
print ([x for x in a_string if not x.isdigit()][-1])
# filter
print ([x for x in filter(lambda x: not x.isdigit(), a_string)][-1])
answered Nov 27 '18 at 14:43
usr2564301usr2564301
17.8k73370
answered Nov 27 '18 at 14:43
usr2564301usr2564301
17.8k73370
answered Nov 27 '18 at 14:43
usr2564301usr2564301
17.8k73370
answered Nov 27 '18 at 14:43
usr2564301usr2564301
17.8k73370
17.8k73370
add a comment |
add a comment |
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Yes, the list comprehension is my winner. Thank you!
– user3765030
Dec 4 '18 at 10:04