Converting unified version in python using regular expression
I'm trying to convert a unified version format to the following in python
Examples:
<2.2.1 || >=4.0.0 < 4.1.9
should be<2.2.1 || (>=4.0.0 && <4.1.9)>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12
should be(>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6
should be(>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
tried the following, it works but messy:
def rchop(thestring, ending):
if thestring.endswith(ending):
return thestring[:-len(ending)]
return thestring
ver = "<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6"
split_ver = ver.split('||')
list_data =
for version in split_ver:
version = version.rstrip()
version = version.lstrip()
vv = version.replace(" ", " && ")
list_data.append(vv)
print(list_data)
new_list =
for data in list_data:
if "&&" not in data and "=0" not in data and ">=" not in data:
new_data = "(>=0 && " + data + ")"
new_list.append(new_data)
else:
new_data1 = new_list.append("("+data+")")
final_list =
for items in new_list:
data = final_list.append(items + " || ")
now_data = [''.join(final_list[:])]
data1 = rchop(now_data[0], ' || ')
print(data1)
python regex
asked Nov 13 '18 at 22:12
Jau LJau L
3501413
add a comment |
I'm trying to convert a unified version format to the following in python
Examples:
<2.2.1 || >=4.0.0 < 4.1.9
should be<2.2.1 || (>=4.0.0 && <4.1.9)>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12
should be(>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6
should be(>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
tried the following, it works but messy:
def rchop(thestring, ending):
if thestring.endswith(ending):
return thestring[:-len(ending)]
return thestring
ver = "<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6"
split_ver = ver.split('||')
list_data =
for version in split_ver:
version = version.rstrip()
version = version.lstrip()
vv = version.replace(" ", " && ")
list_data.append(vv)
print(list_data)
new_list =
for data in list_data:
if "&&" not in data and "=0" not in data and ">=" not in data:
new_data = "(>=0 && " + data + ")"
new_list.append(new_data)
else:
new_data1 = new_list.append("("+data+")")
final_list =
for items in new_list:
data = final_list.append(items + " || ")
now_data = [''.join(final_list[:])]
data1 = rchop(now_data[0], ' || ')
print(data1)
python regex
asked Nov 13 '18 at 22:12
Jau LJau L
3501413
May we know what have you tried and what problems you faced?
– boonwj
Nov 13 '18 at 22:32
It is unclear what the requirement is here, would you please put some test code to reproduce the problem?
– Saul Cruz
Nov 13 '18 at 22:46
the requirements here is that the script I'm using produces buggy version format, so I'm trying to fix it using a regular expression by adding proper parentheses and proper && as the above conversions.
– Jau L
Nov 13 '18 at 22:52
1
I added the code @boonwj
– Jau L
Nov 13 '18 at 23:54
add a comment |
I'm trying to convert a unified version format to the following in python
Examples:
<2.2.1 || >=4.0.0 < 4.1.9
should be<2.2.1 || (>=4.0.0 && <4.1.9)>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12
should be(>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6
should be(>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
tried the following, it works but messy:
def rchop(thestring, ending):
if thestring.endswith(ending):
return thestring[:-len(ending)]
return thestring
ver = "<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6"
split_ver = ver.split('||')
list_data =
for version in split_ver:
version = version.rstrip()
version = version.lstrip()
vv = version.replace(" ", " && ")
list_data.append(vv)
print(list_data)
new_list =
for data in list_data:
if "&&" not in data and "=0" not in data and ">=" not in data:
new_data = "(>=0 && " + data + ")"
new_list.append(new_data)
else:
new_data1 = new_list.append("("+data+")")
final_list =
for items in new_list:
data = final_list.append(items + " || ")
now_data = [''.join(final_list[:])]
data1 = rchop(now_data[0], ' || ')
print(data1)
python regex
asked Nov 13 '18 at 22:12
Jau LJau L
3501413
I'm trying to convert a unified version format to the following in python
Examples:
<2.2.1 || >=4.0.0 < 4.1.9
should be<2.2.1 || (>=4.0.0 && <4.1.9)>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12
should be(>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6
should be(>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
tried the following, it works but messy:
def rchop(thestring, ending):
if thestring.endswith(ending):
return thestring[:-len(ending)]
return thestring
ver = "<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6"
split_ver = ver.split('||')
list_data =
for version in split_ver:
version = version.rstrip()
version = version.lstrip()
vv = version.replace(" ", " && ")
list_data.append(vv)
print(list_data)
new_list =
for data in list_data:
if "&&" not in data and "=0" not in data and ">=" not in data:
new_data = "(>=0 && " + data + ")"
new_list.append(new_data)
else:
new_data1 = new_list.append("("+data+")")
final_list =
for items in new_list:
data = final_list.append(items + " || ")
now_data = [''.join(final_list[:])]
data1 = rchop(now_data[0], ' || ')
print(data1)
python regex
python regex
asked Nov 13 '18 at 22:12
Jau LJau L
3501413
asked Nov 13 '18 at 22:12
Jau LJau L
3501413
edited Nov 14 '18 at 0:07
Jau L
asked Nov 13 '18 at 22:12
Jau LJau L
3501413
asked Nov 13 '18 at 22:12
Jau LJau L
3501413
asked Nov 13 '18 at 22:12
Jau LJau L
3501413
3501413
May we know what have you tried and what problems you faced?
– boonwj
Nov 13 '18 at 22:32
It is unclear what the requirement is here, would you please put some test code to reproduce the problem?
– Saul Cruz
Nov 13 '18 at 22:46
the requirements here is that the script I'm using produces buggy version format, so I'm trying to fix it using a regular expression by adding proper parentheses and proper && as the above conversions.
– Jau L
Nov 13 '18 at 22:52
1
I added the code @boonwj
– Jau L
Nov 13 '18 at 23:54
add a comment |
May we know what have you tried and what problems you faced?
– boonwj
Nov 13 '18 at 22:32
It is unclear what the requirement is here, would you please put some test code to reproduce the problem?
– Saul Cruz
Nov 13 '18 at 22:46
the requirements here is that the script I'm using produces buggy version format, so I'm trying to fix it using a regular expression by adding proper parentheses and proper && as the above conversions.
– Jau L
Nov 13 '18 at 22:52
1
I added the code @boonwj
– Jau L
Nov 13 '18 at 23:54
May we know what have you tried and what problems you faced?
– boonwj
Nov 13 '18 at 22:32
May we know what have you tried and what problems you faced?
– boonwj
Nov 13 '18 at 22:32
It is unclear what the requirement is here, would you please put some test code to reproduce the problem?
– Saul Cruz
Nov 13 '18 at 22:46
It is unclear what the requirement is here, would you please put some test code to reproduce the problem?
– Saul Cruz
Nov 13 '18 at 22:46
the requirements here is that the script I'm using produces buggy version format, so I'm trying to fix it using a regular expression by adding proper parentheses and proper && as the above conversions.
– Jau L
Nov 13 '18 at 22:52
the requirements here is that the script I'm using produces buggy version format, so I'm trying to fix it using a regular expression by adding proper parentheses and proper && as the above conversions.
– Jau L
Nov 13 '18 at 22:52
1
1
I added the code @boonwj
– Jau L
Nov 13 '18 at 23:54
I added the code @boonwj
– Jau L
Nov 13 '18 at 23:54
add a comment |
1 Answer
1
active
oldest
votes
Not sure you need regex for this. Seems like you could just split the strings a couple times, reformat the parts and then join them back together to match your example outputs (assuming that the example output where you show <2.2.1 || (>=4.0.0 && <4.1.9) is a typo and it should actually follow the pattern of the other similar example and be (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9) instead.
Maybe there are more edge cases that don't follow the example patterns, but the below should at least give you a simpler starting place to work from.
def version_formatter(text):
raw = [t.strip().split() for t in text.split('||')]
formatted = [f'({r[0]} && {r[1]})' if len(r) == 2 else f'(>=0 && {r[0]})' for r in raw]
return ' || '.join(formatted)
tests = ['<2.2.1 || >=4.0.0 <4.1.9', '>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12', '<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6']
for test in tests:
result = version_formatter(test)
print(result)
# OUTPUT
# (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9)
# (>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)
# (>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
answered Nov 15 '18 at 4:09
benvcbenvc
5,1471623
thanks a lot, @benvc. this works :)
– Jau L
Nov 15 '18 at 18:51
add a comment |
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1 Answer
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1 Answer
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active
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Not sure you need regex for this. Seems like you could just split the strings a couple times, reformat the parts and then join them back together to match your example outputs (assuming that the example output where you show <2.2.1 || (>=4.0.0 && <4.1.9) is a typo and it should actually follow the pattern of the other similar example and be (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9) instead.
Maybe there are more edge cases that don't follow the example patterns, but the below should at least give you a simpler starting place to work from.
def version_formatter(text):
raw = [t.strip().split() for t in text.split('||')]
formatted = [f'({r[0]} && {r[1]})' if len(r) == 2 else f'(>=0 && {r[0]})' for r in raw]
return ' || '.join(formatted)
tests = ['<2.2.1 || >=4.0.0 <4.1.9', '>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12', '<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6']
for test in tests:
result = version_formatter(test)
print(result)
# OUTPUT
# (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9)
# (>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)
# (>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
answered Nov 15 '18 at 4:09
benvcbenvc
5,1471623
thanks a lot, @benvc. this works :)
– Jau L
Nov 15 '18 at 18:51
add a comment |
Not sure you need regex for this. Seems like you could just split the strings a couple times, reformat the parts and then join them back together to match your example outputs (assuming that the example output where you show <2.2.1 || (>=4.0.0 && <4.1.9) is a typo and it should actually follow the pattern of the other similar example and be (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9) instead.
Maybe there are more edge cases that don't follow the example patterns, but the below should at least give you a simpler starting place to work from.
def version_formatter(text):
raw = [t.strip().split() for t in text.split('||')]
formatted = [f'({r[0]} && {r[1]})' if len(r) == 2 else f'(>=0 && {r[0]})' for r in raw]
return ' || '.join(formatted)
tests = ['<2.2.1 || >=4.0.0 <4.1.9', '>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12', '<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6']
for test in tests:
result = version_formatter(test)
print(result)
# OUTPUT
# (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9)
# (>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)
# (>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
answered Nov 15 '18 at 4:09
benvcbenvc
5,1471623
thanks a lot, @benvc. this works :)
– Jau L
Nov 15 '18 at 18:51
add a comment |
Not sure you need regex for this. Seems like you could just split the strings a couple times, reformat the parts and then join them back together to match your example outputs (assuming that the example output where you show <2.2.1 || (>=4.0.0 && <4.1.9) is a typo and it should actually follow the pattern of the other similar example and be (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9) instead.
Maybe there are more edge cases that don't follow the example patterns, but the below should at least give you a simpler starting place to work from.
def version_formatter(text):
raw = [t.strip().split() for t in text.split('||')]
formatted = [f'({r[0]} && {r[1]})' if len(r) == 2 else f'(>=0 && {r[0]})' for r in raw]
return ' || '.join(formatted)
tests = ['<2.2.1 || >=4.0.0 <4.1.9', '>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12', '<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6']
for test in tests:
result = version_formatter(test)
print(result)
# OUTPUT
# (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9)
# (>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)
# (>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
answered Nov 15 '18 at 4:09
benvcbenvc
5,1471623
Not sure you need regex for this. Seems like you could just split the strings a couple times, reformat the parts and then join them back together to match your example outputs (assuming that the example output where you show <2.2.1 || (>=4.0.0 && <4.1.9) is a typo and it should actually follow the pattern of the other similar example and be (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9) instead.
Maybe there are more edge cases that don't follow the example patterns, but the below should at least give you a simpler starting place to work from.
def version_formatter(text):
raw = [t.strip().split() for t in text.split('||')]
formatted = [f'({r[0]} && {r[1]})' if len(r) == 2 else f'(>=0 && {r[0]})' for r in raw]
return ' || '.join(formatted)
tests = ['<2.2.1 || >=4.0.0 <4.1.9', '>=7.0.23 <7.0.91 || >=8.5.0 <8.5.34 || >=9.0.0 <9.0.12', '<2.7.9.4 || >=2.8.0 <2.8.11.2 || >=2.9.0 <2.9.6']
for test in tests:
result = version_formatter(test)
print(result)
# OUTPUT
# (>=0 && <2.2.1) || (>=4.0.0 && <4.1.9)
# (>=7.0.23 && <7.0.91) || (>=8.5.0 && <8.5.34) || (>=9.0.0 && <9.0.12)
# (>=0 && <2.7.9.4) || (>=2.8.0 && <2.8.11.2) || (>=2.9.0 && <2.9.6)
answered Nov 15 '18 at 4:09
benvcbenvc
5,1471623
edited Nov 15 '18 at 13:12
answered Nov 15 '18 at 4:09
benvcbenvc
5,1471623
answered Nov 15 '18 at 4:09
benvcbenvc
5,1471623
answered Nov 15 '18 at 4:09
benvcbenvc
5,1471623
5,1471623
thanks a lot, @benvc. this works :)
– Jau L
Nov 15 '18 at 18:51
add a comment |
thanks a lot, @benvc. this works :)
– Jau L
Nov 15 '18 at 18:51
thanks a lot, @benvc. this works :)
– Jau L
Nov 15 '18 at 18:51
thanks a lot, @benvc. this works :)
– Jau L
Nov 15 '18 at 18:51
add a comment |
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May we know what have you tried and what problems you faced?
– boonwj
Nov 13 '18 at 22:32
It is unclear what the requirement is here, would you please put some test code to reproduce the problem?
– Saul Cruz
Nov 13 '18 at 22:46
the requirements here is that the script I'm using produces buggy version format, so I'm trying to fix it using a regular expression by adding proper parentheses and proper && as the above conversions.
– Jau L
Nov 13 '18 at 22:52
1
I added the code @boonwj
– Jau L
Nov 13 '18 at 23:54