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Parsing a list of single numbers and number ranges
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$begingroup$
I have input of a string containing a single number (like: $3$) or a range (like: $1-5$). Sample input, all together, looks like: "1-5,3,15-16", and sample output for that input looks like "1,2,3,4,5,15". Output doesn't need to be sorted.
I built something to parse this, but it's ugly. How can I improve this?
from itertools import chain
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
def unpackNums(numString:str):
rList=[]
rList.extend(set(chain(*map(giveRange,numString.split(",")))))
return rList
unpackNums("1-2,30-50,1-10")
python parsing python-3.x interval
edited Sep 5 '15 at 20:21
Jamal♦
30.3k11120227
asked Aug 20 '15 at 18:09
CarbonCarbon
1646
$endgroup$
add a comment |
$begingroup$
I have input of a string containing a single number (like: $3$) or a range (like: $1-5$). Sample input, all together, looks like: "1-5,3,15-16", and sample output for that input looks like "1,2,3,4,5,15". Output doesn't need to be sorted.
I built something to parse this, but it's ugly. How can I improve this?
from itertools import chain
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
def unpackNums(numString:str):
rList=[]
rList.extend(set(chain(*map(giveRange,numString.split(",")))))
return rList
unpackNums("1-2,30-50,1-10")
python parsing python-3.x interval
edited Sep 5 '15 at 20:21
Jamal♦
30.3k11120227
asked Aug 20 '15 at 18:09
CarbonCarbon
1646
$endgroup$
add a comment |
$begingroup$
I have input of a string containing a single number (like: $3$) or a range (like: $1-5$). Sample input, all together, looks like: "1-5,3,15-16", and sample output for that input looks like "1,2,3,4,5,15". Output doesn't need to be sorted.
I built something to parse this, but it's ugly. How can I improve this?
from itertools import chain
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
def unpackNums(numString:str):
rList=[]
rList.extend(set(chain(*map(giveRange,numString.split(",")))))
return rList
unpackNums("1-2,30-50,1-10")
python parsing python-3.x interval
edited Sep 5 '15 at 20:21
Jamal♦
30.3k11120227
asked Aug 20 '15 at 18:09
CarbonCarbon
1646
$endgroup$
I have input of a string containing a single number (like: $3$) or a range (like: $1-5$). Sample input, all together, looks like: "1-5,3,15-16", and sample output for that input looks like "1,2,3,4,5,15". Output doesn't need to be sorted.
I built something to parse this, but it's ugly. How can I improve this?
from itertools import chain
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
def unpackNums(numString:str):
rList=[]
rList.extend(set(chain(*map(giveRange,numString.split(",")))))
return rList
unpackNums("1-2,30-50,1-10")
python parsing python-3.x interval
python parsing python-3.x interval
edited Sep 5 '15 at 20:21
Jamal♦
30.3k11120227
asked Aug 20 '15 at 18:09
CarbonCarbon
1646
edited Sep 5 '15 at 20:21
Jamal♦
30.3k11120227
asked Aug 20 '15 at 18:09
CarbonCarbon
1646
edited Sep 5 '15 at 20:21
Jamal♦
30.3k11120227
edited Sep 5 '15 at 20:21
Jamal♦
30.3k11120227
edited Sep 5 '15 at 20:21
Jamal♦
30.3k11120227
30.3k11120227
asked Aug 20 '15 at 18:09
CarbonCarbon
1646
asked Aug 20 '15 at 18:09
CarbonCarbon
1646
asked Aug 20 '15 at 18:09
CarbonCarbon
1646
1646
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Disclaimer: I'm not a Python programmer!
Your code isn't that bad. It's readable enough for me to understand it. B+ for readability!
Currently, you have this function:
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Why don't you simply store the length in a variable?
Like this:
length=len(z)
if(length==1):
return [int(z[0])]
elif(length==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Now, you don't have to calculate the length twice, only once.
The name z is a really bad name. Better names would be numbers, pieces or something similar.
Looking at the definition of chain(), it seems to accept any iterable, which a range() happens to be. So, you probably don't need that list(), leaving this:
return range(int(z[0]),int(z[1])+1)
On your function unpackNums instead of creating an empty set(), you could use the a set comprehension:
def unpackNums(numString:str):
return {x for x in set(chain(*map(giveRange,numString.split(","))))}
If you notice any inaccuracies, please comment.
edited Aug 21 '15 at 0:49
Quill
10.4k53387
answered Aug 20 '15 at 18:37
Ismael MiguelIsmael Miguel
4,31611453
$endgroup$
add a comment |
$begingroup$
Since the whole exercise is a string-transformation problem, I suggest performing it using a regex substitution.
import re
def expand_ranges(s):
return re.sub(
r'(d+)-(d+)',
lambda match: ','.join(
str(i) for i in range(
int(match.group(1)),
int(match.group(2)) + 1
)
),
s
)
I think that expand_ranges would be a more descriptive name than unpackNums.
answered Aug 20 '15 at 18:27
200_success200_success
130k16153417
$endgroup$
add a comment |
$begingroup$
There's a python library called ifilters built by me:
from ifilters import IntSeqPredicate as isp
print(list(filter(isp('1-5,3,15:16'), range(20))))
will output: [1, 2, 3, 4, 5, 15].
Note that in that library, 15:16 matches [15,16), whereas 15-16 matches [15,16].
answered 21 mins ago
kkewkkew
1
New contributor
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Disclaimer: I'm not a Python programmer!
Your code isn't that bad. It's readable enough for me to understand it. B+ for readability!
Currently, you have this function:
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Why don't you simply store the length in a variable?
Like this:
length=len(z)
if(length==1):
return [int(z[0])]
elif(length==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Now, you don't have to calculate the length twice, only once.
The name z is a really bad name. Better names would be numbers, pieces or something similar.
Looking at the definition of chain(), it seems to accept any iterable, which a range() happens to be. So, you probably don't need that list(), leaving this:
return range(int(z[0]),int(z[1])+1)
On your function unpackNums instead of creating an empty set(), you could use the a set comprehension:
def unpackNums(numString:str):
return {x for x in set(chain(*map(giveRange,numString.split(","))))}
If you notice any inaccuracies, please comment.
edited Aug 21 '15 at 0:49
Quill
10.4k53387
answered Aug 20 '15 at 18:37
Ismael MiguelIsmael Miguel
4,31611453
$endgroup$
add a comment |
$begingroup$
Disclaimer: I'm not a Python programmer!
Your code isn't that bad. It's readable enough for me to understand it. B+ for readability!
Currently, you have this function:
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Why don't you simply store the length in a variable?
Like this:
length=len(z)
if(length==1):
return [int(z[0])]
elif(length==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Now, you don't have to calculate the length twice, only once.
The name z is a really bad name. Better names would be numbers, pieces or something similar.
Looking at the definition of chain(), it seems to accept any iterable, which a range() happens to be. So, you probably don't need that list(), leaving this:
return range(int(z[0]),int(z[1])+1)
On your function unpackNums instead of creating an empty set(), you could use the a set comprehension:
def unpackNums(numString:str):
return {x for x in set(chain(*map(giveRange,numString.split(","))))}
If you notice any inaccuracies, please comment.
edited Aug 21 '15 at 0:49
Quill
10.4k53387
answered Aug 20 '15 at 18:37
Ismael MiguelIsmael Miguel
4,31611453
$endgroup$
add a comment |
$begingroup$
Disclaimer: I'm not a Python programmer!
Your code isn't that bad. It's readable enough for me to understand it. B+ for readability!
Currently, you have this function:
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Why don't you simply store the length in a variable?
Like this:
length=len(z)
if(length==1):
return [int(z[0])]
elif(length==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Now, you don't have to calculate the length twice, only once.
The name z is a really bad name. Better names would be numbers, pieces or something similar.
Looking at the definition of chain(), it seems to accept any iterable, which a range() happens to be. So, you probably don't need that list(), leaving this:
return range(int(z[0]),int(z[1])+1)
On your function unpackNums instead of creating an empty set(), you could use the a set comprehension:
def unpackNums(numString:str):
return {x for x in set(chain(*map(giveRange,numString.split(","))))}
If you notice any inaccuracies, please comment.
edited Aug 21 '15 at 0:49
Quill
10.4k53387
answered Aug 20 '15 at 18:37
Ismael MiguelIsmael Miguel
4,31611453
$endgroup$
Disclaimer: I'm not a Python programmer!
Your code isn't that bad. It's readable enough for me to understand it. B+ for readability!
Currently, you have this function:
def giveRange(numString:str):
z=numString.split("-")
if(len(z)==1):
return [int(z[0])]
elif(len(z)==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Why don't you simply store the length in a variable?
Like this:
length=len(z)
if(length==1):
return [int(z[0])]
elif(length==2):
return list(range(int(z[0]),int(z[1])+1))
else:
raise IndexError("TOO MANY VALS!")
Now, you don't have to calculate the length twice, only once.
The name z is a really bad name. Better names would be numbers, pieces or something similar.
Looking at the definition of chain(), it seems to accept any iterable, which a range() happens to be. So, you probably don't need that list(), leaving this:
return range(int(z[0]),int(z[1])+1)
On your function unpackNums instead of creating an empty set(), you could use the a set comprehension:
def unpackNums(numString:str):
return {x for x in set(chain(*map(giveRange,numString.split(","))))}
If you notice any inaccuracies, please comment.
edited Aug 21 '15 at 0:49
Quill
10.4k53387
answered Aug 20 '15 at 18:37
Ismael MiguelIsmael Miguel
4,31611453
edited Aug 21 '15 at 0:49
Quill
10.4k53387
edited Aug 21 '15 at 0:49
Quill
10.4k53387
edited Aug 21 '15 at 0:49
Quill
10.4k53387
10.4k53387
answered Aug 20 '15 at 18:37
Ismael MiguelIsmael Miguel
4,31611453
answered Aug 20 '15 at 18:37
Ismael MiguelIsmael Miguel
4,31611453
answered Aug 20 '15 at 18:37
Ismael MiguelIsmael Miguel
4,31611453
4,31611453
add a comment |
add a comment |
$begingroup$
Since the whole exercise is a string-transformation problem, I suggest performing it using a regex substitution.
import re
def expand_ranges(s):
return re.sub(
r'(d+)-(d+)',
lambda match: ','.join(
str(i) for i in range(
int(match.group(1)),
int(match.group(2)) + 1
)
),
s
)
I think that expand_ranges would be a more descriptive name than unpackNums.
answered Aug 20 '15 at 18:27
200_success200_success
130k16153417
$endgroup$
add a comment |
$begingroup$
Since the whole exercise is a string-transformation problem, I suggest performing it using a regex substitution.
import re
def expand_ranges(s):
return re.sub(
r'(d+)-(d+)',
lambda match: ','.join(
str(i) for i in range(
int(match.group(1)),
int(match.group(2)) + 1
)
),
s
)
I think that expand_ranges would be a more descriptive name than unpackNums.
answered Aug 20 '15 at 18:27
200_success200_success
130k16153417
$endgroup$
add a comment |
$begingroup$
Since the whole exercise is a string-transformation problem, I suggest performing it using a regex substitution.
import re
def expand_ranges(s):
return re.sub(
r'(d+)-(d+)',
lambda match: ','.join(
str(i) for i in range(
int(match.group(1)),
int(match.group(2)) + 1
)
),
s
)
I think that expand_ranges would be a more descriptive name than unpackNums.
answered Aug 20 '15 at 18:27
200_success200_success
130k16153417
$endgroup$
Since the whole exercise is a string-transformation problem, I suggest performing it using a regex substitution.
import re
def expand_ranges(s):
return re.sub(
r'(d+)-(d+)',
lambda match: ','.join(
str(i) for i in range(
int(match.group(1)),
int(match.group(2)) + 1
)
),
s
)
I think that expand_ranges would be a more descriptive name than unpackNums.
answered Aug 20 '15 at 18:27
200_success200_success
130k16153417
answered Aug 20 '15 at 18:27
200_success200_success
130k16153417
answered Aug 20 '15 at 18:27
200_success200_success
130k16153417
answered Aug 20 '15 at 18:27
200_success200_success
130k16153417
130k16153417
add a comment |
add a comment |
$begingroup$
There's a python library called ifilters built by me:
from ifilters import IntSeqPredicate as isp
print(list(filter(isp('1-5,3,15:16'), range(20))))
will output: [1, 2, 3, 4, 5, 15].
Note that in that library, 15:16 matches [15,16), whereas 15-16 matches [15,16].
answered 21 mins ago
kkewkkew
1
New contributor
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
There's a python library called ifilters built by me:
from ifilters import IntSeqPredicate as isp
print(list(filter(isp('1-5,3,15:16'), range(20))))
will output: [1, 2, 3, 4, 5, 15].
Note that in that library, 15:16 matches [15,16), whereas 15-16 matches [15,16].
answered 21 mins ago
kkewkkew
1
New contributor
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
There's a python library called ifilters built by me:
from ifilters import IntSeqPredicate as isp
print(list(filter(isp('1-5,3,15:16'), range(20))))
will output: [1, 2, 3, 4, 5, 15].
Note that in that library, 15:16 matches [15,16), whereas 15-16 matches [15,16].
answered 21 mins ago
kkewkkew
1
New contributor
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
There's a python library called ifilters built by me:
from ifilters import IntSeqPredicate as isp
print(list(filter(isp('1-5,3,15:16'), range(20))))
will output: [1, 2, 3, 4, 5, 15].
Note that in that library, 15:16 matches [15,16), whereas 15-16 matches [15,16].
answered 21 mins ago
kkewkkew
1
New contributor
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 21 mins ago
kkewkkew
1
New contributor
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 21 mins ago
kkewkkew
1
answered 21 mins ago
kkewkkew
1
1
New contributor
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
kkew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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