How to get the first number from span=(2494, 2516) here?
I want to cut a text from the point where my regex expression is found to the end of the text. The position may vary, so I need that number as a variable.
The position can already be seen in the result of studentnrRegex.search(text):
>>> studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
>>> start = studentnrRegex.search(text)
>>> start
<_sre.SRE_Match object; span=(2494, 2516), match='Studentnr = 1825010243'>
>>> myText = text[2494:]
>>> myText
'Studentnr = 1825010243nTEXT = blablabla
Can I get the start position as a variable directly from my variable start, in this case 2494?
python
edited Nov 24 '18 at 4:42
Craig
2,2061819
asked Nov 24 '18 at 3:59
PedroskiPedroski
1345
add a comment |
I want to cut a text from the point where my regex expression is found to the end of the text. The position may vary, so I need that number as a variable.
The position can already be seen in the result of studentnrRegex.search(text):
>>> studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
>>> start = studentnrRegex.search(text)
>>> start
<_sre.SRE_Match object; span=(2494, 2516), match='Studentnr = 1825010243'>
>>> myText = text[2494:]
>>> myText
'Studentnr = 1825010243nTEXT = blablabla
Can I get the start position as a variable directly from my variable start, in this case 2494?
python
edited Nov 24 '18 at 4:42
Craig
2,2061819
asked Nov 24 '18 at 3:59
PedroskiPedroski
1345
add a comment |
I want to cut a text from the point where my regex expression is found to the end of the text. The position may vary, so I need that number as a variable.
The position can already be seen in the result of studentnrRegex.search(text):
>>> studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
>>> start = studentnrRegex.search(text)
>>> start
<_sre.SRE_Match object; span=(2494, 2516), match='Studentnr = 1825010243'>
>>> myText = text[2494:]
>>> myText
'Studentnr = 1825010243nTEXT = blablabla
Can I get the start position as a variable directly from my variable start, in this case 2494?
python
edited Nov 24 '18 at 4:42
Craig
2,2061819
asked Nov 24 '18 at 3:59
PedroskiPedroski
1345
I want to cut a text from the point where my regex expression is found to the end of the text. The position may vary, so I need that number as a variable.
The position can already be seen in the result of studentnrRegex.search(text):
>>> studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
>>> start = studentnrRegex.search(text)
>>> start
<_sre.SRE_Match object; span=(2494, 2516), match='Studentnr = 1825010243'>
>>> myText = text[2494:]
>>> myText
'Studentnr = 1825010243nTEXT = blablabla
Can I get the start position as a variable directly from my variable start, in this case 2494?
python
python
edited Nov 24 '18 at 4:42
Craig
2,2061819
asked Nov 24 '18 at 3:59
PedroskiPedroski
1345
edited Nov 24 '18 at 4:42
Craig
2,2061819
asked Nov 24 '18 at 3:59
PedroskiPedroski
1345
edited Nov 24 '18 at 4:42
Craig
2,2061819
edited Nov 24 '18 at 4:42
Craig
2,2061819
edited Nov 24 '18 at 4:42
Craig
2,2061819
2,2061819
asked Nov 24 '18 at 3:59
PedroskiPedroski
1345
asked Nov 24 '18 at 3:59
PedroskiPedroski
1345
asked Nov 24 '18 at 3:59
PedroskiPedroski
1345
1345
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The match object returned by calling .search() has .start() and .end() methods that return the starting and ending positions of the match.
studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
m = studentnrRegex.search(text)
start = m.start()
print(mytext[start:])
You can accomplish the same thing with a different regex that matches the student number and everything after it. This will save you the trouble of doing the slice:
studentnrRegex = re.compile(r'(Studentnr = 18d{8}).*', re.DOTALL)
m = studentnrRegex.search(text)
print(m.group())
The {8} matches 8 repeats of the d and the .* matches all remaining characters until the end of the string (including newlines) as long as the re.DOTALL flag is specified. The full match is group 0, which is the default value for the .group() method of the match object. You can access the student number as m.group(1).
answered Nov 24 '18 at 4:34
CraigCraig
2,2061819
Thanks a lot! I see I have a lot to learn!
– Pedroski
Nov 24 '18 at 9:06
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
votes
The match object returned by calling .search() has .start() and .end() methods that return the starting and ending positions of the match.
studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
m = studentnrRegex.search(text)
start = m.start()
print(mytext[start:])
You can accomplish the same thing with a different regex that matches the student number and everything after it. This will save you the trouble of doing the slice:
studentnrRegex = re.compile(r'(Studentnr = 18d{8}).*', re.DOTALL)
m = studentnrRegex.search(text)
print(m.group())
The {8} matches 8 repeats of the d and the .* matches all remaining characters until the end of the string (including newlines) as long as the re.DOTALL flag is specified. The full match is group 0, which is the default value for the .group() method of the match object. You can access the student number as m.group(1).
answered Nov 24 '18 at 4:34
CraigCraig
2,2061819
Thanks a lot! I see I have a lot to learn!
– Pedroski
Nov 24 '18 at 9:06
add a comment |
The match object returned by calling .search() has .start() and .end() methods that return the starting and ending positions of the match.
studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
m = studentnrRegex.search(text)
start = m.start()
print(mytext[start:])
You can accomplish the same thing with a different regex that matches the student number and everything after it. This will save you the trouble of doing the slice:
studentnrRegex = re.compile(r'(Studentnr = 18d{8}).*', re.DOTALL)
m = studentnrRegex.search(text)
print(m.group())
The {8} matches 8 repeats of the d and the .* matches all remaining characters until the end of the string (including newlines) as long as the re.DOTALL flag is specified. The full match is group 0, which is the default value for the .group() method of the match object. You can access the student number as m.group(1).
answered Nov 24 '18 at 4:34
CraigCraig
2,2061819
Thanks a lot! I see I have a lot to learn!
– Pedroski
Nov 24 '18 at 9:06
add a comment |
The match object returned by calling .search() has .start() and .end() methods that return the starting and ending positions of the match.
studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
m = studentnrRegex.search(text)
start = m.start()
print(mytext[start:])
You can accomplish the same thing with a different regex that matches the student number and everything after it. This will save you the trouble of doing the slice:
studentnrRegex = re.compile(r'(Studentnr = 18d{8}).*', re.DOTALL)
m = studentnrRegex.search(text)
print(m.group())
The {8} matches 8 repeats of the d and the .* matches all remaining characters until the end of the string (including newlines) as long as the re.DOTALL flag is specified. The full match is group 0, which is the default value for the .group() method of the match object. You can access the student number as m.group(1).
answered Nov 24 '18 at 4:34
CraigCraig
2,2061819
The match object returned by calling .search() has .start() and .end() methods that return the starting and ending positions of the match.
studentnrRegex = re.compile(r'(Studentnr = 18dddddddd)')
m = studentnrRegex.search(text)
start = m.start()
print(mytext[start:])
You can accomplish the same thing with a different regex that matches the student number and everything after it. This will save you the trouble of doing the slice:
studentnrRegex = re.compile(r'(Studentnr = 18d{8}).*', re.DOTALL)
m = studentnrRegex.search(text)
print(m.group())
The {8} matches 8 repeats of the d and the .* matches all remaining characters until the end of the string (including newlines) as long as the re.DOTALL flag is specified. The full match is group 0, which is the default value for the .group() method of the match object. You can access the student number as m.group(1).
answered Nov 24 '18 at 4:34
CraigCraig
2,2061819
edited Nov 27 '18 at 2:51
answered Nov 24 '18 at 4:34
CraigCraig
2,2061819
answered Nov 24 '18 at 4:34
CraigCraig
2,2061819
answered Nov 24 '18 at 4:34
CraigCraig
2,2061819
2,2061819
Thanks a lot! I see I have a lot to learn!
– Pedroski
Nov 24 '18 at 9:06
add a comment |
Thanks a lot! I see I have a lot to learn!
– Pedroski
Nov 24 '18 at 9:06
Thanks a lot! I see I have a lot to learn!
– Pedroski
Nov 24 '18 at 9:06
Thanks a lot! I see I have a lot to learn!
– Pedroski
Nov 24 '18 at 9:06
add a comment |
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