How to split each individual value between two string in Python
13
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
How can I split each value which is between (; and ~)?
The result will be like CREATEDBY,CREATEDBYNAME,CREATEDBYYOMINAME,...
I have tried the below, but it's giving the first occurrence.
variable[variable.find(";")+1:myString.find("~")]
How do I get the list of strings by using the split?
python python-2.7
edited Dec 18 '18 at 20:05
Peter Mortensen
13.5k1983111
asked Dec 18 '18 at 8:32
Vicky
4841926
add a comment |
13
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
How can I split each value which is between (; and ~)?
The result will be like CREATEDBY,CREATEDBYNAME,CREATEDBYYOMINAME,...
I have tried the below, but it's giving the first occurrence.
variable[variable.find(";")+1:myString.find("~")]
How do I get the list of strings by using the split?
python python-2.7
edited Dec 18 '18 at 20:05
Peter Mortensen
13.5k1983111
asked Dec 18 '18 at 8:32
Vicky
4841926
use split function . mystring.split(';') then mystirng.split('~')
– Hassan ALi
Dec 18 '18 at 8:37
3
Possible duplicate of How to split a string into a list? Note that the second answer indicates how to specify the delimiter.
– jpmc26
Dec 18 '18 at 18:14
add a comment |
13
13
13
1
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
How can I split each value which is between (; and ~)?
The result will be like CREATEDBY,CREATEDBYNAME,CREATEDBYYOMINAME,...
I have tried the below, but it's giving the first occurrence.
variable[variable.find(";")+1:myString.find("~")]
How do I get the list of strings by using the split?
python python-2.7
edited Dec 18 '18 at 20:05
Peter Mortensen
13.5k1983111
asked Dec 18 '18 at 8:32
Vicky
4841926
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
How can I split each value which is between (; and ~)?
The result will be like CREATEDBY,CREATEDBYNAME,CREATEDBYYOMINAME,...
I have tried the below, but it's giving the first occurrence.
variable[variable.find(";")+1:myString.find("~")]
How do I get the list of strings by using the split?
python python-2.7
python python-2.7
edited Dec 18 '18 at 20:05
Peter Mortensen
13.5k1983111
asked Dec 18 '18 at 8:32
Vicky
4841926
edited Dec 18 '18 at 20:05
Peter Mortensen
13.5k1983111
asked Dec 18 '18 at 8:32
Vicky
4841926
edited Dec 18 '18 at 20:05
Peter Mortensen
13.5k1983111
edited Dec 18 '18 at 20:05
Peter Mortensen
13.5k1983111
edited Dec 18 '18 at 20:05
Peter Mortensen
13.5k1983111
13.5k1983111
asked Dec 18 '18 at 8:32
Vicky
4841926
asked Dec 18 '18 at 8:32
Vicky
4841926
asked Dec 18 '18 at 8:32
Vicky
4841926
4841926
use split function . mystring.split(';') then mystirng.split('~')
– Hassan ALi
Dec 18 '18 at 8:37
3
Possible duplicate of How to split a string into a list? Note that the second answer indicates how to specify the delimiter.
– jpmc26
Dec 18 '18 at 18:14
add a comment |
use split function . mystring.split(';') then mystirng.split('~')
– Hassan ALi
Dec 18 '18 at 8:37
3
Possible duplicate of How to split a string into a list? Note that the second answer indicates how to specify the delimiter.
– jpmc26
Dec 18 '18 at 18:14
use split function . mystring.split(';') then mystirng.split('~')
– Hassan ALi
Dec 18 '18 at 8:37
use split function . mystring.split(';') then mystirng.split('~')
– Hassan ALi
Dec 18 '18 at 8:37
3
3
Possible duplicate of How to split a string into a list? Note that the second answer indicates how to specify the delimiter.
– jpmc26
Dec 18 '18 at 18:14
Possible duplicate of How to split a string into a list? Note that the second answer indicates how to specify the delimiter.
– jpmc26
Dec 18 '18 at 18:14
add a comment |
6 Answers
6
active
oldest
votes
19
Using str.split
Ex:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
for i in variable.strip(";").split(";"):
print(i.split("~", 1)[0])
#or
print([i.split("~", 1)[0] for i in variable.strip(";").split(";")])
Output:
CREATEDBY
CREATEDBYNAME
CREATEDBYYOMINAME
CREATEDON
CREATEDONUTC
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:39
Rakesh
37.3k113870
add a comment |
13
We can try using re.findall with the pattern ;(w+)(?=~):
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = re.findall(r';(w+)~', variable)
print(result)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:38
Tim Biegeleisen
217k1386139
It's rejecting if the COLUMN name has '_' (i.e CREATED_BY)
– Vicky
Dec 18 '18 at 9:11
2
@Vicky Then usew+, which includes underscores, to match your words.
– Tim Biegeleisen
Dec 18 '18 at 9:21
@Vicky: good observation. That's why I hate Regex. You never know what they are really doing and what their intension was. It's very easy to get them wrong accidentally.
– Thomas Weller
Dec 18 '18 at 10:26
1
@Thomas I'm not sure that's a reason to hate Regex.[A-Z]should be self-explanatory that it isn't going to match underscores, and the question could have been clearer that underscores should be matched.
– grg
Dec 18 '18 at 17:10
1
@JaredSmith: trivial or not, I can't judge. The spec is not so bad regarding the underscore: between;and~, it did not say that there should be a word (w). So a[^~]would be closer to the spec. Next, I wonder why he says the pattern should be;(w+)(?=~)but then uses;(w+)~in the code.
– Thomas Weller
Dec 18 '18 at 18:51
|
show 1 more comment
5
You can split() the string and then find() the first ~ for each one:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = [item[:item.find('~')] for item in variable.split(';')]
print(result)
answered Dec 18 '18 at 8:39
fixatd
617411
add a comment |
5
Use regular expression with lookahead and lookbehind:
>>> import re
>>> re.findall(r'(?<=;).*?(?=~)', variable)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:44
Martin Frodl
638310
Posted the same answer 5 minutes ago. You don't even need lookarounds if you place the name in a capture group.
– Tim Biegeleisen
Dec 18 '18 at 8:46
3
The two answers are fairly similar, yes, but I wouldn't call them the same.
– Martin Frodl
Dec 18 '18 at 8:54
add a comment |
1
import re
s = 'asdf=5;iwantthis123jasd'
result = re.search('asdf=5;(.*)123jasd', s)
print result.group(1)
answered Dec 18 '18 at 8:38
Test Project
455
How will this give more than 1 occurrence as needed by OP? And how will this fix the additional~s? Try your approach with OPs original data and you'll find out that this is nonsense:s = 'a;iwantthis~oops~' result = re.search(';(.*)~', s)
– Thomas Weller
Dec 18 '18 at 18:58
add a comment |
1
import re
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
pattern = re.compile (";(.+?)~")
matches = re.findall ( pattern, variable )
print matches
answered Dec 18 '18 at 8:42
Raffi
1,04921120
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
19
Using str.split
Ex:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
for i in variable.strip(";").split(";"):
print(i.split("~", 1)[0])
#or
print([i.split("~", 1)[0] for i in variable.strip(";").split(";")])
Output:
CREATEDBY
CREATEDBYNAME
CREATEDBYYOMINAME
CREATEDON
CREATEDONUTC
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:39
Rakesh
37.3k113870
add a comment |
19
Using str.split
Ex:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
for i in variable.strip(";").split(";"):
print(i.split("~", 1)[0])
#or
print([i.split("~", 1)[0] for i in variable.strip(";").split(";")])
Output:
CREATEDBY
CREATEDBYNAME
CREATEDBYYOMINAME
CREATEDON
CREATEDONUTC
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:39
Rakesh
37.3k113870
add a comment |
19
19
19
Using str.split
Ex:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
for i in variable.strip(";").split(";"):
print(i.split("~", 1)[0])
#or
print([i.split("~", 1)[0] for i in variable.strip(";").split(";")])
Output:
CREATEDBY
CREATEDBYNAME
CREATEDBYYOMINAME
CREATEDON
CREATEDONUTC
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:39
Rakesh
37.3k113870
Using str.split
Ex:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
for i in variable.strip(";").split(";"):
print(i.split("~", 1)[0])
#or
print([i.split("~", 1)[0] for i in variable.strip(";").split(";")])
Output:
CREATEDBY
CREATEDBYNAME
CREATEDBYYOMINAME
CREATEDON
CREATEDONUTC
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:39
Rakesh
37.3k113870
answered Dec 18 '18 at 8:39
Rakesh
37.3k113870
answered Dec 18 '18 at 8:39
Rakesh
37.3k113870
answered Dec 18 '18 at 8:39
Rakesh
37.3k113870
37.3k113870
add a comment |
add a comment |
13
We can try using re.findall with the pattern ;(w+)(?=~):
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = re.findall(r';(w+)~', variable)
print(result)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:38
Tim Biegeleisen
217k1386139
It's rejecting if the COLUMN name has '_' (i.e CREATED_BY)
– Vicky
Dec 18 '18 at 9:11
2
@Vicky Then usew+, which includes underscores, to match your words.
– Tim Biegeleisen
Dec 18 '18 at 9:21
@Vicky: good observation. That's why I hate Regex. You never know what they are really doing and what their intension was. It's very easy to get them wrong accidentally.
– Thomas Weller
Dec 18 '18 at 10:26
1
@Thomas I'm not sure that's a reason to hate Regex.[A-Z]should be self-explanatory that it isn't going to match underscores, and the question could have been clearer that underscores should be matched.
– grg
Dec 18 '18 at 17:10
1
@JaredSmith: trivial or not, I can't judge. The spec is not so bad regarding the underscore: between;and~, it did not say that there should be a word (w). So a[^~]would be closer to the spec. Next, I wonder why he says the pattern should be;(w+)(?=~)but then uses;(w+)~in the code.
– Thomas Weller
Dec 18 '18 at 18:51
|
show 1 more comment
13
We can try using re.findall with the pattern ;(w+)(?=~):
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = re.findall(r';(w+)~', variable)
print(result)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:38
Tim Biegeleisen
217k1386139
It's rejecting if the COLUMN name has '_' (i.e CREATED_BY)
– Vicky
Dec 18 '18 at 9:11
2
@Vicky Then usew+, which includes underscores, to match your words.
– Tim Biegeleisen
Dec 18 '18 at 9:21
@Vicky: good observation. That's why I hate Regex. You never know what they are really doing and what their intension was. It's very easy to get them wrong accidentally.
– Thomas Weller
Dec 18 '18 at 10:26
1
@Thomas I'm not sure that's a reason to hate Regex.[A-Z]should be self-explanatory that it isn't going to match underscores, and the question could have been clearer that underscores should be matched.
– grg
Dec 18 '18 at 17:10
1
@JaredSmith: trivial or not, I can't judge. The spec is not so bad regarding the underscore: between;and~, it did not say that there should be a word (w). So a[^~]would be closer to the spec. Next, I wonder why he says the pattern should be;(w+)(?=~)but then uses;(w+)~in the code.
– Thomas Weller
Dec 18 '18 at 18:51
|
show 1 more comment
13
13
13
We can try using re.findall with the pattern ;(w+)(?=~):
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = re.findall(r';(w+)~', variable)
print(result)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:38
Tim Biegeleisen
217k1386139
We can try using re.findall with the pattern ;(w+)(?=~):
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = re.findall(r';(w+)~', variable)
print(result)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:38
Tim Biegeleisen
217k1386139
edited Dec 18 '18 at 9:33
answered Dec 18 '18 at 8:38
Tim Biegeleisen
217k1386139
answered Dec 18 '18 at 8:38
Tim Biegeleisen
217k1386139
answered Dec 18 '18 at 8:38
Tim Biegeleisen
217k1386139
217k1386139
It's rejecting if the COLUMN name has '_' (i.e CREATED_BY)
– Vicky
Dec 18 '18 at 9:11
2
@Vicky Then usew+, which includes underscores, to match your words.
– Tim Biegeleisen
Dec 18 '18 at 9:21
@Vicky: good observation. That's why I hate Regex. You never know what they are really doing and what their intension was. It's very easy to get them wrong accidentally.
– Thomas Weller
Dec 18 '18 at 10:26
1
@Thomas I'm not sure that's a reason to hate Regex.[A-Z]should be self-explanatory that it isn't going to match underscores, and the question could have been clearer that underscores should be matched.
– grg
Dec 18 '18 at 17:10
1
@JaredSmith: trivial or not, I can't judge. The spec is not so bad regarding the underscore: between;and~, it did not say that there should be a word (w). So a[^~]would be closer to the spec. Next, I wonder why he says the pattern should be;(w+)(?=~)but then uses;(w+)~in the code.
– Thomas Weller
Dec 18 '18 at 18:51
|
show 1 more comment
It's rejecting if the COLUMN name has '_' (i.e CREATED_BY)
– Vicky
Dec 18 '18 at 9:11
2
@Vicky Then usew+, which includes underscores, to match your words.
– Tim Biegeleisen
Dec 18 '18 at 9:21
@Vicky: good observation. That's why I hate Regex. You never know what they are really doing and what their intension was. It's very easy to get them wrong accidentally.
– Thomas Weller
Dec 18 '18 at 10:26
1
@Thomas I'm not sure that's a reason to hate Regex.[A-Z]should be self-explanatory that it isn't going to match underscores, and the question could have been clearer that underscores should be matched.
– grg
Dec 18 '18 at 17:10
1
@JaredSmith: trivial or not, I can't judge. The spec is not so bad regarding the underscore: between;and~, it did not say that there should be a word (w). So a[^~]would be closer to the spec. Next, I wonder why he says the pattern should be;(w+)(?=~)but then uses;(w+)~in the code.
– Thomas Weller
Dec 18 '18 at 18:51
It's rejecting if the COLUMN name has '_' (i.e CREATED_BY)
– Vicky
Dec 18 '18 at 9:11
It's rejecting if the COLUMN name has '_' (i.e CREATED_BY)
– Vicky
Dec 18 '18 at 9:11
2
2
@Vicky Then use
w+, which includes underscores, to match your words.– Tim Biegeleisen
Dec 18 '18 at 9:21
@Vicky Then use
w+, which includes underscores, to match your words.– Tim Biegeleisen
Dec 18 '18 at 9:21
@Vicky: good observation. That's why I hate Regex. You never know what they are really doing and what their intension was. It's very easy to get them wrong accidentally.
– Thomas Weller
Dec 18 '18 at 10:26
@Vicky: good observation. That's why I hate Regex. You never know what they are really doing and what their intension was. It's very easy to get them wrong accidentally.
– Thomas Weller
Dec 18 '18 at 10:26
1
1
@Thomas I'm not sure that's a reason to hate Regex.
[A-Z] should be self-explanatory that it isn't going to match underscores, and the question could have been clearer that underscores should be matched.– grg
Dec 18 '18 at 17:10
@Thomas I'm not sure that's a reason to hate Regex.
[A-Z] should be self-explanatory that it isn't going to match underscores, and the question could have been clearer that underscores should be matched.– grg
Dec 18 '18 at 17:10
1
1
@JaredSmith: trivial or not, I can't judge. The spec is not so bad regarding the underscore: between
; and ~, it did not say that there should be a word (w). So a [^~] would be closer to the spec. Next, I wonder why he says the pattern should be ;(w+)(?=~) but then uses ;(w+)~ in the code.– Thomas Weller
Dec 18 '18 at 18:51
@JaredSmith: trivial or not, I can't judge. The spec is not so bad regarding the underscore: between
; and ~, it did not say that there should be a word (w). So a [^~] would be closer to the spec. Next, I wonder why he says the pattern should be ;(w+)(?=~) but then uses ;(w+)~ in the code.– Thomas Weller
Dec 18 '18 at 18:51
|
show 1 more comment
5
You can split() the string and then find() the first ~ for each one:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = [item[:item.find('~')] for item in variable.split(';')]
print(result)
answered Dec 18 '18 at 8:39
fixatd
617411
add a comment |
5
You can split() the string and then find() the first ~ for each one:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = [item[:item.find('~')] for item in variable.split(';')]
print(result)
answered Dec 18 '18 at 8:39
fixatd
617411
add a comment |
5
5
5
You can split() the string and then find() the first ~ for each one:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = [item[:item.find('~')] for item in variable.split(';')]
print(result)
answered Dec 18 '18 at 8:39
fixatd
617411
You can split() the string and then find() the first ~ for each one:
variable=";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
result = [item[:item.find('~')] for item in variable.split(';')]
print(result)
answered Dec 18 '18 at 8:39
fixatd
617411
answered Dec 18 '18 at 8:39
fixatd
617411
answered Dec 18 '18 at 8:39
fixatd
617411
answered Dec 18 '18 at 8:39
fixatd
617411
617411
add a comment |
add a comment |
5
Use regular expression with lookahead and lookbehind:
>>> import re
>>> re.findall(r'(?<=;).*?(?=~)', variable)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:44
Martin Frodl
638310
Posted the same answer 5 minutes ago. You don't even need lookarounds if you place the name in a capture group.
– Tim Biegeleisen
Dec 18 '18 at 8:46
3
The two answers are fairly similar, yes, but I wouldn't call them the same.
– Martin Frodl
Dec 18 '18 at 8:54
add a comment |
5
Use regular expression with lookahead and lookbehind:
>>> import re
>>> re.findall(r'(?<=;).*?(?=~)', variable)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:44
Martin Frodl
638310
Posted the same answer 5 minutes ago. You don't even need lookarounds if you place the name in a capture group.
– Tim Biegeleisen
Dec 18 '18 at 8:46
3
The two answers are fairly similar, yes, but I wouldn't call them the same.
– Martin Frodl
Dec 18 '18 at 8:54
add a comment |
5
5
5
Use regular expression with lookahead and lookbehind:
>>> import re
>>> re.findall(r'(?<=;).*?(?=~)', variable)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:44
Martin Frodl
638310
Use regular expression with lookahead and lookbehind:
>>> import re
>>> re.findall(r'(?<=;).*?(?=~)', variable)
['CREATEDBY', 'CREATEDBYNAME', 'CREATEDBYYOMINAME', 'CREATEDON', 'CREATEDONUTC']
answered Dec 18 '18 at 8:44
Martin Frodl
638310
answered Dec 18 '18 at 8:44
Martin Frodl
638310
answered Dec 18 '18 at 8:44
Martin Frodl
638310
answered Dec 18 '18 at 8:44
Martin Frodl
638310
638310
Posted the same answer 5 minutes ago. You don't even need lookarounds if you place the name in a capture group.
– Tim Biegeleisen
Dec 18 '18 at 8:46
3
The two answers are fairly similar, yes, but I wouldn't call them the same.
– Martin Frodl
Dec 18 '18 at 8:54
add a comment |
Posted the same answer 5 minutes ago. You don't even need lookarounds if you place the name in a capture group.
– Tim Biegeleisen
Dec 18 '18 at 8:46
3
The two answers are fairly similar, yes, but I wouldn't call them the same.
– Martin Frodl
Dec 18 '18 at 8:54
Posted the same answer 5 minutes ago. You don't even need lookarounds if you place the name in a capture group.
– Tim Biegeleisen
Dec 18 '18 at 8:46
Posted the same answer 5 minutes ago. You don't even need lookarounds if you place the name in a capture group.
– Tim Biegeleisen
Dec 18 '18 at 8:46
3
3
The two answers are fairly similar, yes, but I wouldn't call them the same.
– Martin Frodl
Dec 18 '18 at 8:54
The two answers are fairly similar, yes, but I wouldn't call them the same.
– Martin Frodl
Dec 18 '18 at 8:54
add a comment |
1
import re
s = 'asdf=5;iwantthis123jasd'
result = re.search('asdf=5;(.*)123jasd', s)
print result.group(1)
answered Dec 18 '18 at 8:38
Test Project
455
How will this give more than 1 occurrence as needed by OP? And how will this fix the additional~s? Try your approach with OPs original data and you'll find out that this is nonsense:s = 'a;iwantthis~oops~' result = re.search(';(.*)~', s)
– Thomas Weller
Dec 18 '18 at 18:58
add a comment |
1
import re
s = 'asdf=5;iwantthis123jasd'
result = re.search('asdf=5;(.*)123jasd', s)
print result.group(1)
answered Dec 18 '18 at 8:38
Test Project
455
How will this give more than 1 occurrence as needed by OP? And how will this fix the additional~s? Try your approach with OPs original data and you'll find out that this is nonsense:s = 'a;iwantthis~oops~' result = re.search(';(.*)~', s)
– Thomas Weller
Dec 18 '18 at 18:58
add a comment |
1
1
1
import re
s = 'asdf=5;iwantthis123jasd'
result = re.search('asdf=5;(.*)123jasd', s)
print result.group(1)
answered Dec 18 '18 at 8:38
Test Project
455
import re
s = 'asdf=5;iwantthis123jasd'
result = re.search('asdf=5;(.*)123jasd', s)
print result.group(1)
answered Dec 18 '18 at 8:38
Test Project
455
answered Dec 18 '18 at 8:38
Test Project
455
answered Dec 18 '18 at 8:38
Test Project
455
answered Dec 18 '18 at 8:38
Test Project
455
455
How will this give more than 1 occurrence as needed by OP? And how will this fix the additional~s? Try your approach with OPs original data and you'll find out that this is nonsense:s = 'a;iwantthis~oops~' result = re.search(';(.*)~', s)
– Thomas Weller
Dec 18 '18 at 18:58
add a comment |
How will this give more than 1 occurrence as needed by OP? And how will this fix the additional~s? Try your approach with OPs original data and you'll find out that this is nonsense:s = 'a;iwantthis~oops~' result = re.search(';(.*)~', s)
– Thomas Weller
Dec 18 '18 at 18:58
How will this give more than 1 occurrence as needed by OP? And how will this fix the additional
~s? Try your approach with OPs original data and you'll find out that this is nonsense: s = 'a;iwantthis~oops~' result = re.search(';(.*)~', s)– Thomas Weller
Dec 18 '18 at 18:58
How will this give more than 1 occurrence as needed by OP? And how will this fix the additional
~s? Try your approach with OPs original data and you'll find out that this is nonsense: s = 'a;iwantthis~oops~' result = re.search(';(.*)~', s)– Thomas Weller
Dec 18 '18 at 18:58
add a comment |
1
import re
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
pattern = re.compile (";(.+?)~")
matches = re.findall ( pattern, variable )
print matches
answered Dec 18 '18 at 8:42
Raffi
1,04921120
add a comment |
1
import re
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
pattern = re.compile (";(.+?)~")
matches = re.findall ( pattern, variable )
print matches
answered Dec 18 '18 at 8:42
Raffi
1,04921120
add a comment |
1
1
1
import re
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
pattern = re.compile (";(.+?)~")
matches = re.findall ( pattern, variable )
print matches
answered Dec 18 '18 at 8:42
Raffi
1,04921120
import re
variable = ";CREATEDBY~string~1~~72~0~0~0~~~0;CREATEDBYNAME~string~1~~800~0~0~0~~~1;CREATEDBYYOMINAME~string~1~~800~0~0~0~~~2;CREATEDON~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~3;CREATEDONUTC~date~1~yyyy-MM-dd HH:mm:ss.SSS~26~0~0~0~~~4"
pattern = re.compile (";(.+?)~")
matches = re.findall ( pattern, variable )
print matches
answered Dec 18 '18 at 8:42
Raffi
1,04921120
edited Dec 18 '18 at 19:04
answered Dec 18 '18 at 8:42
Raffi
1,04921120
answered Dec 18 '18 at 8:42
Raffi
1,04921120
answered Dec 18 '18 at 8:42
Raffi
1,04921120
1,04921120
add a comment |
add a comment |
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use split function . mystring.split(';') then mystirng.split('~')
– Hassan ALi
Dec 18 '18 at 8:37
3
Possible duplicate of How to split a string into a list? Note that the second answer indicates how to specify the delimiter.
– jpmc26
Dec 18 '18 at 18:14