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Splitting and extracting values from a row?
How to concatenate text from multiple rows into a single text string in SQL server?How do you split a list into evenly sized chunks?Extracting extension from filename in PythonHow do I sort a dictionary by value?How to access environment variable values?Use a list of values to select rows from a pandas dataframeDelete column from pandas DataFrame by column name“Large data” work flows using pandasHow to iterate over rows in a DataFrame in Pandas?Select rows from a DataFrame based on values in a column in pandas
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
From the following line, I want to extract the date and time including AM/PM.
But the time part of DateTime is skipped.
6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p=[]
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2
oil pump runing On "
new=re.split(r's',xx)
print(new)
p.append(new.pop(0))
p.append(new.pop(1))
print(p)
python pandas csv
edited Mar 23 at 23:23
Anony-Mousse
59.8k798164
asked Mar 23 at 11:06
imtiaz ul Hassanimtiaz ul Hassan
417
add a comment |
From the following line, I want to extract the date and time including AM/PM.
But the time part of DateTime is skipped.
6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p=[]
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2
oil pump runing On "
new=re.split(r's',xx)
print(new)
p.append(new.pop(0))
p.append(new.pop(1))
print(p)
python pandas csv
edited Mar 23 at 23:23
Anony-Mousse
59.8k798164
asked Mar 23 at 11:06
imtiaz ul Hassanimtiaz ul Hassan
417
add a comment |
From the following line, I want to extract the date and time including AM/PM.
But the time part of DateTime is skipped.
6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p=[]
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2
oil pump runing On "
new=re.split(r's',xx)
print(new)
p.append(new.pop(0))
p.append(new.pop(1))
print(p)
python pandas csv
edited Mar 23 at 23:23
Anony-Mousse
59.8k798164
asked Mar 23 at 11:06
imtiaz ul Hassanimtiaz ul Hassan
417
From the following line, I want to extract the date and time including AM/PM.
But the time part of DateTime is skipped.
6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p=[]
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2
oil pump runing On "
new=re.split(r's',xx)
print(new)
p.append(new.pop(0))
p.append(new.pop(1))
print(p)
python pandas csv
python pandas csv
edited Mar 23 at 23:23
Anony-Mousse
59.8k798164
asked Mar 23 at 11:06
imtiaz ul Hassanimtiaz ul Hassan
417
edited Mar 23 at 23:23
Anony-Mousse
59.8k798164
asked Mar 23 at 11:06
imtiaz ul Hassanimtiaz ul Hassan
417
edited Mar 23 at 23:23
Anony-Mousse
59.8k798164
edited Mar 23 at 23:23
Anony-Mousse
59.8k798164
edited Mar 23 at 23:23
Anony-Mousse
59.8k798164
59.8k798164
asked Mar 23 at 11:06
imtiaz ul Hassanimtiaz ul Hassan
417
asked Mar 23 at 11:06
imtiaz ul Hassanimtiaz ul Hassan
417
asked Mar 23 at 11:06
imtiaz ul Hassanimtiaz ul Hassan
417
417
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
No, time part is there, pop function is the problem, your regex is fine, as can be seen by running source code below (there is no need for pop in this case tbh):
Simple solution (without pop):
import re
xx = (
"6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On"
)
new = re.split(r"s", xx)
print(new[:3])
Which returns as expected:
['6/1/2018', '12:01:11.490', 'AM']
Why it didn't work?
When you pop element it is removed from the list. You remove 0 element ('6/1/2018'), new 0 element became the missing date and afterwards you popped first element which is actually AM.
With popping you would do that to get all three (assumming new is already created as before):
for _ in range(3):
print(new.pop(0))
answered Mar 23 at 11:32
Szymon MaszkeSzymon Maszke
3,1691729
isn't there any other way I can retrieve those values without poping?
– imtiaz ul Hassan
Mar 23 at 11:37
1
Of course there is,updated my answer. It would work withpopas well, though it's totally not needed here IMO.
– Szymon Maszke
Mar 23 at 11:39
add a comment |
I am not great at regex and there is certainly scope to tighten this regex up but as a first stab assuming your datetime strings have a fixed format. It does not validate the date.
import re
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p1 = re.compile('(?:d1,2/)2d4s+(?:d2:)2d2.d3s+[AaPp][Mm]')
p = re.findall(p1, xx)
print(p)
answered Mar 23 at 11:28
QHarrQHarr
41.7k82447
You missed the actual problem with his code and created highly complex regex which does not work any better (only clutters the whole thing and makes it much less readable/understandable by others).
– Szymon Maszke
Mar 23 at 11:34
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
No, time part is there, pop function is the problem, your regex is fine, as can be seen by running source code below (there is no need for pop in this case tbh):
Simple solution (without pop):
import re
xx = (
"6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On"
)
new = re.split(r"s", xx)
print(new[:3])
Which returns as expected:
['6/1/2018', '12:01:11.490', 'AM']
Why it didn't work?
When you pop element it is removed from the list. You remove 0 element ('6/1/2018'), new 0 element became the missing date and afterwards you popped first element which is actually AM.
With popping you would do that to get all three (assumming new is already created as before):
for _ in range(3):
print(new.pop(0))
answered Mar 23 at 11:32
Szymon MaszkeSzymon Maszke
3,1691729
isn't there any other way I can retrieve those values without poping?
– imtiaz ul Hassan
Mar 23 at 11:37
1
Of course there is,updated my answer. It would work withpopas well, though it's totally not needed here IMO.
– Szymon Maszke
Mar 23 at 11:39
add a comment |
No, time part is there, pop function is the problem, your regex is fine, as can be seen by running source code below (there is no need for pop in this case tbh):
Simple solution (without pop):
import re
xx = (
"6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On"
)
new = re.split(r"s", xx)
print(new[:3])
Which returns as expected:
['6/1/2018', '12:01:11.490', 'AM']
Why it didn't work?
When you pop element it is removed from the list. You remove 0 element ('6/1/2018'), new 0 element became the missing date and afterwards you popped first element which is actually AM.
With popping you would do that to get all three (assumming new is already created as before):
for _ in range(3):
print(new.pop(0))
answered Mar 23 at 11:32
Szymon MaszkeSzymon Maszke
3,1691729
isn't there any other way I can retrieve those values without poping?
– imtiaz ul Hassan
Mar 23 at 11:37
1
Of course there is,updated my answer. It would work withpopas well, though it's totally not needed here IMO.
– Szymon Maszke
Mar 23 at 11:39
add a comment |
No, time part is there, pop function is the problem, your regex is fine, as can be seen by running source code below (there is no need for pop in this case tbh):
Simple solution (without pop):
import re
xx = (
"6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On"
)
new = re.split(r"s", xx)
print(new[:3])
Which returns as expected:
['6/1/2018', '12:01:11.490', 'AM']
Why it didn't work?
When you pop element it is removed from the list. You remove 0 element ('6/1/2018'), new 0 element became the missing date and afterwards you popped first element which is actually AM.
With popping you would do that to get all three (assumming new is already created as before):
for _ in range(3):
print(new.pop(0))
answered Mar 23 at 11:32
Szymon MaszkeSzymon Maszke
3,1691729
No, time part is there, pop function is the problem, your regex is fine, as can be seen by running source code below (there is no need for pop in this case tbh):
Simple solution (without pop):
import re
xx = (
"6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On"
)
new = re.split(r"s", xx)
print(new[:3])
Which returns as expected:
['6/1/2018', '12:01:11.490', 'AM']
Why it didn't work?
When you pop element it is removed from the list. You remove 0 element ('6/1/2018'), new 0 element became the missing date and afterwards you popped first element which is actually AM.
With popping you would do that to get all three (assumming new is already created as before):
for _ in range(3):
print(new.pop(0))
answered Mar 23 at 11:32
Szymon MaszkeSzymon Maszke
3,1691729
edited Mar 23 at 11:38
answered Mar 23 at 11:32
Szymon MaszkeSzymon Maszke
3,1691729
answered Mar 23 at 11:32
Szymon MaszkeSzymon Maszke
3,1691729
answered Mar 23 at 11:32
Szymon MaszkeSzymon Maszke
3,1691729
3,1691729
isn't there any other way I can retrieve those values without poping?
– imtiaz ul Hassan
Mar 23 at 11:37
1
Of course there is,updated my answer. It would work withpopas well, though it's totally not needed here IMO.
– Szymon Maszke
Mar 23 at 11:39
add a comment |
isn't there any other way I can retrieve those values without poping?
– imtiaz ul Hassan
Mar 23 at 11:37
1
Of course there is,updated my answer. It would work withpopas well, though it's totally not needed here IMO.
– Szymon Maszke
Mar 23 at 11:39
isn't there any other way I can retrieve those values without poping?
– imtiaz ul Hassan
Mar 23 at 11:37
isn't there any other way I can retrieve those values without poping?
– imtiaz ul Hassan
Mar 23 at 11:37
1
1
Of course there is,updated my answer. It would work with
pop as well, though it's totally not needed here IMO.– Szymon Maszke
Mar 23 at 11:39
Of course there is,updated my answer. It would work with
pop as well, though it's totally not needed here IMO.– Szymon Maszke
Mar 23 at 11:39
add a comment |
I am not great at regex and there is certainly scope to tighten this regex up but as a first stab assuming your datetime strings have a fixed format. It does not validate the date.
import re
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p1 = re.compile('(?:d1,2/)2d4s+(?:d2:)2d2.d3s+[AaPp][Mm]')
p = re.findall(p1, xx)
print(p)
answered Mar 23 at 11:28
QHarrQHarr
41.7k82447
You missed the actual problem with his code and created highly complex regex which does not work any better (only clutters the whole thing and makes it much less readable/understandable by others).
– Szymon Maszke
Mar 23 at 11:34
add a comment |
I am not great at regex and there is certainly scope to tighten this regex up but as a first stab assuming your datetime strings have a fixed format. It does not validate the date.
import re
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p1 = re.compile('(?:d1,2/)2d4s+(?:d2:)2d2.d3s+[AaPp][Mm]')
p = re.findall(p1, xx)
print(p)
answered Mar 23 at 11:28
QHarrQHarr
41.7k82447
You missed the actual problem with his code and created highly complex regex which does not work any better (only clutters the whole thing and makes it much less readable/understandable by others).
– Szymon Maszke
Mar 23 at 11:34
add a comment |
I am not great at regex and there is certainly scope to tighten this regex up but as a first stab assuming your datetime strings have a fixed format. It does not validate the date.
import re
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p1 = re.compile('(?:d1,2/)2d4s+(?:d2:)2d2.d3s+[AaPp][Mm]')
p = re.findall(p1, xx)
print(p)
answered Mar 23 at 11:28
QHarrQHarr
41.7k82447
I am not great at regex and there is certainly scope to tighten this regex up but as a first stab assuming your datetime strings have a fixed format. It does not validate the date.
import re
xx = "6/1/2018 12:01:11.490 AM HEP.U02.OIL.GOV.P2_RUN <Unit #2>No.2 oil pump runing On "
p1 = re.compile('(?:d1,2/)2d4s+(?:d2:)2d2.d3s+[AaPp][Mm]')
p = re.findall(p1, xx)
print(p)
answered Mar 23 at 11:28
QHarrQHarr
41.7k82447
answered Mar 23 at 11:28
QHarrQHarr
41.7k82447
answered Mar 23 at 11:28
QHarrQHarr
41.7k82447
answered Mar 23 at 11:28
QHarrQHarr
41.7k82447
41.7k82447
You missed the actual problem with his code and created highly complex regex which does not work any better (only clutters the whole thing and makes it much less readable/understandable by others).
– Szymon Maszke
Mar 23 at 11:34
add a comment |
You missed the actual problem with his code and created highly complex regex which does not work any better (only clutters the whole thing and makes it much less readable/understandable by others).
– Szymon Maszke
Mar 23 at 11:34
You missed the actual problem with his code and created highly complex regex which does not work any better (only clutters the whole thing and makes it much less readable/understandable by others).
– Szymon Maszke
Mar 23 at 11:34
You missed the actual problem with his code and created highly complex regex which does not work any better (only clutters the whole thing and makes it much less readable/understandable by others).
– Szymon Maszke
Mar 23 at 11:34
add a comment |
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