creating duration into separate half an hour bands Pandas datetime











up vote
0
down vote

favorite












Need a small help. Working on the following. Separating rows.



enter image description here



Input: 



Name,  Channel,  Duration, Start_Time   

John, A, 2, 15:55:00    

John, A,    3, 15:57:00 

John,  A,  5, 16:00:00  

Joseph, B, 10, 15:25:00


Output



Name, Channel,  TB, Count, Duration

John, A, 15:30:00-16:00:00,1,5

John,  A,  16:00:00-16:30:00,  1, 5

Joseph, B, 15:00:00-15:30:00, 1,    5

Joseph, B, 15:30:00-16:00:00, 1,    5


Thank you in advance






python pandas datetime pandas-groupby timedelta







share|improve this question









New contributor




Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    Can you please explain the logic? Also please don't add pictures of data. Give a reproducible example
    – Sotos
    yesterday















up vote
0
down vote

favorite












Need a small help. Working on the following. Separating rows.



enter image description here



Input: 



Name,  Channel,  Duration, Start_Time   

John, A, 2, 15:55:00    

John, A,    3, 15:57:00 

John,  A,  5, 16:00:00  

Joseph, B, 10, 15:25:00


Output



Name, Channel,  TB, Count, Duration

John, A, 15:30:00-16:00:00,1,5

John,  A,  16:00:00-16:30:00,  1, 5

Joseph, B, 15:00:00-15:30:00, 1,    5

Joseph, B, 15:30:00-16:00:00, 1,    5


Thank you in advance






python pandas datetime pandas-groupby timedelta







share|improve this question









New contributor




Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    Can you please explain the logic? Also please don't add pictures of data. Give a reproducible example
    – Sotos
    yesterday













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Need a small help. Working on the following. Separating rows.



enter image description here



Input: 



Name,  Channel,  Duration, Start_Time   

John, A, 2, 15:55:00    

John, A,    3, 15:57:00 

John,  A,  5, 16:00:00  

Joseph, B, 10, 15:25:00


Output



Name, Channel,  TB, Count, Duration

John, A, 15:30:00-16:00:00,1,5

John,  A,  16:00:00-16:30:00,  1, 5

Joseph, B, 15:00:00-15:30:00, 1,    5

Joseph, B, 15:30:00-16:00:00, 1,    5


Thank you in advance






python pandas datetime pandas-groupby timedelta







share|improve this question









New contributor




Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Need a small help. Working on the following. Separating rows.



enter image description here



Input: 



Name,  Channel,  Duration, Start_Time   

John, A, 2, 15:55:00    

John, A,    3, 15:57:00 

John,  A,  5, 16:00:00  

Joseph, B, 10, 15:25:00


Output



Name, Channel,  TB, Count, Duration

John, A, 15:30:00-16:00:00,1,5

John,  A,  16:00:00-16:30:00,  1, 5

Joseph, B, 15:00:00-15:30:00, 1,    5

Joseph, B, 15:30:00-16:00:00, 1,    5


Thank you in advance






python pandas datetime pandas-groupby timedelta




python pandas datetime pandas-groupby timedelta






share|improve this question









New contributor




Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited yesterday









Sotos

26.9k51540




26.9k51540






New contributor




Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









Srikanth Ayithy

83




83




New contributor




Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Srikanth Ayithy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    Can you please explain the logic? Also please don't add pictures of data. Give a reproducible example
    – Sotos
    yesterday














  • 3




    Can you please explain the logic? Also please don't add pictures of data. Give a reproducible example
    – Sotos
    yesterday








3




3




Can you please explain the logic? Also please don't add pictures of data. Give a reproducible example
– Sotos
yesterday




Can you please explain the logic? Also please don't add pictures of data. Give a reproducible example
– Sotos
yesterday












1 Answer
1






active

oldest

votes

















up vote
0
down vote













Use - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min'))


Output



    Name    Channel Duration    Start_Time  Start_time  TB

0   John    A   2   15:55:00    2018-11-19 15:55:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

1   John    A   3   15:57:00    2018-11-19 15:57:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

2   John    A   5   16:00:00    2018-11-19 16:00:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

3   Joseph  B   10  15:25:00    2018-11-19 15:25:00 (2018-11-19 15:00:00, 2018-11-19 15:30:00]


If you want the exact format, do - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min')).apply(lambda x: ' - '.join(str(x).replace('(','').replace(']','').split(',')))


This will yield - 



    Name    Channel Duration    Start_Time  TB

0   John    A   2   15:55:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

1   John    A   3   15:57:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

2   John    A   5   16:00:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

3   Joseph  B   10  15:25:00    2018-11-19 15:00:00 - 2018-11-19 15:30:00





share|improve this answer





















  • df=df['Start_time'].astype('datetime64[D]').dtype df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='20:30:00', end='21:00:00', freq='30min')) Getting an Error - 'There are no fields in dtype datetime64[ns].' Issue is with the data types it appears to be. I tried converting into datetime formats. Still it is showing different datatype error everytime.
    – Srikanth Ayithy
    yesterday











Your Answer






StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});






Srikanth Ayithy is a new contributor. Be nice, and check out our Code of Conduct.










 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53372704%2fcreating-duration-into-separate-half-an-hour-bands-pandas-datetime%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













Use - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min'))


Output



    Name    Channel Duration    Start_Time  Start_time  TB

0   John    A   2   15:55:00    2018-11-19 15:55:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

1   John    A   3   15:57:00    2018-11-19 15:57:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

2   John    A   5   16:00:00    2018-11-19 16:00:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

3   Joseph  B   10  15:25:00    2018-11-19 15:25:00 (2018-11-19 15:00:00, 2018-11-19 15:30:00]


If you want the exact format, do - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min')).apply(lambda x: ' - '.join(str(x).replace('(','').replace(']','').split(',')))


This will yield - 



    Name    Channel Duration    Start_Time  TB

0   John    A   2   15:55:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

1   John    A   3   15:57:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

2   John    A   5   16:00:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

3   Joseph  B   10  15:25:00    2018-11-19 15:00:00 - 2018-11-19 15:30:00





share|improve this answer





















  • df=df['Start_time'].astype('datetime64[D]').dtype df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='20:30:00', end='21:00:00', freq='30min')) Getting an Error - 'There are no fields in dtype datetime64[ns].' Issue is with the data types it appears to be. I tried converting into datetime formats. Still it is showing different datatype error everytime.
    – Srikanth Ayithy
    yesterday















up vote
0
down vote













Use - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min'))


Output



    Name    Channel Duration    Start_Time  Start_time  TB

0   John    A   2   15:55:00    2018-11-19 15:55:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

1   John    A   3   15:57:00    2018-11-19 15:57:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

2   John    A   5   16:00:00    2018-11-19 16:00:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

3   Joseph  B   10  15:25:00    2018-11-19 15:25:00 (2018-11-19 15:00:00, 2018-11-19 15:30:00]


If you want the exact format, do - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min')).apply(lambda x: ' - '.join(str(x).replace('(','').replace(']','').split(',')))


This will yield - 



    Name    Channel Duration    Start_Time  TB

0   John    A   2   15:55:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

1   John    A   3   15:57:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

2   John    A   5   16:00:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

3   Joseph  B   10  15:25:00    2018-11-19 15:00:00 - 2018-11-19 15:30:00





share|improve this answer





















  • df=df['Start_time'].astype('datetime64[D]').dtype df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='20:30:00', end='21:00:00', freq='30min')) Getting an Error - 'There are no fields in dtype datetime64[ns].' Issue is with the data types it appears to be. I tried converting into datetime formats. Still it is showing different datatype error everytime.
    – Srikanth Ayithy
    yesterday













up vote
0
down vote










up vote
0
down vote









Use - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min'))


Output



    Name    Channel Duration    Start_Time  Start_time  TB

0   John    A   2   15:55:00    2018-11-19 15:55:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

1   John    A   3   15:57:00    2018-11-19 15:57:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

2   John    A   5   16:00:00    2018-11-19 16:00:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

3   Joseph  B   10  15:25:00    2018-11-19 15:25:00 (2018-11-19 15:00:00, 2018-11-19 15:30:00]


If you want the exact format, do - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min')).apply(lambda x: ' - '.join(str(x).replace('(','').replace(']','').split(',')))


This will yield - 



    Name    Channel Duration    Start_Time  TB

0   John    A   2   15:55:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

1   John    A   3   15:57:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

2   John    A   5   16:00:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

3   Joseph  B   10  15:25:00    2018-11-19 15:00:00 - 2018-11-19 15:30:00





share|improve this answer












Use - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min'))


Output



    Name    Channel Duration    Start_Time  Start_time  TB

0   John    A   2   15:55:00    2018-11-19 15:55:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

1   John    A   3   15:57:00    2018-11-19 15:57:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

2   John    A   5   16:00:00    2018-11-19 16:00:00 (2018-11-19 15:30:00, 2018-11-19 16:00:00]

3   Joseph  B   10  15:25:00    2018-11-19 15:25:00 (2018-11-19 15:00:00, 2018-11-19 15:30:00]


If you want the exact format, do - 



df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='15:00:00', end='16:30:00', freq='30min')).apply(lambda x: ' - '.join(str(x).replace('(','').replace(']','').split(',')))


This will yield - 



    Name    Channel Duration    Start_Time  TB

0   John    A   2   15:55:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

1   John    A   3   15:57:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

2   John    A   5   16:00:00    2018-11-19 15:30:00 - 2018-11-19 16:00:00

3   Joseph  B   10  15:25:00    2018-11-19 15:00:00 - 2018-11-19 15:30:00






share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









Vivek Kalyanarangan

3,8821725




3,8821725












  • df=df['Start_time'].astype('datetime64[D]').dtype df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='20:30:00', end='21:00:00', freq='30min')) Getting an Error - 'There are no fields in dtype datetime64[ns].' Issue is with the data types it appears to be. I tried converting into datetime formats. Still it is showing different datatype error everytime.
    – Srikanth Ayithy
    yesterday


















  • df=df['Start_time'].astype('datetime64[D]').dtype df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='20:30:00', end='21:00:00', freq='30min')) Getting an Error - 'There are no fields in dtype datetime64[ns].' Issue is with the data types it appears to be. I tried converting into datetime formats. Still it is showing different datatype error everytime.
    – Srikanth Ayithy
    yesterday
















df=df['Start_time'].astype('datetime64[D]').dtype df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='20:30:00', end='21:00:00', freq='30min')) Getting an Error - 'There are no fields in dtype datetime64[ns].' Issue is with the data types it appears to be. I tried converting into datetime formats. Still it is showing different datatype error everytime.
– Srikanth Ayithy
yesterday




df=df['Start_time'].astype('datetime64[D]').dtype df['TB'] = pd.cut(df['Start_time'], bins=pd.date_range(start='20:30:00', end='21:00:00', freq='30min')) Getting an Error - 'There are no fields in dtype datetime64[ns].' Issue is with the data types it appears to be. I tried converting into datetime formats. Still it is showing different datatype error everytime.
– Srikanth Ayithy
yesterday










Srikanth Ayithy is a new contributor. Be nice, and check out our Code of Conduct.










 

draft saved


draft discarded


















Srikanth Ayithy is a new contributor. Be nice, and check out our Code of Conduct.













Srikanth Ayithy is a new contributor. Be nice, and check out our Code of Conduct.












Srikanth Ayithy is a new contributor. Be nice, and check out our Code of Conduct.















 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53372704%2fcreating-duration-into-separate-half-an-hour-bands-pandas-datetime%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more