time_difference | missing Ruby method to print out time difference | Date Time Utils library

by tmlee Ruby Version: Current License: MIT

X-Ray Key Features Code Snippets Community Discussions(10)Vulnerabilities Install Support

kandi X-RAY | time_difference Summary

time_difference is a Ruby library typically used in Utilities, Date Time Utils applications. time_difference has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

This latest version of the gem works with ActiveSupport 5.1. For prior version, check out v0.6.x-activesupport42.

Support

Quality

Security

License

Reuse

Support

time_difference has a low active ecosystem.

It has 284 star(s) with 84 fork(s). There are 11 watchers for this library.

It had no major release in the last 6 months.

There are 8 open issues and 13 have been closed. On average issues are closed in 125 days. There are 17 open pull requests and 0 closed requests.

It has a neutral sentiment in the developer community.

The latest version of time_difference is current.

Quality

time_difference has 0 bugs and 0 code smells.

Security

time_difference has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

time_difference code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

time_difference is licensed under the MIT License. This license is Permissive.

Permissive licenses have the least restrictions, and you can use them in most projects.

Reuse

time_difference releases are not available. You will need to build from source code and install.

Installation instructions, examples and code snippets are available.

time_difference saves you 89 person hours of effort in developing the same functionality from scratch.

It has 227 lines of code, 15 functions and 3 files.

It has high code complexity. Code complexity directly impacts maintainability of the code.

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time_difference Key Features

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time_difference Examples and Code Snippets

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Community Discussions

Trending Discussions on time_difference

Data wrangling, keep rows based on time threshold

Finding a difference in time in minutes for values in the same column

Inserting a concatenated value into a table

Perl: Calculate time differences from a list of datetime

Time multiplication per hourly wage

Delete row n and rows from n to n+ x in dataframe in R

Django - Bypass loading code in the apps.ready function when making migrations or migrating to the database

Add a tolerance to a gap with Pandas

How do I find the difference in seconds for a column of datetime object in a group?

Calculate how many n.months between two dates

QUESTION

Data wrangling, keep rows based on time threshold

Asked 2022-Mar-17 at 16:12

I have a data frame where I have a time variable (seconds) and a site variable, but basically the issue is just regarding the time. I want to select which rows are "good" or "bad" based on a specific time threshold.

Hypothetically, let's assume that my threshold is time_threshold >= 250

Here's an example dataset:

...

ANSWER

Answered 2022-Mar-17 at 16:12

I think this loop captures your logic.

Source https://stackoverflow.com/questions/71515081

QUESTION

Finding a difference in time in minutes for values in the same column

Asked 2022-Mar-15 at 07:04

What I want to do here is to find the time difference between StatusID = 'Processed' and StatusID = 'NEW' according to its owner ID. and after getting the difference for each owner ID, I want to find the maximum, minimum and average time difference. This is the sample data for it.

OWNER_ID CREATED_TIME STATUS_ID 1 2020-07-16 08:29:29.000 NEW 1 2022-02-21 04:38:01.000 PROCESSED 3 2022-02-28 14:24:28.000 1 3 2022-02-28 14:27:32.000 CONVERTED 4 2022-02-28 14:33:06.000 NEW 4 2022-02-28 14:33:19.000 IN_PROCESS 5 2022-03-01 12:01:48.000 NEW 5 2022-03-01 12:02:00.000 IN_PROCESS

I have tried out this code to get the time difference but my code is not working.

...

ANSWER

Answered 2022-Mar-15 at 05:59

Use a CASE expression with aggregate

Source https://stackoverflow.com/questions/71477350

QUESTION

Inserting a concatenated value into a table

Asked 2021-Dec-28 at 08:51

With MYSQL, I can calculate the time difference and get it as a concatenated output. I am using queries sourced from here.

...

ANSWER

Answered 2021-Dec-28 at 08:51

DEMO.

Source https://stackoverflow.com/questions/70504413

QUESTION

Perl: Calculate time differences from a list of datetime

Asked 2021-Nov-29 at 04:27

Need help on how to calculate time differences from a list of datetime.
my list of datetime DATA as below, I want to calculate time differences between t1-t2 and following with next line t2 -t1, t1-t2 until the end of list :-

DATA
t1: 2021-Sep-29 18:52:49 - begin here t1(this line)- t1(this line)
t2: 2021-Sep-29 18:52:50 - t1(above line) -t2 (this Line)
t1: 2021-Oct-08 16:09:36 -t2 (above line) - t1 (this line)
t2: 2021-Oct-08 16:49:23 - t1 (above line) - t2 (this line(
t1: 2021-Oct-08 16:55:01
t2: 2021-Oct-08 17:00:00
....
...
...
tn:2021-Nov-11 12:36:40

expected output
t1: 2021-Sep-29 18:52:49
time: 0 seconds
t2: 2021-Sep-29 18:52:50
time: 1 seconds
t1: 2021-Oct-08 16:09:36
time: 1 seconds
t2: 2021-Oct-08 16:49:23
time : 767,806 seconds
t1: 2021-Oct-08 16:55:01
time : 338 seconds
t2: 2021-Oct-08 17:00:00
time : 299 seconds
....

my script as below , I've tried to split all data into array of t1 & t2 and calculate time difference of t1-t2.

...

ANSWER

Answered 2021-Nov-29 at 04:27

use 5.014;
use warnings;

use Time::Piece qw( );

sub parse_time { Time::Piece->strptime($_[0], '%Y-%b-%d %H:%M:%S')->seconds } 

my $prev;
while (<>) {
   say;

   s/^[^:]*:\s*//;
   my $time = parse_time($_);

   if ($prev) {
      my $diff = $time - $prev;
      say "time: $diff seconds";
   }

   $prev = $time;
}

Source https://stackoverflow.com/questions/70149497

QUESTION

Time multiplication per hourly wage

Asked 2021-Sep-19 at 14:41

I am writing a code for a flight school management software. Basically, I have to enter the Departure and Arrival flight time, make the difference between the two times and make the multiplication for the hourly flight instructor wage. My code is something like this:

...

ANSWER

Answered 2021-Sep-19 at 14:38

Firstly, your time_difference variable is not defined.
For you to calculate the total wage: You can take the difference of the times in seconds and convert it to hours(because you want to multiply it by hourly wage).
To convert from seconds, simply divide by 3600 (60*60)
To summarize, you can do:

Source https://stackoverflow.com/questions/69244040

QUESTION

Delete row n and rows from n to n+ x in dataframe in R

Asked 2021-Sep-07 at 17:50

I have data representing stock prices, with 1 minute bars. I need to delete the row corresponding to the first minute of each day and the following 29 rows. The first row of each day always has value >60 at the time_difference variable. If I write del<- df[which(df$time_difference>60),] , then df_new=anti_join(df, sel, by= "Time")I select the first row of each day. However, I need to remove the next 29 rows as well.

Here is a sample of the df, I also added a time_difference vector computed as difference between each row and the next row for the Time variable (not displayed here). Df file can be downloaded from here

Time Open High Low Close Volume Wap Gap Count 1 1536154200 234.61 234.95 234.57 234.76 302 234.600 0 31 2 1536154260 234.76 235.23 234.76 235.16 135 235.008 0 94 3 1536154320 235.09 235.33 234.88 235.33 121 235.010 0 109 4 1536154380 235.24 235.35 235.08 235.35 24 235.203 0 22 5 1536154440 235.27 235.47 235.22 235.42 62 235.340 0 35 6 1536154500 235.39 235.81 235.39 235.63 136 235.633 0 110

...

ANSWER

Answered 2021-Sep-07 at 17:50

My original answer only works on one set of rows at a time. Thanks for sharing your data. Below is an updated answer. Note that this is sensitive to your data being in chronological order as we are using row indices rather than the actual time!

Source https://stackoverflow.com/questions/69091097

QUESTION

Django - Bypass loading code in the apps.ready function when making migrations or migrating to the database

Asked 2021-Sep-05 at 12:59

For starters, I have searched for this before with no luck to find something around this.

Background:

I have created an inventory application for work to help my team be able to quickly see stats around our IT infrastructure. I have threads kicked off when the application loads to kick off some scraping functions. That code works great, but it happens whenever I create and apply database migrations (manage.py makemigrations & manage.py migrate).

Goal:

I'd like to only kick off the scraping code when I'm issuing the runserver command (manage.py runserver). This way, I don't have resources competing between the migration activities and the scraping activities. It also often generates lots of errors because sometimes not all of the database models/fields exist in the database yet.

Ideas:

Modify the code in the django repository to introduce a flag that I can then check against before I have the scraping code run. Not recommended This will get overwritten when I update django, and it won't persist between my dev server and prod server.
Find a way to check which command is being run with the manage.py, and introduce a check to only start scraping if that command is run. Recommended This stays in my codebase and can easily be moved around between dev and prod instances.

I'm open to other ideas to accomplish this. If there's a different way to kick off the scraping activities upon starting the application, then that would likely work too. The apps.ready function was the only thing I was able to find that would run something upon the application start.

Edit: Here's what's inside the apps.ready()function:

...

ANSWER

Answered 2021-Aug-31 at 17:48

The fact that you are trying to bypass code means that you have the code in the wrong place. Launching the scraper doesn't belong in apps.ready because you only need it to run in some situations, but not others.

Also, launching scraping code when you do ./manage.py runserver sounds like a bad idea. You should only use runserver for development and not as an actual web service in a production environment. Instead you should deploy with a real web server such as Apache or Nginx.

You can launch the scraping code with a cron job or some other scheduler. If launching this scraping code is complex, then a bash script or management command is a great way to encapsulate what needs to be done.

Source https://stackoverflow.com/questions/69003195

QUESTION

Add a tolerance to a gap with Pandas

Asked 2021-Aug-06 at 11:15

Edit 1: Printed the output from the sample code.

Using pandas to find gaps in a set of syslog files. Example below gives me a True/False value for where the values of real_date are further apart than the 120 seconds I want to tolerate (lines 4029,30)

I've found that a significant number of entries in syslog can have a slight inaccuracy in the timestamp (lines 4027, 4028). I'm guessing that because line 4028 is an earlier value than line 4027 the comparison correctly ends up being true.

How could I make it tolerate a difference of say 1 second?

...

ANSWER

Answered 2021-Aug-02 at 19:10

The accessor dt.seconds always gives you a positive time value. The days will go negative to make this happen. Try dt.total_seconds() instead.

Source https://stackoverflow.com/questions/68600481

QUESTION

How do I find the difference in seconds for a column of datetime object in a group?

Asked 2021-Jul-06 at 10:54

I tried using the following code, but it is not taking the difference in the group and throwing random values.

...

ANSWER

Answered 2021-Jul-06 at 10:54

Sorting before groupby, so possible avoid lambda function and use instead DataFrameGroupBy.diff:

Source https://stackoverflow.com/questions/68269114

QUESTION

Calculate how many n.months between two dates

Asked 2021-May-10 at 01:19

I want to calculate the distance between two dates by month:

Q: start_date + n.months >= end_dates, what is the variable n? ...

ANSWER

Answered 2021-May-08 at 12:24

# find minimum n so that `start_date + n.months >= end_dates`
def calc_diff(start_date, end_date)
 diff = (end_date.yday - start_date.yday) / 30
 return diff if start_date + diff.months >= end_date
 diff + 1
end

calc_diff(Date.parse('2021-01-31'), Date.parse('2021-02-28')) # 1
calc_diff(Date.parse('2021-01-31'), Date.parse('2021-04-30')) # 3
calc_diff(Date.parse('2021-01-31'), Date.parse('2021-05-31')) # 4
calc_diff(Date.parse('2021-02-01'), Date.parse('2021-06-01')) # 4
calc_diff(Date.parse('2021-02-01'), Date.parse('2021-06-02')) # 5

Source https://stackoverflow.com/questions/67445943

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