Django "Latest" filter with multiple values - django

I have a table that looks like this:
Date Value
Oct. 23, 2018 -400
Oct. 23, 2018 -1100
Oct. 23, 2018 -200
Oct. 22, 2018 -400
Oct. 22, 2018 -1100
Oct. 21, 2018 -400
I would like to return the latest value for the date, but with multiple results.
filter().latest() only returns one object. I'd need three in this case.
Thanks!


You can give a try:
filter('some_filter_conditions').order_by('-date')[:3]

Related

Display all dates of November in a template

I'm trying to learn datetime and I'm currently trying to display all dates of November in an html template, in views I have:
year = today.year
month= today.month
num_days = calendar.monthrange(year, month)[1]
days = [datetime.date(year, month, day) for day in range(1, num_days+1)]
for days in days:
days_str = days.strftime('%A, %B, %d, %Y')
print(days_str)
context = {'': }
return render(request, 'template.html', context)
The output of the above is:
Monday, November, 01, 2021
Tuesday, November, 02, 2021
Wednesday, November, 03, 2021
Thursday, November, 04, 2021
Friday, November, 05, 2021
Saturday, November, 06, 2021
Sunday, November, 07, 2021
Monday, November, 08, 2021
Tuesday, November, 09, 2021
Wednesday, November, 10, 2021
Thursday, November, 11, 2021
Friday, November, 12, 2021
Saturday, November, 13, 2021
Sunday, November, 14, 2021
Monday, November, 15, 2021
Tuesday, November, 16, 2021
Wednesday, November, 17, 2021
Thursday, November, 18, 2021
Friday, November, 19, 2021
Saturday, November, 20, 2021
Sunday, November, 21, 2021
Monday, November, 22, 2021
Tuesday, November, 23, 2021
Wednesday, November, 24, 2021
Thursday, November, 25, 2021
Friday, November, 26, 2021
Saturday, November, 27, 2021
Sunday, November, 28, 2021
Monday, November, 29, 2021
Tuesday, November, 30, 2021
How to display above dates in a template?

Simply place all dates in a list and pass the list to the context of the template.
year = today.year
month= today.month
num_days = calendar.monthrange(year, month)[1]
days = [datetime.date(year, month, day) for day in range(1, num_days+1)]
days_list = []
for days in days:
days_str = days.strftime('%A, %B, %d, %Y')
days_list.append(days_str)
context = {'days_list':days_list}
return render(request, 'template.html', context)
Then in your template
{% for day in days_list %}
{{day}}
{% endfor %}
Alternatively, you can pass the dates as datetime objects to the template and format them there using the Django built-in date filter.

Django charts.js of database column?

I have a model database in this format
sno Date Premium Count Time Date_time Close
None Oct. 13, 2021 None 2 2:22 p.m. 38735
None Oct. 13, 2021 None 2 2:23 p.m. 38727
None Oct. 13, 2021 None 2 2:24 p.m. 38739
None Oct. 13, 2021 None 2 2:25 p.m. 38750
None Oct. 13, 2021 None 2 2:26 p.m. 38730
None Oct. 13, 2021 None 2 2:27 p.m. 38723
I want to make a simple bar chart of 'Close' column in the database. I have tried using some manual random values in chart.js code which I have done successfully, but how do I take the database field 'Close' , converting it into a list and plot a chart from it?

display each day's date python from today

I was able to display the week that starts every Saturday by:
today = now().date()
sat_offset = (today.weekday() - 5) % 7
week_start = today - datetime.timedelta(days=sat_offset)
This will display the week from last Saturday but how would I show the dates of each day forward as well? So if the week: Oct. 27, 2018 is display it should say:
Saturday : Oct. 27, 2018
Sunday: Oct. 28, 2018
Monday: Oct. 29, 2018
Tuesday: Oct. 30, 2018
Wednesday: Oct. 31, 2018
Thursday: Nov. 01, 2018
Friday: Nov. 02, 2018
Thank you for your help.

You can iterate through the days of the week using range and time delta like so:
for i in range(7):
week_start += datetime.timedelta(days=1)
print(week_start.strftime("%A %d. %B %Y"))
This will produce a dates like:
Monday : Oct. 28, 2018
Tuesday : Oct. 29, 2018
Wednesday : Oct. 30, 2018
Thursday : Oct. 31, 2018
Friday : Nov. 01, 2018
Saturday : Nov. 02, 2018
Sunday : Nov. 03, 2018
You can format the string how ever you want. Here is some info on dates in python.

get mean of values that fit a specific criteria (pattern matching)

I asked this question before and got a reply that solved it for me. I have a dataframe that looks like this:
id weekdays halflife
241732222300860000 Friday, Aug 31, 2012, 22 0.4166666667
241689170123309000 Friday, Aug 31, 2012, 19 0.3833333333
241686878137512000 Friday, Aug 31, 2012, 19 0.4
241651117396738000 Friday, Aug 31, 2012, 16 1.5666666667
241635163505820000 Friday, Aug 31, 2012, 15 0.95
241633401382265000 Friday, Aug 31, 2012, 15 2.3666666667
And I would like to get average half life of items that were created on Monday, then on Tuesday...etc. (My date range spans over 6 months).
To get the date values I used strptime and difftime. Also, I found the maximum halflife with max(df$halflife), how can I find which id it corresponds to?
Reproducible code:
structure(list(id = c(241732222300860416, 241689170123309056,
241686878137511936, 241651117396738048, 241635163505819648, 241633401382264832
), weekdays = c("Friday, Aug 31, 2012, 22", "Friday, Aug 31, 2012, 19",
"Friday, Aug 31, 2012, 19", "Friday, Aug 31, 2012, 16", "Friday, Aug 31, 2012, 15",
"Friday, Aug 31, 2012, 15"), halflife = structure(c(0.416666666666667,
0.383333333333333, 0.4, 1.56666666666667, 0.95, 2.36666666666667
), class = "difftime", units = "mins")), .Names = c("id",
"weekdays", "halflife"), row.names = c(NA, 6L), class = "data.frame")
So now, I have an average half life value for all mondays, tuesdays...etc. How can I get the average value for all hours within those weekdays, i.e.: Average half life of all items that were created on all Mondays at 9am, then 10am, then 11am..etc. And then Tuesday at 9am, 10am, 11am..etc. The dates in the weekdays column is formatted so that the last number after the comma is the hour it was created at. I am really bad with regular expressions and pattern matching, which is why I am asking this follow-up question.

with base packages you can do following.
> mydf
id weekdays halflife
1 2.417322e+17 Friday, Aug 31, 2012, 22 0.4166667 mins
2 2.416892e+17 Friday, Aug 31, 2012, 19 0.3833333 mins
3 2.416869e+17 Friday, Aug 31, 2012, 19 0.4000000 mins
4 2.416511e+17 Friday, Aug 31, 2012, 16 1.5666667 mins
5 2.416352e+17 Friday, Aug 31, 2012, 15 0.9500000 mins
6 2.416334e+17 Friday, Aug 31, 2012, 15 2.3666667 mins
Instead of using regex, we can just use strsplit on each element of weekdays, unlist the result, and it back in 4 column format as matrix and cbind it back with mydf.
> mydf2 <- cbind(mydf, matrix(unlist(sapply(mydf$weekdays, strsplit, split=',')), byrow=TRUE, ncol=4, dimnames=list(1:nrow(mydf), c('Weekday', 'Day', 'Year', 'Hour'))))
> mydf2
id weekdays halflife Weekday Day Year Hour
1 2.417322e+17 Friday, Aug 31, 2012, 22 0.4166667 mins Friday Aug 31 2012 22
2 2.416892e+17 Friday, Aug 31, 2012, 19 0.3833333 mins Friday Aug 31 2012 19
3 2.416869e+17 Friday, Aug 31, 2012, 19 0.4000000 mins Friday Aug 31 2012 19
4 2.416511e+17 Friday, Aug 31, 2012, 16 1.5666667 mins Friday Aug 31 2012 16
5 2.416352e+17 Friday, Aug 31, 2012, 15 0.9500000 mins Friday Aug 31 2012 15
6 2.416334e+17 Friday, Aug 31, 2012, 15 2.3666667 mins Friday Aug 31 2012 15
Now we have split weekdays column appropriately, we can use aggregate function to calculate mean over desired grouping columns.
> aggregate(halflife ~ Weekday, data=mydf2, FUN = mean)
Weekday halflife
1 Friday 1.013889
If you want to group by Weekday as well as Hour then
> aggregate(halflife ~ Weekday + Hour, data=mydf2, FUN = mean)
Weekday Hour halflife
1 Friday 15 1.6583333
2 Friday 16 1.5666667
3 Friday 19 0.3916667
4 Friday 22 0.4166667
As such first parameter of aggregate function here is a forumla object which supports one ~ one, one ~ many, many ~ one, and many ~ many relationships. See ?aggregate examples to understand how to use it.
I will give brief example of how to many to many relationships.
> set.seed(12345)
> mydf2 <- cbind(mydf2, newvar = rnorm(nrow(mydf2)))
> mydf2
id weekdays halflife Weekday Day Year Hour newvar
1 2.417322e+17 Friday, Aug 31, 2012, 22 0.4166667 mins Friday Aug 31 2012 22 0.5855288
2 2.416892e+17 Friday, Aug 31, 2012, 19 0.3833333 mins Friday Aug 31 2012 19 0.7094660
3 2.416869e+17 Friday, Aug 31, 2012, 19 0.4000000 mins Friday Aug 31 2012 19 -0.1093033
4 2.416511e+17 Friday, Aug 31, 2012, 16 1.5666667 mins Friday Aug 31 2012 16 -0.4534972
5 2.416352e+17 Friday, Aug 31, 2012, 15 0.9500000 mins Friday Aug 31 2012 15 0.6058875
6 2.416334e+17 Friday, Aug 31, 2012, 15 2.3666667 mins Friday Aug 31 2012 15 -1.8179560
> aggregate(cbind(newvar,halflife) ~ Weekday + Hour, data=mydf2, FUN = mean)
Weekday Hour newvar halflife
1 Friday 15 -0.6060343 1.6583333
2 Friday 16 -0.4534972 1.5666667
3 Friday 19 0.3000814 0.3916667
4 Friday 22 0.5855288 0.4166667

javascript regular expression: how do I find date without year or date with year<2010

I need to find date without year, or date with year<2010.
basically,
Feb 15
Feb 20
Feb 20, 2009
Feb 20, 1995
should be accepted
Feb 20, 2010
Feb 20, 2011
should be rejected
How do I do it?
Thanks,
Cheng

Try this:
(Jan|Feb|Mar...Dec)\s\d{1,2},\s([1][0-9][0-9][0-9]|200[0-9])
Note: Expand the month list with proepr names. I was too lazy to spell it all out.