var x1= ee.ImageCollection('LANDSAT/LC08/C01/T1_SR').filterBounds(geometry)
.filterDate('2019-07-01', '2019-10-30')
.sort('CLOUD_COVER');
How could I filter the dates in Sep, Oct, and Nov between 2014 and 2020?
Here is an example of how to filter by year, then specific months within those years.
// Create image collection of S-2 imagery for the perdiod 2015-2020
var S2 = ee.ImageCollection('COPERNICUS/S2')
//filter start and end date
.filterDate('2015-07-01', '2020-10-31')
.filter(ee.Filter.calendarRange(9, 11,'month'))
Related
As we all know, date parsing in Go has it's quirks*.
However, I have now come up against needing to parse a datetime string in CCYY-MM-DDThh:mm:ss[.sss...] to a valid date in Go.
This CCYY format is a format that seems to be ubiquitous in astronomy, essentially the CC is the current century, so although we're in 2022, the century is the 21st century, meaning the date in CCYY format would be 2122.
How do I parse a date string in this format, when we can't specify a coded layout?
Should I just parse in that format, and subtract one "century" e.g., 2106 becomes 2006 in the parsed datetime...?
Has anyone come up against this niche problem before?
*(I for one would never have been able to remember January 2nd, 3:04:05 PM of 2006, UTC-0700 if it wasn't the exact time of my birth! I got lucky)
The time package does not support parsing centuries. You have to handle it yourself.
Also note that a simple subtraction is not enough, as e.g. the 21st century takes place between January 1, 2001 and December 31, 2100 (the year may start with 20 or 21). If the year ends with 00, you do not have to subtract 100 years.
I would write a helper function to parse such dates:
func parse(s string) (t time.Time, err error) {
t, err = time.Parse("2006-01-02T15:04:05[.000]", s)
if err == nil && t.Year()%100 != 0 {
t = t.AddDate(-100, 0, 0)
}
return
}
Testing it:
fmt.Println(parse("2101-12-31T12:13:14[.123]"))
fmt.Println(parse("2122-10-29T12:13:14[.123]"))
fmt.Println(parse("2100-12-31T12:13:14[.123]"))
fmt.Println(parse("2201-12-31T12:13:14[.123]"))
Which outputs (try it on the Go Playground):
2001-12-31 12:13:14.123 +0000 UTC <nil>
2022-10-29 12:13:14.123 +0000 UTC <nil>
2100-12-31 12:13:14.123 +0000 UTC <nil>
2101-12-31 12:13:14.123 +0000 UTC <nil>
As for remembering the layout's time:
January 2, 15:04:05, 2006 (zone: -0700) is a common order in the US, and in this representation parts are in increasing numerical order: January is month 1, 15 hour is 3PM, year 2006 is 6. So the ordinals are 1, 2, 3, 4, 5, 6, 7.
I for one would never have been able to remember January 2nd, 3:04:05 PM of 2006, UTC-0700 if it wasn't the exact time of my birth! I got lucky.
The reason for the Go time package layout is that it is derived from the Unix (and Unix-like) date command format. For example, on Linux,
$ date
Fri Apr 15 08:20:43 AM EDT 2022
$
Now, count from left to right,
Month = 1
Day = 2
Hour = 3 (or 15 = 12 + 3)
Minute = 4
Second = 5
Year = 6
Note: Rob Pike is an author of The Unix Programming Environment
How can I add a decimal/float value to a moment.js date object?
moment('2017-09-20').add(1.234, 'years');
does only change the year and month, not the day. I was expecting that moment will calculate the correct date of '2017-09-20' plus one year plus the nearest day of the 0.234th of the year 2018. But instead it prints out '2017-11-20'.
You can create a moment.duration for 1.234 years and then add using moment().add(Duration);.
EDIT:
1.234 years is equal to 1 year and 2.808 months as you can see using toISOString(). The same duration is equal to 451 days. As the asDays() states:
moment.duration().asDays() gets the length of the duration in days.
So you can use asDays() output as input of add(Number, String);.
Here a live example:
// Create moment duration for 1.234 years
var dur = moment.duration(1.234, 'years');
// 1.234 years is equal to 1 year and 2.808 months
console.log(dur.toISOString()); // P1Y2.808M
// 1.234 years is also equal to 451 days
console.log(dur.asDays()); // 451
// Add duration using moment().add(Duration);
var m1 = moment('2017-09-20').add(dur);
console.log(m1.format()); // 2018-12-20T00:00:00+01:00
// Add number of days
var m2 = moment('2017-09-20').add(dur.asDays(), 'd');
console.log(m2.format()); // 2018-12-15T00:00:00+01:00
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.18.1/moment.min.js"></script>
I am trying to return the month number passing the month name using MomentJS. For example if I pass "July" to moment() I would expect 7 to be returned.
After reading through the docs I tried several different ways, and this way came close...
console.log(moment().month("July"));
In the console, buried in the response I could see this...
_monthsParse: Array[7]
Could anyone please tell me how to return the month number using MomentJS correctly?
Try :
moment().month("July").format("M");
Relevant documentation: http://momentjs.com/docs/#/get-set/month/
alert(moment().month("July").format("M"));
<script src="https://momentjs.com/downloads/moment.min.js"></script>
Anybody looking to get month name from month number then you can try :
const number = 1; // 0 = Jan & 11 = Dec
moment().month(number).format("MMM"); // Feb
Use following to get full month name :
const number = 1; // 0 = January & 11 = December
moment().month(number).format("MMMM"); // February
To use simple month number try this:
const month = 2 //Feb
moment(month, 'M').format('MMMM');
##get month name in moment js with node js
moment() give today date
format("DD-MMMM-YYYY") / output 18-May-2020
format("DD-MM-YYYY") / output 18-05-2020
- sperator you can use /
```
var moment = require('moment');
m_date = moment().format("DD-MMMM-YYYY");
console.log("moment date :", m_date)
```
##output
```
moment date : 18-May-2020
```
Read Officail Docdescription here
I have pattern namely QUARTERDATE and FILENAME inside the file.
Both will have some value as in below eg.
My requirement is, I should rename the file like FILENAME_QUARTERDATE.
My file(myfile.txt) will be as below:
QUARTERDATE: 03/31/14 - 06/29/14
FILENAME : LEAD
field1 field2
34567
20.0 5,678
20.0 5,678
20.0 5,678
20.0 5,678
20.0 5,678
I want the the file name to be as LEAD_201402.txt
Date range in the file is for Quarter 2, so i given as 201402.
Thanks in advance for the replies.
newname=$(awk '/QUARTERDATE/ { split($4, d, "/");
quarter=sprintf("%04d%02d", 2000+d[3], int((d[1]-1)/3)+1); }
/FILENAME/ { fn = $3; print fn "_" quarter; exit; }' "$file")
mv "$file" "$newname"
How is a quarter defined?
As noted in comments to the main question, the problem is as yet ill-defined.
What data would appear in the previous quarter's QUARTERDATE line? Could Q1 ever start with a date in December of the previous year? Could the end date of Q2 ever be in July (or Q1 in April, or Q3 in October, or Q4 in January)? Since the first date of Q2 is in March, these alternatives need to be understood. Could a quarter ever start early and end late simultaneously (a 14 week quarter)?
To which the response was:
QUARTERDATE of Q2 will start as 1st Monday of April and end as last Sunday of June.
Which triggered a counter-response:
2014-03-31 is a Monday, but hardly a Monday in April. What this mainly means is that your definition of a quarter is, as yet, not clear. For example, next year, 2015-03-30 is a Monday, but 'the first Monday in April' is 2015-04-06. The last Sunday in March 2015 is 2015-03-29. So which quarter does the week (Mon) 2015-03-30 to (Sun) 2015-04-05 belong to, and why? If you don't know (both how and why), we can't help you reliably.
Plausible working hypothesis
The lessons of Y2K have been forgotten already (why else are two digits used for the year, dammit!).
Quarters run for an integral number of weeks.
Quarters start on a Monday and end on a Sunday.
Quarters remain aligned with the calendar quarters, rather than drifting around the year. (There are 13 weeks in 91 days, and 4 such quarters in a year, but there's a single extra day in an ordinary year and two extra in a leap year, which mean that occasionally you will get a 14-week quarter, to ensure things stay aligned.)
The date for the first date in a quarter will be near 1st January, 1st April, 1st July or 1st October, but the month might be December, March (as in the question), June or September.
The date for the last date in a quarter will be near 31st March, 30th June, 30th September, 31st December, but the month might be April, July, October or January.
By adding 1 modulo 12 (values in the range 1..12, not 0..11) to the start month, you should end up with a month firmly in the calendar quarter.
By subtracting 1 modulo 12 (values in the range 1..12 again) to the end month, you should end up with a month firmly in calendar quarter.
If the data is valid, the 'start + 1' and 'end - 1' months should be in the same quarter.
The early year might be off-by-one if the start date is in December (but that indicates Q1 of the next year).
The end year might be off-by-one if the end date is in January (but that indicates Q4 of the prior year).
More resilient code
Despite the description above, it is possible to write code that detects the quarter despite any or all of the idiosyncrasies of the quarter start and end dates. This code borrows a little from Barmar's answer, but the algorithm is more resilient to the vagaries of the calendar and the quarter start and end dates.
#!/bin/sh
awk '/QUARTERDATE/ {
split($2, b, "/")
split($4, e, "/")
if (b[1] == 12) { q = 1; y = e[3] }
else if (e[1] == 1) { q = 4; y = b[3] }
else
{
if (b[3] != e[3]) {
print "Year mismatch (" $2 " vs " $4 ") in file " FILENAME
exit 1
}
m = int((b[1] + e[1]) / 2)
q = int((m - 1) / 3) + 1
y = e[3]
}
quarter = sprintf("%.4d%.2d", y + 2000, q)
}
/FILENAME/ {
print $3 "_" quarter
# exit
}' "$#"
The calculation for m adds the start month plus one to the end month minus one and then does integer division by two. With the extreme cases already taken care of, this always yields a month number that is in the correct quarter.
The comment in front of the exit associated with FILENAME allows testing more easily. When processing each file separately, as in Barmar's example, that exit is an important optimization. Note that the error message gives an empty file name if the input comes from standard input. (Offhand, I'm not sure how to print the error message to standard error rather than standard output, other than by a platform-specific technique such as print "message" > "/dev/stderr" or print "message" > "/dev/fd/2".)
Given this sample input data (semi-plausible start and end dates for 6 quarters from 2014Q1 through 2015Q2):
QUARTERDATE: 12/30/13 - 03/30/14
FILENAME : LEAD
QUARTERDATE: 03/31/14 - 06/29/14
FILENAME : LEAD
QUARTERDATE: 06/30/14 - 09/28/14
FILENAME : LEAD
QUARTERDATE: 09/29/14 - 12/28/14
FILENAME : LEAD
QUARTERDATE: 12/29/14 - 03/29/15
FILENAME : LEAD
QUARTERDATE: 03/30/15 - 06/29/15
FILENAME : LEAD
The output from this script is:
LEAD_201401
LEAD_201402
LEAD_201403
LEAD_201404
LEAD_201501
LEAD_201502
You can juggle the start and end dates of the quarters within reason and you should still get the required output. But always be wary of calendrical calculations; they are almost invariably harder than you expect.
OK so for example, today is Tuesday, Feb 02. Well the equivalent "Tuesday" from last year was on Feb 03.
How can I find this out programmatically?
Thanks!!
According to Google, there are 604,800,000 milliseconds in a week. That times 52 should give you the same day of the week a year later (right?).
For example:
var date:Date = new Date(2010, 1, 2);
trace(date);
date.setMilliseconds(date.milliseconds - 604800000 * 52);
trace(date);
Output:
Tue Feb 2 00:00:00 GMT-0800 2010
Tue Feb 3 00:00:00 GMT-0800 2009
Just my two cents. I don't like the idea, that the second answer assumes 52 weeks in a year, it will work for a single year, but is a solution to only this exact problem - eg. if you want to check the same thing moving back 10 years it won't work. I'd do it like this:
var today:Date = new Date();
// Here we store the day of the week
var currentDay:int = today.day;
trace (today);
const milisecondsInADay:uint = 1000*60*60*24;
// Here we move back a year, but we can just as well move back 10 years
// or 2 months
today.fullYear -= 1;
// Find the closest date that is the same day of the week as current day
today.time -= milisecondsInADay*(today.day-currentDay);
trace (today);
returns:
Tue Feb 2 21:13:18 GMT+0100 2010
Tue Feb 3 21:13:18 GMT+0100 2009