Open sasha-cher opened 3 days ago
nickginsberg
commented
3 days ago Thanks @sasha-cher ! I'm trying to replicate, but I'm unable to--it seems to be working fine in my environment, and there are test cases which validate that the time/date conversions work as well. Is it possible there's something different about your environment? Screenshot of the ourput when I run the same code you posted is below for reference.
sasha-cher
commented
1 day ago The problem is that depending on the time zone set in the system, the results will be different. If UTC is greater than 0, the year-month-day will lag behind, and if UTC is less than 0, they will be ahead.
const mins_per_day = 60 * 24;
function noaa_time_to_utc_datetime(noaa_time) {
/**
* Handy function to convert times from the NOAA format
* NOAA times are days since year 0 (not 1900, legit year 0)
* So 738931.25 is Feb 14th, 2024, at 6:00 AM (738,931 full days and .25 days = 6 hours)
* @param {float} noaa_time The days since year 0 including fractions
* @return {string} Returns a string representing the time
*/
const result = new Date(Date.UTC(0, 0, 1, 0, 0, 0));
result.setFullYear(0);
// Note: Subtract 1 from noaa time in days because otherwise we double count Jan 1, 0000
result.setDate(result.getDate() + Math.floor(noaa_time - 1));
// Convert the fraction of a day into hours/minutes
const day_frac = noaa_time - Math.floor(noaa_time);
const minutes_elapsed_in_day = day_frac * mins_per_day;
const hours_elapsed_in_day = Math.floor(minutes_elapsed_in_day / 60);
const minutes_remainder = minutes_elapsed_in_day - hours_elapsed_in_day * 60;
result.setHours(hours_elapsed_in_day);
result.setMinutes(minutes_remainder);
return result.toLocaleString();
}And here's how we ran it and got different results by changing the system time zone settings. Example:
noaa_time_to_utc_datetime(739145.25)Output: UTC-4:
'9/15/2024, 6:00:00 AM'UTC-8:
"9/15/2024, 6:00:22 AM"UTC+9:
"9/16/2023, 6:00:00 AM"UTC+0:
"9/15/2024, 6:00:52 AM"As a result, the program outputs different values depending on the time zone set in the systems.
The Python example outputs the correct results:
from datetime import datetime, timedelta
def calc_date(days_to_add):
return datetime(1, 1, 1) + timedelta(days=days_to_add)
calc_date(739145.25).strftime("%Y-%m-%d %H:%M:%S")Output:
'2024-09-17 06:00:00'Possible fixes:
const mins_per_day = 60 * 24;
function noaa_time_to_utc_datetime(noaa_time) {
/**
* Handy function to convert times from the NOAA format
* NOAA times are days since year 0 (not 1900, legit year 0)
* So 738931.25 is Feb 14th, 2024, at 6:00 AM (738,931 full days and .25 days = 6 hours)
* @param {float} noaa_time The days since year 0 including fractions
* @return {string} Returns a string representing the time
*/
const result = new Date(Date.UTC(0, 0, 1, 0, 0, 0));
result.setUTCFullYear(1);
// Note: Subtract 1 from noaa time in days because otherwise we double count Jan 1, 0000
result.setUTCDate(result.getUTCDate() + Math.floor(noaa_time));
// Convert the fraction of a day into hours/minutes
const day_frac = noaa_time - Math.floor(noaa_time);
const minutes_elapsed_in_day = day_frac * mins_per_day;
const hours_elapsed_in_day = Math.floor(minutes_elapsed_in_day / 60);
const minutes_remainder = minutes_elapsed_in_day - hours_elapsed_in_day * 60;
result.setUTCHours(hours_elapsed_in_day);
result.setUTCMinutes(minutes_remainder);
return result.toUTCString();
}
In Get_gfs_data function, passing argument forecast_date with date 20240915 (2024-09-15) returns incorrect year 2023
javascript:
Result:
In the line of the noaa-gfs-js/index.js:17 (function noaa_time_to_utc_datetime(noaa_time)) -> result.setFullYear(0); As a result, it starts from 1 year BC. Possible correction: result.setUTCFullYear(1);