This code should return the ex-dividend date:
import yfinance as yf
yf.Ticker('AGNC').info['exDividendDate']
but I get this as an output:
1661817600
I am wondering if there is a way to get the date from that number ?
It looks like this number is obtained based on seconds. In order to get the real date, you can use pd.to_datetime to convert the seconds to calendar date.
import pandas as pd
pd.to_datetime(1661817600, unit='s')
Out[6]: Timestamp('2022-08-30 00:00:00')
or you can use the built-in datetime package in Python.
from datetime import datetime
print(datetime.fromtimestamp(1661817600))
2022-08-30 08:00:00
Related
I have gone through some similar questions but those solutions didn't worked for me I am having a date field which is String of timestamp "1631898440" I tried converting this string into date using tMap but got this error - java.lang.RuntimeException: java.text.ParseException: Unparseable date: "1631898440".
The function I am using -
row5.mydatecolumn!=null && !"".equalsIgnoreCase(row5.mydatecolumn)? TalendDate.parseDateLocale("EEE MMM dd HH:mm:ss zzz yyyy", row5.mydatecolumn, "EN") :null
Also tried -
TalendDate.parseDate("ddMMyyyy",row5.mydatecolumn)
In this I am getting this err- timestamp out of range: "898442-07-16 00:00:00+05:30"ERROR
How to resolve this issue is there anything wrong with the format of date?
In your user routine just create a fonction like this :
public static Date Convert_String_To_Date(String String_Timestamp) {
SimpleDateFormat sf = new SimpleDateFormat("ddMMyyyy");
Date date = new Date(Long.parseLong(String_Timestamp));
System.out.println("*** Date Converted to this patter ddMMyyyy : "+sf.format(date));
return TalendDate.parseDate("ddMMyyyy",sf.format(date)) ;
}
don't forget the import
import java.text.SimpleDateFormat;
import java.util.Date;
then for me i just put a tjava component where i called my fonction like below
String str = "1631898440";
System.out.println(Format_String_Date.Convert_String_To_Date(str)) ;
So , in your case you would call this fonction in your tMap like this i guess :
row5.mydatecolumn!=null && !"".equalsIgnoreCase(row5.mydatecolumn)?
Format_String_Date.Convert_String_To_Date(row5.mydatecolumn) :null
Here is the output
[statistics] connected
*** Date Converted to this patter ddMMyyyy : 19011970
Mon Jan 19 00:00:00 CET 1970
[statistics] disconnected
I can't reduce one day from current
def now = new Date();
print(now); // print Fri Sep 06 13:10:03 EEST 2019
print(now - 1.days); // not working
print(now - 1); // not working
Please help me. Thanks in advance
the solution works. There might be 2 problems though:
- the snippet you wrote has to be included in a script if you plan to execute it in a stage
- the DateGroovyMethods is not allowed to be used by default. You need administrator rights and to check the build log to allow the execution of that stuff.
The error will look like this:
Scripts not permitted to use staticMethod org.codehaus.groovy.runtime.DateGroovyMethods minus java.util.Date int. Administrators can decide whether to approve or reject this signature.
This is my test example:
pipeline {
agent any
stages {
stage('MyDate test') {
steps {
script {
def date = new Date()
print date
print date - 1
}
}
}
}
}
EDIT:
If you are not an administrator, you can replace the script block with sh 'date -d "-1 days"'
You can also use minus(1) instead of - 1:
def now = new Date();
print(now);
print(now.minus(1))
The best thing to do is to skip the use of Date entirely. java.util.Date is literally the oldest java implementation of date and time. The newest comes with Java 8. You can do it like this:
groovy:000> java.time.LocalDateTime.now().minusDays(1)
===> 2019-09-08T12:07:30.835557
groovy:000>
You can convert from Date to LocalDateTime as well if needed.
(Java syntax used here, as I do not know Groovy.)
tl;dr
Subtract 24-hours.
Instant.now().minus( Duration.ofHours( 24 ) ) // UTC.
…or…
Subtract one calendar day.
ZonedDateTime.now( ZoneId.of( "America/New_York" ) ).minusDays( 1 ) ) // Time zone for Toledo, Ohio, US.
java.time
Never use java.util.Date. That terrible class was supplanted years ago by the modern java.time classes with the adoption of JSR 310. Specifically replaced by Instant.
I can't reduce one day from current
What do you mean by “one day”?
Generic 24-hour days
Do you mean to subtract 24-hours?
Duration d = Duration.ofHours( 24 ) ;
Instant instant = Instant.now() ;
Instant twentyFourHoursAgo = instant.minus( d ) ;
The Instant class represents a moment in UTC with a resolution of nanoseconds.
Run this code live at IdeOne.com.
instant.now().toString(): 2019-09-09T18:48:17.106438Z
twentyFourHoursAgo.toString(): 2019-09-08T18:48:17.106438Z
Calendar days
Do you mean to subtract one calendar day?
This requires a time zone. For any given moment, the date varies around the globe by time zone. It may be “tomorrow” in Tokyo Japan while still “yesterday” in Toledo Ohio US.
Specify a time zone with ZoneId to capture the current moment as seen through the wall-clock time used by the people of a particular region in a ZonedDateTime object.
ZoneId z = ZoneId.of( "Asia/Tokyo" ) ;
ZonedDateTime zdt = ZonedDateTime.now( z ) ;
ZonedDateTime oneDayAgo = zdt.minusDays( 1 ) ;
Run this code live at IdeOne.com.
zdt.toString(): 2019-09-10T03:48:17.147539+09:00[Asia/Tokyo]
oneDayAgo.toString(): 2019-09-09T03:48:17.147539+09:00[Asia/Tokyo]
Convert
If you must have a java.util.Date object to interoperate with old code not yet updated to java.time, you can convert. See the new to…/from… conversion methods added to the old classes.
java.util.Date javaUtilDate =
Date.from( Instant.now().minus( Duration.ofHours( 24 ) ) ) ;
…or…
java.util.Date javaUtilDate =
Date.from( ZonedDateTime.now( ZoneId.of( "Asia/Tokyo" ) ).minusDays( 1 ) ) ) ;
Keep in mind that java.util.Date.toString method tells a lie, dynamically applying the JVM’s current default time zone while generating the text. One of many reasons to avoid this badly-designed class.
I am interested in rounding off timestamps to full hours. What I got so far is to round to the nearest hour. For example with this:
df.withColumn("Full Hour", hour((round(unix_timestamp("Timestamp")/3600)*3600).cast("timestamp")))
But this "round" function uses HALF_UP rounding. This means: 23:56 results in 00:00 but I would instead prefer to have 23:00. Is this possible? I didn't find an option field how to set the rounding behaviour in the function.
I think you're overcomplicating things. Hour function returns by default an hour component of a timestamp.
from pyspark.sql.functions import to_timestamp
from pyspark.sql import Row
df = (sc
.parallelize([Row(Timestamp='2016_08_21 11_59_08')])
.toDF()
.withColumn("parsed", to_timestamp("Timestamp", "yyyy_MM_dd hh_mm_ss")))
df2 = df.withColumn("Full Hour", hour(unix_timestamp("parsed").cast("timestamp")))
df2.show()
Output:
+-------------------+-------------------+---------+
| Timestamp| parsed|Full Hour|
+-------------------+-------------------+---------+
|2016_08_21 11_59_08|2016-08-21 11:59:08| 11|
+-------------------+-------------------+---------+
I have a number of quite large cvs files (1,000,000 rows each) which contain a DateTime column. I am using Pandas pivot tables to summarise them. Part of what this involves is splitting out this DateTime variable into hours and minutes. I am using the following code, which is working fine, but it is taking quite a lot of time (around 4-5 minutes).
My question is: Is this just because the files are so large/my laptop to slow, or is there a more efficient code that allows me to split out hours and minutes from a DateTime variable?
Thanks
df['hours'], df['minutes'] = pd.DatetimeIndex(df['DateTime']).hour, pd.DatetimeIndex(df['DateTime']).minute
If dtypes of column Datetime is not datetime, first convert it to_datetime. Then use dt.hour and dt.minute:
df['DateTime'] = pd.to_datetime(df['DateTime'])
df['hours'], df['minutes'] = df['DateTime'].dt.hour, df['DateTime'].dt.minute
Sample:
import pandas as pd
df = pd.DataFrame({'DateTime': ['2014-06-17 11:09:20', '2014-06-18 10:02:10']})
print (df)
DateTime
0 2014-06-17 11:09:20
1 2014-06-18 10:02:10
print (df.dtypes)
DateTime object
dtype: object
df['DateTime'] = pd.to_datetime(df['DateTime'])
df['hours'], df['minutes'] = df['DateTime'].dt.hour, df['DateTime'].dt.minute
print (df)
DateTime hours minutes
0 2014-06-17 11:09:20 11 9
1 2014-06-18 10:02:10 10 2
I have a csv file that looks like this:
"06/09/2013 14:08:34.930","7.2680542849633447","1.6151231744362988","0","0","21","1546964992","15.772567829158248","1577332736","8360","21.400382061280961","0","15","0","685","0","0","0","0","0","0","0","4637","0"
the csv includes 1 month daily values (24 hrs)
I have a need to load it to pandas and then get some stats on data (min, max) but I need the data to include data records for all days only working hours (between 8:00 to 18:00)
I am very new to pandas library
Load your data:
import pandas as pd
from datetime import datetime
df = pd.read_csv('data.csv', header=None, index_col=0)
Filter your data for working hours from 8:00 to 18:00:
work_hours = lambda d: datetime.strptime(d, '%d/%m/%Y %H:%M:%S.%f').hour in range(8, 18)
df = df[map(work_hours, df.index)]
Get the min and max of the first data column:
min, max = df[1].min(), df[1].max()