I have a pySpark DataFrame Column with Julian Dates. I tried to convert the date to Calender Date.
number
julian_date
1
17196
2
17199
3
17281
I tried with the below code:
spdf = spdf.withColumn('date_new',functions.to_date(functions.from_unixtime("julian_date")))
However, I am getting output as:
number
julian_date
date_new
1
17196
1970-01-01
2
17199
1970-01-01
3
17281
1970-01-01
Please help. Thanks in advance
Julian date is consists of 2 year numbers and 3 digits of day-of-year.
For example: 17196 is year 2017's 196th day, which is 2017-07-15.
Thus, you can use to_date with using year (y) and day-of-year (D) format. (ref: date pattern)
df.withColumn('date_new', functions.to_date(df.julian_date, 'yyDDD'))
# If julian_date is not String type.
# df.julian_date.cast(StringType())
Related
I have dataframe df with column name ILDGL which record date in jde julian date format. I tried to convert that julian date into calender date and store in column ILDGL_Normal But I am not sucessful.
df = df.withColumn("ILDGL_Normal", to_date(concat(lit("20"), col("ILDGL")), "yyMMdd"))
Julian date 123002 mean 2023-01-02, 000001 mean 1900-01-01. How can i convert jde enterprise julian date into normal date format of YYYY-MM-DD?
The JDE Julian date format is CYYDDD
C - Century, Y - Year, D -Day of year
We can ignore Century and convert date to yyyy-MM-dd using to_date function. After that add years by checking centruy.
spark.conf.set("spark.sql.legacy.timeParserPolicy", "LEGACY")
df = spark.createDataFrame([('123002',), ('201002',), ('301002',)], ['jde_julian_date'])
df.withColumn("std_date",
add_months(to_date(substring("jde_julian_date", 2, 5), 'yyDDD'),
when(substring("jde_julian_date", 0, 1) > 1,
(substring("jde_julian_date", 0, 1) - 1) * 100 * 12).
otherwise(0))).show()
+---------------+----------+
|jde_julian_date| std_date|
+---------------+----------+
| 123002|2023-01-02|
| 201002|2101-01-02|
| 301002|2201-01-02|
+---------------+----------+
So I have a column say Date1 which has date in datetime stamp. I want to subtract 10 days from Date1 column and keep in another column say Date2. I only want to subtract ten days from date not from datetime.
How to remove the time stamp. Read many solutions online but could not find for excel
Input table
Date1
26-03-2000 21:00:00
25-04-2000 00:00:00
21-03-2000 01:00:00
31-03-2000 13:00:00
05-03-2012 12:00:00
Expected output
Date1 Date2 Date1_no_timestamp
26-03-2000 21:00:00 16-03-2000 26-03-2000
25-04-2000 00:00:00 15-04-2000 25-04-2000
21-03-2000 01:00:00 11-03-2000 21-03-2000
31-03-2000 13:00:00 21-03-2000 31-03-2000
05-03-2012 12:00:00 24-02-2012 05-03-2012 and so on
You could use the TEXT() function.
=TEXT(B2, "DD-MM-YYYY")
Alternatively, as the above solution could cause issue based on timezone formatting, you could remove anything past the first space:
=LEFT(B2, FIND(" ",A2,1)-1)
Place either the following in C2 (assuming those headers exist) and drag down.
You could use:
Method 1:
Date1_no_timestamp:
=TEXT(A2,"dd-mm-yyyy")
Date2:
=TEXT(A2-10,"dd-mm-yyyy")
Method 2
Date1_no_timestamp:
=RIGHT("0"&DAY(A2),2)&"-"&RIGHT("0"&MONTH(A2),2) & "-" & YEAR(A2)
Date2:
=TEXT(DATEVALUE(E2)-10,"dd-mm-yyyy")
Results:
You can also use the INT() and TRUNC() functions:
=INT(A2)
=TRUNC(A2)
Their behavior is identical for positive numbers - the decimal part is sliced off.
I would like to convert all day in the data-frame into day/feb/2020 format
here date field consist only day
from first one convert the date field like this
My current approach is:
import datetime
y=[]
for day in planned_ds.Date:
x=datetime.datetime(2020, 5, day)
print(x)
Is there any easy method to convert all day data-frame to d/m/y format?
One way as assuming you have data like
df = pd.DataFrame([1,2,3,4,5], columns=["date"])
is to convert them to dates and then shift them to start when you need them to:
pd.to_datetime(df["date"], unit="D") - pd.datetime(1970,1,1) + pd.datetime(2020,1,31)
this results in
0 2020-02-01
1 2020-02-02
2 2020-02-03
3 2020-02-04
4 2020-02-05
Date
19112018
19112016
19112015
19112013
I have a column named Date.
I want to convert 19112018 to 2019-11-20 18:00
19 means year 2019
11 month
20 days
18 is hour
Thanks .
Use the following formula:
=DATE(20&LEFT(A2,2),MID(A2,3,2),MID(A2,5,2))+RIGHT(A2,2)/24
Or:
=--(20&REPLACE(REPLACE(REPLACE(A2,7,0," "),5,0,"-"),3,0,"-")&":00")
Then format the output cell:
yyyy-mm-dd hh:mm
I have an object column which contains date. I have extracted these dates from a text column. So all these dates are in different format. Which are mentioned below . But all the date are in mm/dd/yyy or mm/dd/yy or similar formats where month/date/year.
How can I convert this column in mm/dd/yyyy format. Most of the values are in mm/dd/yyyy format only but there are also number of values in other format as mentioned.
date_df =pd.DataFrame(data =['01/14/2019',
'1/14/2019',
'1/3/2019',
'1/03/2018',
'01/09/19',
'1/09/17',
'1/9/19',
'1/09/13'])
date_df:
01/14/2019
1/14/2019
1/3/2019
1/03/2018
01/09/19
1/09/17
1/9/19
1/09/13
Expected result :
01/14/2019
01/14/2019
01/03/2019
01/03/2018
01/09/2019
01/09/2017
01/09/2019
01/09/2013
Use to_datetime with Series.dt.strftime for custom format in strings (objects), if need datetimes only omit dt.strftime:
df['col'] = pd.to_datetime(df['col']).dt.strftime('%m/%d/%Y')
print (df)
col
0 01/14/2019
1 01/14/2019
2 01/03/2019
3 01/03/2019
4 01/09/2019
5 01/09/2019
6 01/09/2019
7 01/09/2019