I am currently attempting to convert a column "datetime" which has values that are dates/times in string form, and I want to convert the column such that all of the strings are converted to timestamps.
The date/time strings are of the form "10/11/2015 0:41", and I'd like to convert the string to a timestamp of form YYYY-MM-DD HH:MM:SS. At first I attempted to cast the column to timestamp in the following way:
df=df.withColumn("datetime", df["datetime"].cast("timestamp"))
Though when I did so, I received null for every value, which lead me to believe that the input dates needed to be formatted somehow. I have looked into numerous other possible remedies such as to_timestamp(), though this also gives the same null results for all of the values. How can a string of this format be converted into a timestamp?
Any insights or guidance are greatly appreciated.
Try:
import datetime
def to_timestamp(date_string):
return datetime.datetime.strptime(date_string, "%m/%d/%Y %H:%M")
df = df.withColumn("datetime", to_timestamp(df.datetime))
You can use the to_timestamp function. See Datetime Patterns for valid date and time format patterns.
df = df.withColumn('datetime', F.to_timestamp('datetime', 'M/d/y H:m'))
df.show(truncate=False)
You were doing it in the right way, except you missed to add the format ofstring type which is in this case MM/dd/yyyy HH:mm. Here M is used for months and m is used to detect minutes. Having said that, see the code below for reference -
df = spark.createDataFrame([('10/11/2015 0:41',), ('10/11/2013 10:30',), ('12/01/2016 15:56',)], ("String_Timestamp", ))
from pyspark.sql.functions import *
df.withColumn("Timestamp_Format", to_timestamp(col("String_Timestamp"), "MM/dd/yyyy HH:mm")).show(truncate=False)
+----------------+-------------------+
|String_Timestamp| Timestamp_Format|
+----------------+-------------------+
| 10/11/2015 0:41|2015-10-11 00:41:00|
|10/11/2013 10:30|2013-10-11 10:30:00|
|12/01/2016 15:56|2016-12-01 15:56:00|
+----------------+-------------------+
Related
Using a Python script, I need to read a CVS file where dates are formated as DD/MM/YYYY, and convert them to YYYY-MM-DD before saving this into a SQLite database.
This almost works, but fails because I don't provide time:
from datetime import datetime
lastconnection = datetime.strptime("21/12/2008", "%Y-%m-%d")
#ValueError: time data did not match format: data=21/12/2008 fmt=%Y-%m-%d
print lastconnection
I assume there's a method in the datetime object to perform this conversion very easily, but I can't find an example of how to do it. Thank you.
Your example code is wrong. This works:
import datetime
datetime.datetime.strptime("21/12/2008", "%d/%m/%Y").strftime("%Y-%m-%d")
The call to strptime() parses the first argument according to the format specified in the second, so those two need to match. Then you can call strftime() to format the result into the desired final format.
you first would need to convert string into datetime tuple, and then convert that datetime tuple to string, it would go like this:
lastconnection = datetime.strptime("21/12/2008", "%d/%m/%Y").strftime('%Y-%m-%d')
I am new to programming. I wanted to convert from yyyy-mm-dd to dd/mm/yyyy to print out a date in the format that people in my part of the world use and recognise.
The accepted answer above got me on the right track.
The answer I ended up with to my problem is:
import datetime
today_date = datetime.date.today()
print(today_date)
new_today_date = today_date.strftime("%d/%m/%Y")
print (new_today_date)
The first two lines after the import statement gives today's date in the USA format (2017-01-26). The last two lines convert this to the format recognised in the UK and other countries (26/01/2017).
You can shorten this code, but I left it as is because it is helpful to me as a beginner. I hope this helps other beginner programmers starting out!
Does anyone else else think it's a waste to convert these strings to date/time objects for what is, in the end, a simple text transformation? If you're certain the incoming dates will be valid, you can just use:
>>> ddmmyyyy = "21/12/2008"
>>> yyyymmdd = ddmmyyyy[6:] + "-" + ddmmyyyy[3:5] + "-" + ddmmyyyy[:2]
>>> yyyymmdd
'2008-12-21'
This will almost certainly be faster than the conversion to and from a date.
#case_date= 03/31/2020
#Above is the value stored in case_date in format(mm/dd/yyyy )
demo=case_date.split("/")
new_case_date = demo[1]+"-"+demo[0]+"-"+demo[2]
#new format of date is (dd/mm/yyyy) test by printing it
print(new_case_date)
If you need to convert an entire column (from pandas DataFrame), first convert it (pandas Series) to the datetime format using to_datetime and then use .dt.strftime:
def conv_dates_series(df, col, old_date_format, new_date_format):
df[col] = pd.to_datetime(df[col], format=old_date_format).dt.strftime(new_date_format)
return df
Sample usage:
import pandas as pd
test_df = pd.DataFrame({"Dates": ["1900-01-01", "1999-12-31"]})
old_date_format='%Y-%m-%d'
new_date_format='%d/%m/%Y'
conv_dates_series(test_df, "Dates", old_date_format, new_date_format)
Dates
0 01/01/1900
1 31/12/1999
The most simplest way
While reading the csv file, put an argument parse_dates
df = pd.read_csv("sample.csv", parse_dates=['column_name'])
This will convert the dates of mentioned column to YYYY-MM-DD format
Convert date format DD/MM/YYYY to YYYY-MM-DD according to your question, you can use this:
from datetime import datetime
lastconnection = datetime.strptime("21/12/2008", "%d/%m/%Y").strftime("%Y-%m-%d")
print(lastconnection)
df is your data frame
Dateclm is the column that you want to change
This column should be in DateTime datatype.
df['Dateclm'] = pd.to_datetime(df['Dateclm'])
df.dtypes
#Here is the solution to change the format of the column
df["Dateclm"] = pd.to_datetime(df["Dateclm"]).dt.strftime('%Y-%m-%d')
print(df)
I wanted to transform the date string into an integer so I can classify if tweets were made in the morning, midday or evening.
I transformed f.e. the username and location with the encoder, but it doesn't work for the date.
df['user.name'] = encoder.fit_transform(df['user.name'])
df['user.location'] = encoder.fit_transform(df['user.location'])
do I need to transform it into 'datetime' first?
date format from Twitter-data
I think the best way to do it is to convert the date string into datetime first
new_date = datetime.datetime.strptime(twitter_date_string, '%Y-%m-%dT%H:%M:%S.%fZ')
and then use either
new_date.year
new_date.month
new_date.day
new_date.hour
new_date.minute
new_date.second
to get what you need as an integer.
I've been trying to convert a timestamp that is a string to a datetime object. The problem is the timestamps formatting. I haven't been able to properly parse the timestamp using datetime.datetime.strptime. I could write my own little parser as its a simple problem but I was hoping to use strptime function, I just need help on the formatting.
Example
import datetime
formater = "%y-%m-%dT%H:%M:%SZ"
str_timestamp = "2021-03-13T18:27:37.60918Z"
timestamp = datetime.datetime.strptime(str_timestamp, formater)
print (timestamp)
Output
builtins.ValueError: time data '2021-03-13T18:27:37.60918Z' does not match format '%y-%m-%dT%H:%M:%SZ'
I'm clearly not symbolizing the formatter properly, the T and Z parts are what I can't account for.
y should be Y. y is for 2 digits year.
You should also take care for the milliseconds with .%f:
%Y-%m-%dT%H:%M:%S.%fZ
This format works:
formater = "%Y-%m-%dT%H:%M:%S.%fZ"
output:
2021-03-13 18:27:37.609180
Please can someone help me with the correct way to convert "hh:mm:ss AM/PM" Object column to Timestamp.
E.g. Input = "06:12:39 PM"
Expected Output = 18:12:39
I tried the below already:
df['col'] = pd.to_datetime(df.col,format='%H:%M:%S %p').dt.time
However I am getting output = 06:12:39 with the datatype unchanged
Not sure where I am going wrong here.
Thank you.
Can you try this
For single datetime value,
tme='06:12:39 PM'
pd.to_datetime(tme).strftime('%H:%M:%S')
For the column you can do this. This will give you the time in string that you need to convert.
pd.to_datetime(df['col']).dt.strftime('%H:%M:%S')
Using a Python script, I need to read a CVS file where dates are formated as DD/MM/YYYY, and convert them to YYYY-MM-DD before saving this into a SQLite database.
This almost works, but fails because I don't provide time:
from datetime import datetime
lastconnection = datetime.strptime("21/12/2008", "%Y-%m-%d")
#ValueError: time data did not match format: data=21/12/2008 fmt=%Y-%m-%d
print lastconnection
I assume there's a method in the datetime object to perform this conversion very easily, but I can't find an example of how to do it. Thank you.
Your example code is wrong. This works:
import datetime
datetime.datetime.strptime("21/12/2008", "%d/%m/%Y").strftime("%Y-%m-%d")
The call to strptime() parses the first argument according to the format specified in the second, so those two need to match. Then you can call strftime() to format the result into the desired final format.
you first would need to convert string into datetime tuple, and then convert that datetime tuple to string, it would go like this:
lastconnection = datetime.strptime("21/12/2008", "%d/%m/%Y").strftime('%Y-%m-%d')
I am new to programming. I wanted to convert from yyyy-mm-dd to dd/mm/yyyy to print out a date in the format that people in my part of the world use and recognise.
The accepted answer above got me on the right track.
The answer I ended up with to my problem is:
import datetime
today_date = datetime.date.today()
print(today_date)
new_today_date = today_date.strftime("%d/%m/%Y")
print (new_today_date)
The first two lines after the import statement gives today's date in the USA format (2017-01-26). The last two lines convert this to the format recognised in the UK and other countries (26/01/2017).
You can shorten this code, but I left it as is because it is helpful to me as a beginner. I hope this helps other beginner programmers starting out!
Does anyone else else think it's a waste to convert these strings to date/time objects for what is, in the end, a simple text transformation? If you're certain the incoming dates will be valid, you can just use:
>>> ddmmyyyy = "21/12/2008"
>>> yyyymmdd = ddmmyyyy[6:] + "-" + ddmmyyyy[3:5] + "-" + ddmmyyyy[:2]
>>> yyyymmdd
'2008-12-21'
This will almost certainly be faster than the conversion to and from a date.
#case_date= 03/31/2020
#Above is the value stored in case_date in format(mm/dd/yyyy )
demo=case_date.split("/")
new_case_date = demo[1]+"-"+demo[0]+"-"+demo[2]
#new format of date is (dd/mm/yyyy) test by printing it
print(new_case_date)
If you need to convert an entire column (from pandas DataFrame), first convert it (pandas Series) to the datetime format using to_datetime and then use .dt.strftime:
def conv_dates_series(df, col, old_date_format, new_date_format):
df[col] = pd.to_datetime(df[col], format=old_date_format).dt.strftime(new_date_format)
return df
Sample usage:
import pandas as pd
test_df = pd.DataFrame({"Dates": ["1900-01-01", "1999-12-31"]})
old_date_format='%Y-%m-%d'
new_date_format='%d/%m/%Y'
conv_dates_series(test_df, "Dates", old_date_format, new_date_format)
Dates
0 01/01/1900
1 31/12/1999
The most simplest way
While reading the csv file, put an argument parse_dates
df = pd.read_csv("sample.csv", parse_dates=['column_name'])
This will convert the dates of mentioned column to YYYY-MM-DD format
Convert date format DD/MM/YYYY to YYYY-MM-DD according to your question, you can use this:
from datetime import datetime
lastconnection = datetime.strptime("21/12/2008", "%d/%m/%Y").strftime("%Y-%m-%d")
print(lastconnection)
df is your data frame
Dateclm is the column that you want to change
This column should be in DateTime datatype.
df['Dateclm'] = pd.to_datetime(df['Dateclm'])
df.dtypes
#Here is the solution to change the format of the column
df["Dateclm"] = pd.to_datetime(df["Dateclm"]).dt.strftime('%Y-%m-%d')
print(df)