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
Related
I am trying to convert a string to datetime object using the strptime function.
I am encountering a ValueError that says format doesn't match, so I did double checking and confirmed that the format in the string matches the format I am passing as the parameter for strptime.
I have also referenced this question: time data does not match format but there the month and year were swapped.
So does this only work with the '%y-%m-%d %H:%M:%S' format or is it dynamic as per the user input like in my case '%y-%m-%d-%H:%M:%S' ?
input:-
from datetime import datetime
stg = "2022-10-31-01:17:46"
do = datetime.strptime(stg, '%y-%m-%d-%H:%M:%S')
output
ValueError: time data '2022-09-31-01:17:46' does not match format '%y-%m-%d-%H:%M:%S'
Expected output:
#while printing 'do'
2020-09-31-01:17:46
You're almost there. You need %Y instead of %y since you're providing the year with the century (2022 instead of 22).
Your code would be
from datetime import datetime
stg = "2022-10-31-01:17:46"
do = datetime.strptime(stg, '%Y-%m-%d-%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)
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|
+----------------+-------------------+
I try to convert a date in english (2019-10-07) in french (07/10/2016)
I try
dat = '07/10/2019'
dat = time.strftime('%Y-%m-%d')
but got the result '2019-10-16' instead of '2019-10-07'
using datetime you can decide the format in which the source date is provided, and the target format you want.
from datetime import datetime
dat = '07/10/2019'
datetime.strptime(dat, "%d/%m/%Y").strftime("%Y-%m-%d")
out[6]: '2019-10-07'
strftime needs a time/date to convert, and it will use the current date and time if you don't provide one. The previous value of dat is not relevant - this information is not seen by strftime.
You need to provide the time information that strftime will format, as a tuple that you can get by parsing the original string. For this, use strptime (f for format, p for parse).
So:
dmy = '07/10/2019'
ymd = time.strftime('%Y-%m-%d', time.strptime(dmy, '%d/%m/%Y'))
# ^^^^^^^^ ^^^^^^^^
# output schema input schema
# now ymd is '2019-10-07'
(Or you can use the datetime module as in the other answer. This way, the parsing gives you an object, which has a method to format back - so you can write the whole operation "in order" on the line. But the general principle is the same: you need to parse, then format, and you need to specify the schema on each side.)
with :
dat = time.strftime('%Y-%m-%d')
you recover your actual date.
you need to make :
from datetime import datetime
dat = '07/10/2019'
dat = datetime.strptime(dat, '%m/%d/%Y')
print(dat.strftime('%Y-%m-%d') )
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)