I'm trying to use dt.datetime to split my data by date. However, my date structure is ' date2013-07-04' see attached image. Therefore, it doesn't fit with the traditional '%m/%d/%Y %H:%M:%S.%f' format: I've tried re-formatting but I get 2 errors;
does not match format
redefinition of group name 'd' as group 4; was group 1.
The line of code I have is:
x = 'date2013-07-04'
df['Date'] = filename['Date'].apply(lambda x: dt.datetime.strptime(x,'%date%YYYY-%mm-%dd'))
Then the errors pull up. Can anyone help? Cheers!
The reason it doesn't work is that your format string is invalid, you can find the reference for python's date formatting here.
In the meantime you can try this snippet:
import datetime as dt
x = "date2013-07-04"
print(dt.datetime.strptime(x, "date%Y-%m-%d"))
It should do it.
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 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'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
I'm preprocessing data and one column represents dates such as '6/1/51'
I'm trying to convert the string to a date object and so far what I have is:
date = row[2].strip()
format = "%m/%d/%y"
datetime_object = datetime.strptime(date, format)
date_object = datetime_object.date()
print(date_object)
print(type(date_object))
The problem I'm facing is changing 2051 to 1951.
I tried writing
format = "%m/%d/19%y"
But it gives me a ValueError.
ValueError: time data '6/1/51' does not match format '%m/%d/19%y'
I couldn't easily find the answer online so I'm asking here. Can anyone please help me with this?
Thanks.
Parse the date without the century using '%m/%d/%y', then:
year_1900 = datetime_object.year - 100
datetime_object = datetime_object.replace(year=year_1900)
You should put conditionals around that so you only do it on dates that are actually in the 1900's, for example anything later than today.
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)