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 calling some financial data from an API which is storing the time values as (I think) UTC (example below):
enter image description here
I cannot seem to convert the entire column into a useable date, I can do it for a single value using the following code so I know this works, but I have 1000's of rows with this problem and thought pandas would offer an easier way to update all the values.
from datetime import datetime
tx = int('1645804609719')/1000
print(datetime.utcfromtimestamp(tx).strftime('%Y-%m-%d %H:%M:%S'))
Any help would be greatly appreciated.
Simply use pandas.DataFrame.apply:
df['date'] = df.date.apply(lambda x: datetime.utcfromtimestamp(int(x)/1000).strftime('%Y-%m-%d %H:%M:%S'))
Another way to do it is by using pd.to_datetime as recommended by Panagiotos in the comments:
df['date'] = pd.to_datetime(df['date'],unit='ms')
You can use "to_numeric" to convert the column in integers, "div" to divide it by 1000 and finally a loop to iterate the dataframe column with datetime to get the format you want.
import pandas as pd
import datetime
df = pd.DataFrame({'date': ['1584199972000', '1645804609719'], 'values': [30,40]})
df['date'] = pd.to_numeric(df['date']).div(1000)
for i in range(len(df)):
df.iloc[i,0] = datetime.utcfromtimestamp(df.iloc[i,0]).strftime('%Y-%m-%d %H:%M:%S')
print(df)
Output:
date values
0 2020-03-14 15:32:52 30
1 2022-02-25 15:56:49 40
One column in dataframe is like this:
2018-01-23 23:55:07
I want to convert the values in this column to unix time.
Below is my code:
def convert_to_unix(s):
return float(time.mktime(datetime.datetime.strptime(s, "%Y-%m-%d %H:%M:%S").timetuple()))
pd.set_option('display.max_columns', None)
fields=['JOB_START_TIMESTAMP','JOB_END_TIMESTAMP','JOB_RUNTIME_SECONDS', 'JOB_NODES_USED']
df_temp=pd.read_csv('a.csv',usecols=fields)
df_temp['JOB_START_TIMESTAMP']=df_temp['JOB_START_TIMESTAMP'].apply(convert_to_unix)
Then it shows errorTypeError: must be string, not float.
error_ image
Can anybody help me? Thanks very much!
Code below converts a date column (datetime64[ns]) to unix time (float64).
Import libraries
import pandas as pd
import numpy as np
from datetime import datetime
from time import mktime
Create sample dataframe
df = pd.DataFrame({'Date': ['2018-01-23 23:55:07', '2017-01-23 23:55:07', '2015-11-23 11:50:07',
'2013-01-03 13:55:07', '2007-01-24 23:55:07', '2017-12-23 12:55:07']})
df['Date'] = pd.to_datetime(df['Date'])
df
Function that converts to unix time
def convert_to_unix(s):
return df.apply(lambda x: mktime((x['Date']).timetuple()),axis=1)
Get unix time
df['unix_time'] = convert_to_unix(df)
df
df.dtypes
Alternative without using function
df['unix_time'] = df.apply(lambda x: mktime((x['Date']).timetuple()),axis=1)
df
Thanks Kunar. My problem is there is NaTType in my data.
His answer works and is concise since it is in the comments and hided, I just put it here.
df_temp['JOB_START_TIMESTAMP']=df_temp['JOB_START_TIMESTAMP'].apply(pd.Timestamp).apply(pd.Timestamp.timestamp)
Today I was positively surprised by the fact that while reading data from a data file (for example) pandas is able to recognize types of values:
df = pandas.read_csv('test.dat', delimiter=r"\s+", names=['col1','col2','col3'])
For example it can be checked in this way:
for i, r in df.iterrows():
print type(r['col1']), type(r['col2']), type(r['col3'])
In particular integer, floats and strings were recognized correctly. However, I have a column that has dates in the following format: 2013-6-4. These dates were recognized as strings (not as python date-objects). Is there a way to "learn" pandas to recognized dates?
You should add parse_dates=True, or parse_dates=['column name'] when reading, thats usually enough to magically parse it. But there are always weird formats which need to be defined manually. In such a case you can also add a date parser function, which is the most flexible way possible.
Suppose you have a column 'datetime' with your string, then:
from datetime import datetime
dateparse = lambda x: datetime.strptime(x, '%Y-%m-%d %H:%M:%S')
df = pd.read_csv(infile, parse_dates=['datetime'], date_parser=dateparse)
This way you can even combine multiple columns into a single datetime column, this merges a 'date' and a 'time' column into a single 'datetime' column:
dateparse = lambda x: datetime.strptime(x, '%Y-%m-%d %H:%M:%S')
df = pd.read_csv(infile, parse_dates={'datetime': ['date', 'time']}, date_parser=dateparse)
You can find directives (i.e. the letters to be used for different formats) for strptime and strftime in this page.
Perhaps the pandas interface has changed since #Rutger answered, but in the version I'm using (0.15.2), the date_parser function receives a list of dates instead of a single value. In this case, his code should be updated like so:
from datetime import datetime
import pandas as pd
dateparse = lambda dates: [datetime.strptime(d, '%Y-%m-%d %H:%M:%S') for d in dates]
df = pd.read_csv('test.dat', parse_dates=['datetime'], date_parser=dateparse)
Since the original question asker said he wants dates and the dates are in 2013-6-4 format, the dateparse function should really be:
dateparse = lambda dates: [datetime.strptime(d, '%Y-%m-%d').date() for d in dates]
You could use pandas.to_datetime() as recommended in the documentation for pandas.read_csv():
If a column or index contains an unparseable date, the entire column
or index will be returned unaltered as an object data type. For
non-standard datetime parsing, use pd.to_datetime after pd.read_csv.
Demo:
>>> D = {'date': '2013-6-4'}
>>> df = pd.DataFrame(D, index=[0])
>>> df
date
0 2013-6-4
>>> df.dtypes
date object
dtype: object
>>> df['date'] = pd.to_datetime(df.date, format='%Y-%m-%d')
>>> df
date
0 2013-06-04
>>> df.dtypes
date datetime64[ns]
dtype: object
When merging two columns into a single datetime column, the accepted answer generates an error (pandas version 0.20.3), since the columns are sent to the date_parser function separately.
The following works:
def dateparse(d,t):
dt = d + " " + t
return pd.datetime.strptime(dt, '%d/%m/%Y %H:%M:%S')
df = pd.read_csv(infile, parse_dates={'datetime': ['date', 'time']}, date_parser=dateparse)
pandas read_csv method is great for parsing dates. Complete documentation at http://pandas.pydata.org/pandas-docs/stable/generated/pandas.io.parsers.read_csv.html
you can even have the different date parts in different columns and pass the parameter:
parse_dates : boolean, list of ints or names, list of lists, or dict
If True -> try parsing the index. If [1, 2, 3] -> try parsing columns 1, 2, 3 each as a
separate date column. If [[1, 3]] -> combine columns 1 and 3 and parse as a single date
column. {‘foo’ : [1, 3]} -> parse columns 1, 3 as date and call result ‘foo’
The default sensing of dates works great, but it seems to be biased towards north american Date formats. If you live elsewhere you might occasionally be caught by the results. As far as I can remember 1/6/2000 means 6 January in the USA as opposed to 1 Jun where I live. It is smart enough to swing them around if dates like 23/6/2000 are used. Probably safer to stay with YYYYMMDD variations of date though. Apologies to pandas developers,here but i have not tested it with local dates recently.
you can use the date_parser parameter to pass a function to convert your format.
date_parser : function
Function to use for converting a sequence of string columns to an array of datetime
instances. The default uses dateutil.parser.parser to do the conversion.
Yes - according to the pandas.read_csv documentation:
Note: A fast-path exists for iso8601-formatted dates.
So if your csv has a column named datetime and the dates looks like 2013-01-01T01:01 for example, running this will make pandas (I'm on v0.19.2) pick up the date and time automatically:
df = pd.read_csv('test.csv', parse_dates=['datetime'])
Note that you need to explicitly pass parse_dates, it doesn't work without.
Verify with:
df.dtypes
You should see the datatype of the column is datetime64[ns]
While loading csv file contain date column.We have two approach to to make pandas to
recognize date column i.e
Pandas explicit recognize the format by arg date_parser=mydateparser
Pandas implicit recognize the format by agr infer_datetime_format=True
Some of the date column data
01/01/18
01/02/18
Here we don't know the first two things It may be month or day. So in this case we have to use
Method 1:-
Explicit pass the format
mydateparser = lambda x: pd.datetime.strptime(x, "%m/%d/%y")
df = pd.read_csv(file_name, parse_dates=['date_col_name'],
date_parser=mydateparser)
Method 2:- Implicit or Automatically recognize the format
df = pd.read_csv(file_name, parse_dates=[date_col_name],infer_datetime_format=True)
In addition to what the other replies said, if you have to parse very large files with hundreds of thousands of timestamps, date_parser can prove to be a huge performance bottleneck, as it's a Python function called once per row. You can get a sizeable performance improvements by instead keeping the dates as text while parsing the CSV file and then converting the entire column into dates in one go:
# For a data column
df = pd.read_csv(infile, parse_dates={'mydatetime': ['date', 'time']})
df['mydatetime'] = pd.to_datetime(df['mydatetime'], exact=True, cache=True, format='%Y-%m-%d %H:%M:%S')
# For a DateTimeIndex
df = pd.read_csv(infile, parse_dates={'mydatetime': ['date', 'time']}, index_col='mydatetime')
df.index = pd.to_datetime(df.index, exact=True, cache=True, format='%Y-%m-%d %H:%M:%S')
# For a MultiIndex
df = pd.read_csv(infile, parse_dates={'mydatetime': ['date', 'time']}, index_col=['mydatetime', 'num'])
idx_mydatetime = df.index.get_level_values(0)
idx_num = df.index.get_level_values(1)
idx_mydatetime = pd.to_datetime(idx_mydatetime, exact=True, cache=True, format='%Y-%m-%d %H:%M:%S')
df.index = pd.MultiIndex.from_arrays([idx_mydatetime, idx_num])
For my use case on a file with 200k rows (one timestamp per row), that cut down processing time from about a minute to less than a second.
Read the existing string columns in date and time format respectively
pd.read_csv('CGMData.csv', parse_dates=['Date', 'Time'])
Resulted Columns
Concat string columns of date and time and add new column of datetype object - Remove Original columns
if want to rename the new column name then pass as dictionary as
show in below example and the new column name will be the key name,
if pass as list of column, new column name will be concate of column name passed in the list separated by _ e.g Date_Time
# parse_dates={'given_name': ['Date', 'Time']}
pd.read_csv("InsulinData.csv",low_memory=False,
parse_dates=[['Date', 'Time']])
pd.read_csv("InsulinData.csv",low_memory=False,
parse_dates={'date_time': ['Date', 'Time']})
Concat string columns of date and time and add new column of datetype object and Keep the Original columns
pd.read_csv("InsulinData.csv",low_memory=False,
parse_dates=[['Date', 'Time']], keep_date_col=True)
Want to change the format of date and time when read from csv
parser = lambda x: pd.to_datetime(x, format='%Y-%m-%d %H:%M:%S')
pd.read_csv('path', date_parser=parser, parse_dates=['date', 'time'])
If performance matters to you make sure you time:
import sys
import timeit
import pandas as pd
print('Python %s on %s' % (sys.version, sys.platform))
print('Pandas version %s' % pd.__version__)
repeat = 3
numbers = 100
def time(statement, _setup=None):
print (min(
timeit.Timer(statement, setup=_setup or setup).repeat(
repeat, numbers)))
print("Format %m/%d/%y")
setup = """import pandas as pd
import io
data = io.StringIO('''\
ProductCode,Date
''' + '''\
x1,07/29/15
x2,07/29/15
x3,07/29/15
x4,07/30/15
x5,07/29/15
x6,07/29/15
x7,07/29/15
y7,08/05/15
x8,08/05/15
z3,08/05/15
''' * 100)"""
time('pd.read_csv(data); data.seek(0)')
time('pd.read_csv(data, parse_dates=["Date"]); data.seek(0)')
time('pd.read_csv(data, parse_dates=["Date"],'
'infer_datetime_format=True); data.seek(0)')
time('pd.read_csv(data, parse_dates=["Date"],'
'date_parser=lambda x: pd.datetime.strptime(x, "%m/%d/%y")); data.seek(0)')
print("Format %Y-%m-%d %H:%M:%S")
setup = """import pandas as pd
import io
data = io.StringIO('''\
ProductCode,Date
''' + '''\
x1,2016-10-15 00:00:43
x2,2016-10-15 00:00:56
x3,2016-10-15 00:00:56
x4,2016-10-15 00:00:12
x5,2016-10-15 00:00:34
x6,2016-10-15 00:00:55
x7,2016-10-15 00:00:06
y7,2016-10-15 00:00:01
x8,2016-10-15 00:00:00
z3,2016-10-15 00:00:02
''' * 1000)"""
time('pd.read_csv(data); data.seek(0)')
time('pd.read_csv(data, parse_dates=["Date"]); data.seek(0)')
time('pd.read_csv(data, parse_dates=["Date"],'
'infer_datetime_format=True); data.seek(0)')
time('pd.read_csv(data, parse_dates=["Date"],'
'date_parser=lambda x: pd.datetime.strptime(x, "%Y-%m-%d %H:%M:%S")); data.seek(0)')
prints:
Python 3.7.1 (v3.7.1:260ec2c36a, Oct 20 2018, 03:13:28)
[Clang 6.0 (clang-600.0.57)] on darwin
Pandas version 0.23.4
Format %m/%d/%y
0.19123052499999993
8.20691274
8.143124389
1.2384357139999977
Format %Y-%m-%d %H:%M:%S
0.5238807110000039
0.9202787830000005
0.9832778819999959
12.002349824999996
So with iso8601-formatted date (%Y-%m-%d %H:%M:%S is apparently an iso8601-formatted date, I guess the T can be dropped and replaced by a space) you should not specify infer_datetime_format (which does not make a difference with more common ones either apparently) and passing your own parser in just cripples performance. On the other hand, date_parser does make a difference with not so standard day formats. Be sure to time before you optimize, as usual.
You can use the parameter date_parser with a function for converting a sequence of string columns to an array of datetime instances:
parser = lambda x: pd.to_datetime(x, format='%Y-%m-%d %H:%M:%S')
pd.read_csv('path', date_parser=parser, parse_dates=['date_col1', 'date_col2'])
Yes, this code works like breeze. Here index 0 refers to the index of the date column.
df = pd.read_csv(filepath, parse_dates=[0], infer_datetime_format = True)
No, there is no way in pandas to automatically recognize date columns.
Pandas does a poor job at type inference. It basically puts most columns as the generic object type, unless you manually work around it eg. using the abovementioned parse_dates parameter.
If you want to automatically detect columns types, you'd have to use a separate data profiling tool, eg. visions, and then cast or feed the inferred types back into your DataFrame constructor (eg. for dates and from_csv, using the parse_dates parameter).
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)