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).
Related
I am trying to split a column from a CSV file. The first column contains a date (YYmmdd) and then time (HHmmss) so the string looks like 20221001131245. I want to split this so it reads 2022 10 01 in one column and then 13:12:45 in another.
I have tried the str.split but I recognise my data isn't in a string so this isn't working.
Here is my code so far:
import pandas as pd
CSVPath = "/Desktop/Test Data.csv"
data = pd.read_csv(CSVPath)
print(data)
To answer the question from your comment:
You can use df.drop(['COLUMN_1', 'COLUMN_2'], axis=1) to drop unwanted columns.
I am guessing you want to write the data back to a .csv file? Use the following snippet to only write specific columns:
df[['COLUMN_1', 'COLUMN_2']].to_csv("/Desktop/Test Data Edited.csv")
Use to_datetime combined with strftime:
# convert to datetime
s = pd.to_datetime(df['col'], format='%Y%m%d%H%M%S')
# or if integer as input
# s = pd.to_datetime(df['col'].astype(str), format='%Y%m%d%H%M%S')
# format strings
df['date'] = s.dt.strftime('%Y %m %d')
df['time'] = s.dt.strftime('%H:%M:%S')
Output:
col date time
0 20221001131245 2022 10 01 13:12:45
alternative
using string slicing and concatenation
s = df['col'].str
df['date'] = s[:4]+' '+s[4:6]+' '+s[6:8]
df['time'] = s[8:10]+':'+s[10:12]+':'+s[12:]
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
How can I convert a DataFrame column of strings (in dd/mm/yyyy format) to datetime dtype?
The easiest way is to use to_datetime:
df['col'] = pd.to_datetime(df['col'])
It also offers a dayfirst argument for European times (but beware this isn't strict).
Here it is in action:
In [11]: pd.to_datetime(pd.Series(['05/23/2005']))
Out[11]:
0 2005-05-23 00:00:00
dtype: datetime64[ns]
You can pass a specific format:
In [12]: pd.to_datetime(pd.Series(['05/23/2005']), format="%m/%d/%Y")
Out[12]:
0 2005-05-23
dtype: datetime64[ns]
If your date column is a string of the format '2017-01-01'
you can use pandas astype to convert it to datetime.
df['date'] = df['date'].astype('datetime64[ns]')
or use datetime64[D] if you want Day precision and not nanoseconds
print(type(df_launath['date'].iloc[0]))
yields
<class 'pandas._libs.tslib.Timestamp'>
the same as when you use pandas.to_datetime
You can try it with other formats then '%Y-%m-%d' but at least this works.
You can use the following if you want to specify tricky formats:
df['date_col'] = pd.to_datetime(df['date_col'], format='%d/%m/%Y')
More details on format here:
Python 2 https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior
Python 3 https://docs.python.org/3.7/library/datetime.html#strftime-strptime-behavior
If you have a mixture of formats in your date, don't forget to set infer_datetime_format=True to make life easier.
df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)
Source: pd.to_datetime
or if you want a customized approach:
def autoconvert_datetime(value):
formats = ['%m/%d/%Y', '%m-%d-%y'] # formats to try
result_format = '%d-%m-%Y' # output format
for dt_format in formats:
try:
dt_obj = datetime.strptime(value, dt_format)
return dt_obj.strftime(result_format)
except Exception as e: # throws exception when format doesn't match
pass
return value # let it be if it doesn't match
df['date'] = df['date'].apply(autoconvert_datetime)
Try this solution:
Change '2022–12–31 00:00:00' to '2022–12–31 00:00:01'
Then run this code: pandas.to_datetime(pandas.Series(['2022–12–31 00:00:01']))
Output: 2022–12–31 00:00:01
Multiple datetime columns
If you want to convert multiple string columns to datetime, then using apply() would be useful.
df[['date1', 'date2']] = df[['date1', 'date2']].apply(pd.to_datetime)
You can pass parameters to to_datetime as kwargs.
df[['start_date', 'end_date']] = df[['start_date', 'end_date']].apply(pd.to_datetime, format="%m/%d/%Y")
Use format= to speed up
If the column contains a time component and you know the format of the datetime/time, then passing the format explicitly would significantly speed up the conversion. There's barely any difference if the column is only date, though. In my project, for a column with 5 millions rows, the difference was huge: ~2.5 min vs 6s.
It turns out explicitly specifying the format is about 25x faster. The following runtime plot shows that there's a huge gap in performance depending on whether you passed format or not.
The code used to produce the plot:
import perfplot
import random
mdYHM = range(1, 13), range(1, 29), range(2000, 2024), range(24), range(60)
perfplot.show(
kernels=[lambda x: pd.to_datetime(x), lambda x: pd.to_datetime(x, format='%m/%d/%Y %H:%M')],
labels=['pd.to_datetime(x)', "pd.to_datetime(x, format='%m/%d/%Y %H:%M')"],
n_range=[2**k for k in range(19)],
setup=lambda n: pd.Series([f"{m}/{d}/{Y} {H}:{M}"
for m,d,Y,H,M in zip(*[random.choices(e, k=n) for e in mdYHM])]),
equality_check=pd.Series.equals,
xlabel='len(df)'
)
[PYTHON]
having a column in my dataframe where date has following format:
'JAN07' or '200701'
Is there a simple way to transform this type of date to :
'01-2007' for example?
Maybe using regular expression?
First convert your string column to proper datetimes:
df['colname'] = pd.to_datetime(df['colname'], format='%Y%m') # e.g. 200701
Then back to string in the format you want:
df['colname'] = df['colname'].dt.strftime('%m-%Y')
This two-step process ensures you have no invalid values, yet is relatively fast because it is vectorized (no Python loops).
This might help.
import pandas as pd
from datetime import datetime
df = pd.DataFrame({"A": ['JAN07', '200701']})
print df["A"].apply(lambda x: datetime.strptime(x, '%Y%m').strftime('%m-%Y') if x[0].isdigit() else datetime.strptime(x, '%b%y').strftime('%m-%Y'))
Output:
0 01-2007
1 01-2007
Name: A, dtype: object
How can I convert a DataFrame column of strings (in dd/mm/yyyy format) to datetime dtype?
The easiest way is to use to_datetime:
df['col'] = pd.to_datetime(df['col'])
It also offers a dayfirst argument for European times (but beware this isn't strict).
Here it is in action:
In [11]: pd.to_datetime(pd.Series(['05/23/2005']))
Out[11]:
0 2005-05-23 00:00:00
dtype: datetime64[ns]
You can pass a specific format:
In [12]: pd.to_datetime(pd.Series(['05/23/2005']), format="%m/%d/%Y")
Out[12]:
0 2005-05-23
dtype: datetime64[ns]
If your date column is a string of the format '2017-01-01'
you can use pandas astype to convert it to datetime.
df['date'] = df['date'].astype('datetime64[ns]')
or use datetime64[D] if you want Day precision and not nanoseconds
print(type(df_launath['date'].iloc[0]))
yields
<class 'pandas._libs.tslib.Timestamp'>
the same as when you use pandas.to_datetime
You can try it with other formats then '%Y-%m-%d' but at least this works.
You can use the following if you want to specify tricky formats:
df['date_col'] = pd.to_datetime(df['date_col'], format='%d/%m/%Y')
More details on format here:
Python 2 https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior
Python 3 https://docs.python.org/3.7/library/datetime.html#strftime-strptime-behavior
If you have a mixture of formats in your date, don't forget to set infer_datetime_format=True to make life easier.
df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)
Source: pd.to_datetime
or if you want a customized approach:
def autoconvert_datetime(value):
formats = ['%m/%d/%Y', '%m-%d-%y'] # formats to try
result_format = '%d-%m-%Y' # output format
for dt_format in formats:
try:
dt_obj = datetime.strptime(value, dt_format)
return dt_obj.strftime(result_format)
except Exception as e: # throws exception when format doesn't match
pass
return value # let it be if it doesn't match
df['date'] = df['date'].apply(autoconvert_datetime)
Try this solution:
Change '2022–12–31 00:00:00' to '2022–12–31 00:00:01'
Then run this code: pandas.to_datetime(pandas.Series(['2022–12–31 00:00:01']))
Output: 2022–12–31 00:00:01
Multiple datetime columns
If you want to convert multiple string columns to datetime, then using apply() would be useful.
df[['date1', 'date2']] = df[['date1', 'date2']].apply(pd.to_datetime)
You can pass parameters to to_datetime as kwargs.
df[['start_date', 'end_date']] = df[['start_date', 'end_date']].apply(pd.to_datetime, format="%m/%d/%Y")
Use format= to speed up
If the column contains a time component and you know the format of the datetime/time, then passing the format explicitly would significantly speed up the conversion. There's barely any difference if the column is only date, though. In my project, for a column with 5 millions rows, the difference was huge: ~2.5 min vs 6s.
It turns out explicitly specifying the format is about 25x faster. The following runtime plot shows that there's a huge gap in performance depending on whether you passed format or not.
The code used to produce the plot:
import perfplot
import random
mdYHM = range(1, 13), range(1, 29), range(2000, 2024), range(24), range(60)
perfplot.show(
kernels=[lambda x: pd.to_datetime(x), lambda x: pd.to_datetime(x, format='%m/%d/%Y %H:%M')],
labels=['pd.to_datetime(x)', "pd.to_datetime(x, format='%m/%d/%Y %H:%M')"],
n_range=[2**k for k in range(19)],
setup=lambda n: pd.Series([f"{m}/{d}/{Y} {H}:{M}"
for m,d,Y,H,M in zip(*[random.choices(e, k=n) for e in mdYHM])]),
equality_check=pd.Series.equals,
xlabel='len(df)'
)