I have one field in a pandas DataFrame that was imported as string format.
It should be a datetime variable. How do I convert it to a datetime column and then filter based on date.
Example:
df = pd.DataFrame({'date': ['05SEP2014:00:00:00.000']})
Use the to_datetime function, specifying a format to match your data.
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], format='%d%b%Y:%H:%M:%S.%f')
If you have more than one column to be converted you can do the following:
df[["col1", "col2", "col3"]] = df[["col1", "col2", "col3"]].apply(pd.to_datetime)
You can use the DataFrame method .apply() to operate on the values in Mycol:
>>> df = pd.DataFrame(['05SEP2014:00:00:00.000'],columns=['Mycol'])
>>> df
Mycol
0 05SEP2014:00:00:00.000
>>> import datetime as dt
>>> df['Mycol'] = df['Mycol'].apply(lambda x:
dt.datetime.strptime(x,'%d%b%Y:%H:%M:%S.%f'))
>>> df
Mycol
0 2014-09-05
Use the pandas to_datetime function to parse the column as DateTime. Also, by using infer_datetime_format=True, it will automatically detect the format and convert the mentioned column to DateTime.
import pandas as pd
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], infer_datetime_format=True)
chrisb's answer works:
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], format='%d%b%Y:%H:%M:%S.%f')
however it results in a Python warning of
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
I would guess this is due to some chaining indexing.
Time Saver:
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'])
To silence SettingWithCopyWarning
If you got this warning, then that means your dataframe was probably created by filtering another dataframe. Make a copy of your dataframe before any assignment and you're good to go.
df = df.copy()
df['date'] = pd.to_datetime(df['date'], format='%d%b%Y:%H:%M:%S.%f')
errors='coerce' is useful
If some rows are not in the correct format or not datetime at all, errors= parameter is very useful, so that you can convert the valid rows and handle the rows that contained invalid values later.
df['date'] = pd.to_datetime(df['date'], format='%d%b%Y:%H:%M:%S.%f', errors='coerce')
# for multiple columns
df[['start', 'end']] = df[['start', 'end']].apply(pd.to_datetime, format='%d%b%Y:%H:%M:%S.%f', errors='coerce')
Setting the correct format= is much faster than letting pandas find out1
Long story short, passing the correct format= from the beginning as in chrisb's post is much faster than letting pandas figure out the format, especially if the format contains time component. The runtime difference for dataframes greater than 10k rows is huge (~25 times faster, so we're talking like a couple minutes vs a few seconds). All valid format options can be found at https://strftime.org/.
1 Code used to produce the timeit test plot.
import perfplot
from random import choices
from datetime import datetime
mdYHMSf = range(1,13), range(1,29), range(2000,2024), range(24), *[range(60)]*2, range(1000)
perfplot.show(
kernels=[lambda x: pd.to_datetime(x),
lambda x: pd.to_datetime(x, format='%m/%d/%Y %H:%M:%S.%f'),
lambda x: pd.to_datetime(x, infer_datetime_format=True),
lambda s: s.apply(lambda x: datetime.strptime(x, '%m/%d/%Y %H:%M:%S.%f'))],
labels=["pd.to_datetime(df['date'])",
"pd.to_datetime(df['date'], format='%m/%d/%Y %H:%M:%S.%f')",
"pd.to_datetime(df['date'], infer_datetime_format=True)",
"df['date'].apply(lambda x: datetime.strptime(x, '%m/%d/%Y %H:%M:%S.%f'))"],
n_range=[2**k for k in range(20)],
setup=lambda n: pd.Series([f"{m}/{d}/{Y} {H}:{M}:{S}.{f}"
for m,d,Y,H,M,S,f in zip(*[choices(e, k=n) for e in mdYHMSf])]),
equality_check=pd.Series.equals,
xlabel='len(df)'
)
Just like we convert object data type to float or int. Use astype()
raw_data['Mycol']=raw_data['Mycol'].astype('datetime64[ns]')
Related
I want to get rid of the hours and minutes in the pandas dataframe and convert them to days. The value type in the data is datetime.datetime but when I use the .dt.date function it gives an error.here is the code
df = pd.DataFrame({'id': ['45259191000','45488870311'], 'time': ['2022-10-04 08:57:00', '2022-10-07 11:17:00']})
print(type(df.iat[0, 0]))
df['new'] = df['time'].dt.date
display(df)
this code returns Can only use .dt accessor with datetimelike value
and my datatype <class 'datetime.datetime'> thank you in advance. i hope answer is not very obvious.
try to convert the 'time' column to pandas datetime before you work with it:
df['time'] = pd.to_datetime(df['time'])
I’m trying to read a CSV into Pandas, and then write it to Parquet. The challenge is that the CSV has a date column with a value of 3000-12-31, and apparently Pandas has no way to store that value as an actual date. Because of that, PyArrow fails to read the date value.
An example file and code to reproduce is
test.csv
t
3000-12-31
import pandas as pd
import pyarrow as pa
df = pd.read_csv("test.csv", parse_dates=["t"])
schema = pa.schema([pa.field("t", pa.date64())])
table = pa.Table.from_pandas(df, schema=schema)
This gives (a somewhat unhelpful error)
TypeError: an integer is required (got type str)
What's the right way to do this?
Pandas datetime columns (which use the datetime64[ns] data type) indeed cannot store such dates.
One possible workaround to convert the strings to datetime.datetime objects in an object dtype column. And then pyarrow should be able to accept them to create a date column.
This conversion could eg be done with dateutil:
>>> import dateutil
>>> df['t'] = df['t'].apply(dateutil.parser.parse)
>>> df
t
0 3000-12-31 00:00:00
>>> table = pa.Table.from_pandas(df, schema=schema)
>>> table
pyarrow.Table
t: date64[ms]
or if you use a fixed format, using datetime.date.strptime is probably more reliable:
>>> import datetime
>>> df['t'] = df['t'].apply(lambda s: datetime.datetime.strptime(s, "%Y-%m-%d"))
>>> table = pa.Table.from_pandas(df, schema=schema)
>>> table
pyarrow.Table
t: date64[ms]
Is there a simple way to ignore all even/odd rows when reading a csv using pandas?
I know skiprows argument in pd.read_csv but for that I'll need to know the number of rows in advance.
The pd.read_csv skiprows argument accepts a callable, so you could use a lambda function. E.g.:
df = pd.read_csv(some_path, skiprows=lambda x: x%2 == 0)
A possible solution after reading would be:
import pandas as pd
df = pd.read_csv(some_path)
# remove odd rows:
df = df.iloc[::2]
# remove even rows:
df = df.iloc[1::2]
I have one field in a pandas DataFrame that was imported as string format.
It should be a datetime variable. How do I convert it to a datetime column and then filter based on date.
Example:
df = pd.DataFrame({'date': ['05SEP2014:00:00:00.000']})
Use the to_datetime function, specifying a format to match your data.
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], format='%d%b%Y:%H:%M:%S.%f')
If you have more than one column to be converted you can do the following:
df[["col1", "col2", "col3"]] = df[["col1", "col2", "col3"]].apply(pd.to_datetime)
You can use the DataFrame method .apply() to operate on the values in Mycol:
>>> df = pd.DataFrame(['05SEP2014:00:00:00.000'],columns=['Mycol'])
>>> df
Mycol
0 05SEP2014:00:00:00.000
>>> import datetime as dt
>>> df['Mycol'] = df['Mycol'].apply(lambda x:
dt.datetime.strptime(x,'%d%b%Y:%H:%M:%S.%f'))
>>> df
Mycol
0 2014-09-05
Use the pandas to_datetime function to parse the column as DateTime. Also, by using infer_datetime_format=True, it will automatically detect the format and convert the mentioned column to DateTime.
import pandas as pd
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], infer_datetime_format=True)
chrisb's answer works:
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], format='%d%b%Y:%H:%M:%S.%f')
however it results in a Python warning of
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
I would guess this is due to some chaining indexing.
Time Saver:
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'])
To silence SettingWithCopyWarning
If you got this warning, then that means your dataframe was probably created by filtering another dataframe. Make a copy of your dataframe before any assignment and you're good to go.
df = df.copy()
df['date'] = pd.to_datetime(df['date'], format='%d%b%Y:%H:%M:%S.%f')
errors='coerce' is useful
If some rows are not in the correct format or not datetime at all, errors= parameter is very useful, so that you can convert the valid rows and handle the rows that contained invalid values later.
df['date'] = pd.to_datetime(df['date'], format='%d%b%Y:%H:%M:%S.%f', errors='coerce')
# for multiple columns
df[['start', 'end']] = df[['start', 'end']].apply(pd.to_datetime, format='%d%b%Y:%H:%M:%S.%f', errors='coerce')
Setting the correct format= is much faster than letting pandas find out1
Long story short, passing the correct format= from the beginning as in chrisb's post is much faster than letting pandas figure out the format, especially if the format contains time component. The runtime difference for dataframes greater than 10k rows is huge (~25 times faster, so we're talking like a couple minutes vs a few seconds). All valid format options can be found at https://strftime.org/.
1 Code used to produce the timeit test plot.
import perfplot
from random import choices
from datetime import datetime
mdYHMSf = range(1,13), range(1,29), range(2000,2024), range(24), *[range(60)]*2, range(1000)
perfplot.show(
kernels=[lambda x: pd.to_datetime(x),
lambda x: pd.to_datetime(x, format='%m/%d/%Y %H:%M:%S.%f'),
lambda x: pd.to_datetime(x, infer_datetime_format=True),
lambda s: s.apply(lambda x: datetime.strptime(x, '%m/%d/%Y %H:%M:%S.%f'))],
labels=["pd.to_datetime(df['date'])",
"pd.to_datetime(df['date'], format='%m/%d/%Y %H:%M:%S.%f')",
"pd.to_datetime(df['date'], infer_datetime_format=True)",
"df['date'].apply(lambda x: datetime.strptime(x, '%m/%d/%Y %H:%M:%S.%f'))"],
n_range=[2**k for k in range(20)],
setup=lambda n: pd.Series([f"{m}/{d}/{Y} {H}:{M}:{S}.{f}"
for m,d,Y,H,M,S,f in zip(*[choices(e, k=n) for e in mdYHMSf])]),
equality_check=pd.Series.equals,
xlabel='len(df)'
)
Just like we convert object data type to float or int. Use astype()
raw_data['Mycol']=raw_data['Mycol'].astype('datetime64[ns]')
I have one field in a pandas DataFrame that was imported as string format.
It should be a datetime variable. How do I convert it to a datetime column and then filter based on date.
Example:
df = pd.DataFrame({'date': ['05SEP2014:00:00:00.000']})
Use the to_datetime function, specifying a format to match your data.
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], format='%d%b%Y:%H:%M:%S.%f')
If you have more than one column to be converted you can do the following:
df[["col1", "col2", "col3"]] = df[["col1", "col2", "col3"]].apply(pd.to_datetime)
You can use the DataFrame method .apply() to operate on the values in Mycol:
>>> df = pd.DataFrame(['05SEP2014:00:00:00.000'],columns=['Mycol'])
>>> df
Mycol
0 05SEP2014:00:00:00.000
>>> import datetime as dt
>>> df['Mycol'] = df['Mycol'].apply(lambda x:
dt.datetime.strptime(x,'%d%b%Y:%H:%M:%S.%f'))
>>> df
Mycol
0 2014-09-05
Use the pandas to_datetime function to parse the column as DateTime. Also, by using infer_datetime_format=True, it will automatically detect the format and convert the mentioned column to DateTime.
import pandas as pd
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], infer_datetime_format=True)
chrisb's answer works:
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'], format='%d%b%Y:%H:%M:%S.%f')
however it results in a Python warning of
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
I would guess this is due to some chaining indexing.
Time Saver:
raw_data['Mycol'] = pd.to_datetime(raw_data['Mycol'])
To silence SettingWithCopyWarning
If you got this warning, then that means your dataframe was probably created by filtering another dataframe. Make a copy of your dataframe before any assignment and you're good to go.
df = df.copy()
df['date'] = pd.to_datetime(df['date'], format='%d%b%Y:%H:%M:%S.%f')
errors='coerce' is useful
If some rows are not in the correct format or not datetime at all, errors= parameter is very useful, so that you can convert the valid rows and handle the rows that contained invalid values later.
df['date'] = pd.to_datetime(df['date'], format='%d%b%Y:%H:%M:%S.%f', errors='coerce')
# for multiple columns
df[['start', 'end']] = df[['start', 'end']].apply(pd.to_datetime, format='%d%b%Y:%H:%M:%S.%f', errors='coerce')
Setting the correct format= is much faster than letting pandas find out1
Long story short, passing the correct format= from the beginning as in chrisb's post is much faster than letting pandas figure out the format, especially if the format contains time component. The runtime difference for dataframes greater than 10k rows is huge (~25 times faster, so we're talking like a couple minutes vs a few seconds). All valid format options can be found at https://strftime.org/.
1 Code used to produce the timeit test plot.
import perfplot
from random import choices
from datetime import datetime
mdYHMSf = range(1,13), range(1,29), range(2000,2024), range(24), *[range(60)]*2, range(1000)
perfplot.show(
kernels=[lambda x: pd.to_datetime(x),
lambda x: pd.to_datetime(x, format='%m/%d/%Y %H:%M:%S.%f'),
lambda x: pd.to_datetime(x, infer_datetime_format=True),
lambda s: s.apply(lambda x: datetime.strptime(x, '%m/%d/%Y %H:%M:%S.%f'))],
labels=["pd.to_datetime(df['date'])",
"pd.to_datetime(df['date'], format='%m/%d/%Y %H:%M:%S.%f')",
"pd.to_datetime(df['date'], infer_datetime_format=True)",
"df['date'].apply(lambda x: datetime.strptime(x, '%m/%d/%Y %H:%M:%S.%f'))"],
n_range=[2**k for k in range(20)],
setup=lambda n: pd.Series([f"{m}/{d}/{Y} {H}:{M}:{S}.{f}"
for m,d,Y,H,M,S,f in zip(*[choices(e, k=n) for e in mdYHMSf])]),
equality_check=pd.Series.equals,
xlabel='len(df)'
)
Just like we convert object data type to float or int. Use astype()
raw_data['Mycol']=raw_data['Mycol'].astype('datetime64[ns]')