Consider the following Polars dataframe:
import polars as pl
df = pl.DataFrame({'year': [2023], 'month': [2], 'day': [1]})
I want to construct a date column from year, month and day. I know this can be done by first concatenating into a string and then parsing this:
df.with_column(
pl.concat_str([pl.col('year'), pl.col('month'), pl.col('day')], sep='-')
.str.strptime(pl.Date).alias('date')
)
But that seems like a detour. Is it possible to construct it directly with the three inputs? Something like this (that doesn't work):
import datetime
df.with_column(
datetime.date(pl.col('year'), pl.col('month'), pl.col('day')).alias('date')
)
polars has a pl.datetime and pl.date function that works the same as base datetime except that it takes expressions so you can just do
df.with_columns(pl.date(pl.col('year'), pl.col('month'), pl.col('day')).alias('date'))
Related
I have a big excel file with a datetime format column which are in strings. The column looks like this:
ingezameldop
2022-10-10 15:51:18
2022-10-10 15:56:19
I have found two ways of trying to do this, however they do not work.
First (nice way):
import pandas as pd
from datetime import datetime
from datetime import date
dagStart = datetime.strptime(str(date.today())+' 06:00:00', '%Y-%m-%d %H:%M:%S')
dagEind = datetime.strptime(str(date.today())+' 23:00:00', '%Y-%m-%d %H:%M:%S')
data = pd.read_excel('inzamelbestand.xlsx', index_col=9)
data = data.loc[pd.to_datetime(data['ingezameldop']).dt.time.between(dagStart.time(), dagEind.time())]
data.to_excel("oefenexcel.xlsx")
However, this returns me with an excel file identical to the original one. I cant seem to fix this.
Second way (sketchy):
import pandas as pd
from datetime import datetime
from datetime import date
df = pd.read_excel('inzamelbestand.xlsx', index_col=9)
# uitfilteren dag van vandaag
dag = str(date.today())
dag1 = dag[8]+dag[9]
vgl = df['ingezameldop']
vgl2 = vgl.str[8]+vgl.str[9]
df = df.loc[vgl2 == dag1]
# uitfilteren vanaf 6 uur 's ochtends
# str11 str12 = uur
df.to_excel("oefenexcel.xlsx")
This one works for filtering out the exact day. But when I want to filter out the hours it does not. Because I use the same way (getting the 11nd and 12th character from the string) but I cant use logic operators (>=) on strings, so I cant filter out for times >6
You can modify this line of code
data = data.loc[pd.to_datetime(data['ingezameldop']).dt.time.between(dagStart.time(), dagEind.time())]
as
(dagStart.hour, dagStart.minute) <= (data['ingezameldop'].hour, data['ingezameldop'].minute) < (dagEind.hour, dagEind.minute)
to get boolean values that are only true for records within the date range.
dagStart, dagEind and data['ingezameldop'] must be in datetime format.
In order to apply it on individual element of the column, wrap it in a function and use apply as follows
def filter(ingezameldop, dagStart, dagEind):
return (dagStart.hour, dagStart.minute) <= (data['ingezameldop'].hour, data['ingezameldop'].minute) < (dagEind.hour, dagEind.minute)
then apply the filter on the column in this way
data['filter'] = data['ingezameldop'].apply(filter, dagStart=dagStart, dagEind=dagEind)
That will apply the function on individual series element which must be in datetime format
I am using python to do some data cleaning and i've used the datetime module to split date time and tried to create another column with just the time.
My script works but it just takes the last value of the data frame.
Here is the code:
import datetime
i = 0
for index, row in df.iterrows():
date = datetime.datetime.strptime(df.iloc[i, 0], "%Y-%m-%dT%H:%M:%SZ")
df['minutes'] = date.minute
i = i + 1
This is the dataframe :
Output
df['minutes'] = date.minute reassigns the entire 'minutes' column with the scalar value date.minute from the last iteration.
You don't need a loop, as 99% of the cases when using pandas.
You can use vectorized assignment, just replace 'source_column_name' with the name of the column with the source data.
df['minutes'] = pd.to_datetime(df['source_column_name'], format='%Y-%m-%dT%H:%M:%SZ').dt.minute
It is also most likely that you won't need to specify format as pd.to_datetime is fairly smart.
Quick example:
df = pd.DataFrame({'a': ['2020.1.13', '2019.1.13']})
df['year'] = pd.to_datetime(df['a']).dt.year
print(df)
outputs
a year
0 2020.1.13 2020
1 2019.1.13 2019
Seems like you're trying to get the time column from the datetime which is in string format. That's what I understood from your post.
Could you give this a shot?
from datetime import datetime
import pandas as pd
def get_time(date_cell):
dt = datetime.strptime(date_cell, "%Y-%m-%dT%H:%M:%SZ")
return datetime.strftime(dt, "%H:%M:%SZ")
df['time'] = df['date_time'].apply(get_time)
I want to convert the date in a column in a dataframe to a different format. Currently, it has this format: '2019-11-20T01:04:18'. I want it to have this format: 20-11-19 1:04.
I think I need to develop a loop and generate a new column for the new date format. So essentially, in the loop, I would refer to the initial column and then generate the variable for the new column in the format I want.
Can someone help me out to complete this task?
The following code works for one occasion:
import datetime
d = datetime.datetime.strptime('2019-11-20T01:04:18', '%Y-%m-%dT%H:%M:%S')
print d.strftime('%d-%m-%y %H:%M')
From a previous answer in this site , this should be able to help you, comments give explanation
You can read your data into pandas from csv or database or create some test data as shown below for testing.
>>> import pandas as pd
>>> df = pd.DataFrame({'column': {0: '26/1/2016', 1: '26/1/2016'}})
>>> # First convert the column to datetime datatype
>>> df['column'] = pd.to_datetime(df.column)
>>> # Then call the datetime object format() method, set the modifiers you want here
>>> df['column'] = df['column'].dt.strftime('%Y-%m-%dT%H:%M:%S')
>>> df
column
0 2016-01-26T00:00:00
1 2016-01-26T00:00:00
NB. Check to ensure that all your columns have similar date strings
You can either achieve it like this:
from datetime import datetime
df['your_column'] = df['your_column'].apply(lambda x: datetime.strptime(x, '%Y-%m-%dT%H:%M:%S').strftime('%d-%m-%y %H:%M'))
I am trying to apply a simple function to extract the month from a string column in a pandas dataframe, where the string is of the form m/d/yyyy.
The dataframe is called data, the date column is called transaction date, and my new proposed month column I wish to call transaction month.
The below works just fine:
data['transaction month']=data['transaction date'].map(lambda x: x[0:x.index('/')])
However, if I try to do the same thing with a named function, it just returns a column where every value is None
def extract_month_from_date(date):
return date[0:date.index('/')]
data['transaction month 2']=data['transaction date'].map(extract_month_from_date)
I've stared at the code for long enough that I think I'm going crazy, what's wrong with the
You can extract the month via pd.Series.dt.month:
import pandas as pd
df = pd.DataFrame({'date': ['8/2/2018']})
df['date'] = pd.to_datetime(df['date'])
df['month'] = df['date'].dt.month
# date month
# 0 2018-08-02 8
I have a dataframe full of dates and I would like to select all dates where the month==12 and the day==25 and add replace the zero in the xmas column with a 1.
Anyway to do this? the second line of my code errors out.
df = DataFrame({'date':[datetime(2013,1,1).date() + timedelta(days=i) for i in range(0,365*2)], 'xmas':np.zeros(365*2)})
df[df['date'].month==12 and df['date'].day==25] = 1
Pandas Series with datetime now behaves differently. See .dt accessor.
This is how it should be done now:
df.loc[(df['date'].dt.day==25) & (cust_df['date'].dt.month==12), 'xmas'] = 1
Basically what you tried won't work as you need to use the & to compare arrays, additionally you need to use parentheses due to operator precedence. On top of this you should use loc to perform the indexing:
df.loc[(df['date'].month==12) & (df['date'].day==25), 'xmas'] = 1
An update was needed in reply to this question. As of today, there's a slight difference in how you extract months from datetime objects in a pd.Series.
So from the very start, incase you have a raw date column, first convert it to datetime objects by using a simple function:
import datetime as dt
def read_as_datetime(str_date):
# replace %Y-%m-%d with your own date format
return dt.datetime.strptime(str_date,'%Y-%m-%d')
then apply this function to your dates column and save results in a new column namely datetime:
df['datetime'] = df.dates.apply(read_as_datetime)
finally in order to extract dates by day and month, use the same piece of code that #Shayan RC explained, with this slight change; notice the dt.datetime after calling the datetime column:
df.loc[(df['datetime'].dt.datetime.month==12) &(df['datetime'].dt.datetime.day==25),'xmas'] =1