I have a dataframe full of Mondays or Tuesdays as dates and another dataframe full of Mondays, Tuesdays and Wednesdays. I'd like to match each of the dates in the second dataframe with the Monday or Tuesday in the first dataframe of the same week:
import pandas as pd
df1 = pd.DataFrame(['01-25-2022','01-17-2022'])
df2 = pd.DataFrame(['01-26-2022','01-27-2022','01-20-2022'])
So in that example I would like a third dataframe as output which combines df1 and df2:
df3 = pd.DataFrame([['01-25-2022','01-25-2022','01-17-2022'],['01-26-2022','01-27-2022','01-20-2022']]).T
You can get the week (Mon-Sun) by using .dt.to_period('W') (by default .dt.to_period('W-SUN') for Sunday as last week day):
df1 = pd.DataFrame({'A': ['01-25-2022','01-17-2022']},
dtype='datetime64[s]')
df2 = pd.DataFrame({'B': ['01-26-2022','01-27-2022','01-20-2022']},
dtype='datetime64[s]')
df1.merge(df2,
left_on=df1['A'].dt.to_period('W'),
right_on=df2['B'].dt.to_period('W'),
how='right'
).drop(columns='key_0')
output:
A B
0 2022-01-25 2022-01-26
1 2022-01-25 2022-01-27
2 2022-01-17 2022-01-20
Related
I'd like to subtract dates if the next row's id is the same. I'm able to subtract dates, but stuck on creating the condition to check if the next row has the same id.
d = {'date':['2021-01', '2020-01', '2020-05', '2021-01'], 'id':['a', 'a', 'b', 'b']}
df = pd.DataFrame(data=d)
date id
2021-01 a
2020-01 a
2020-05 b
2021-01 b
My code
df = df.sort_values(by=['id', 'date'])
df['date_diff'] = pd.to_datetime(df['date']) - pd.to_datetime(df['date'].shift())
result
date id date_diff
2020-01 a NaT
2021-01 a 366 days
2020-05 b -245 days
2021-01 b 245 days
Expected result should as below, which the dates only be subtracted when the ids are the same.
Chain with groupby
df['date'] = pd.to_datetime(df['date'])
df['date_diff'] = df.groupby('id')['date'].diff()
df['date']=pd.to_datetime(df['date'])
df['date_diff']=df.groupby('id')['date'].diff()
My Pandas df:
import pandas as pd
import io
data = """date value
"2015-09-01" 71.925000
"2015-09-06" 71.625000
"2015-09-11" 71.333333
"2015-09-12" 64.571429
"2015-09-21" 72.285714
"""
df = pd.read_table(io.StringIO(data), delim_whitespace=True)
df.date = pd.to_datetime(df.date)
I Given a user input date ( 01-09-2015).
I would like to keep only those date where difference between date and input date is multiple of 5.
Expected output:
input = 01-09-2015
df:
date value
0 2015-09-01 71.925000
1 2015-09-06 71.625000
2 2015-09-11 71.333333
3 2015-09-21 72.285714
My Approach so far:
I am taking the delta between input_date and date in pandas and saving this delta in separate column.
If delta%5 == 0, keep the row else drop. Is this the best that can be done?
Use boolean indexing for filter by mask, here convert input values to datetimes and then timedeltas to days by Series.dt.days:
input1 = '01-09-2015'
df = df[df.date.sub(pd.to_datetime(input1)).dt.days % 5 == 0]
print (df)
date value
0 2015-09-01 71.925000
1 2015-09-06 71.625000
2 2015-09-11 71.333333
4 2015-09-21 72.285714
I want to categorize data by month column
e.g.
date Month
2009-05-01==>May
I want to check outcomes by monthly
In this table I am excluding years and only want to keep months.
This is simple when using pd.Series.dt.month_name (https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.dt.month_name.html):
import pandas as pd
df = pd.DataFrame({
'date': pd.date_range('2000-01-01', '2010-01-01', freq='1M')
})
df['month'] = df.date.dt.month_name()
df.head()
Output
date month
0 2000-01-31 January
1 2000-02-29 February
2 2000-03-31 March
3 2000-04-30 April
4 2000-05-31 May
I have a dataframe in pandas called 'munged_data' with two columns 'entry_date' and 'dob' which i have converted to Timestamps using pd.to_timestamp.I am trying to figure out how to calculate ages of people based on the time difference between 'entry_date' and 'dob' and to do this i need to get the difference in days between the two columns ( so that i can then do somehting like round(days/365.25). I do not seem to be able to find a way to do this using a vectorized operation. When I do munged_data.entry_date-munged_data.dob i get the following :
internal_quote_id
2 15685977 days, 23:54:30.457856
3 11651985 days, 23:49:15.359744
4 9491988 days, 23:39:55.621376
7 11907004 days, 0:10:30.196224
9 15282164 days, 23:30:30.196224
15 15282227 days, 23:50:40.261632
However i do not seem to be able to extract the days as an integer so that i can continue with my calculation.
Any help appreciated.
Using the Pandas type Timedelta available since v0.15.0 you also can do:
In[1]: import pandas as pd
In[2]: df = pd.DataFrame([ pd.Timestamp('20150111'),
pd.Timestamp('20150301') ], columns=['date'])
In[3]: df['today'] = pd.Timestamp('20150315')
In[4]: df
Out[4]:
date today
0 2015-01-11 2015-03-15
1 2015-03-01 2015-03-15
In[5]: (df['today'] - df['date']).dt.days
Out[5]:
0 63
1 14
dtype: int64
You need 0.11 for this (0.11rc1 is out, final prob next week)
In [9]: df = DataFrame([ Timestamp('20010101'), Timestamp('20040601') ])
In [10]: df
Out[10]:
0
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00
In [11]: df = DataFrame([ Timestamp('20010101'),
Timestamp('20040601') ],columns=['age'])
In [12]: df
Out[12]:
age
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00
In [13]: df['today'] = Timestamp('20130419')
In [14]: df['diff'] = df['today']-df['age']
In [16]: df['years'] = df['diff'].apply(lambda x: float(x.item().days)/365)
In [17]: df
Out[17]:
age today diff years
0 2001-01-01 00:00:00 2013-04-19 00:00:00 4491 days, 00:00:00 12.304110
1 2004-06-01 00:00:00 2013-04-19 00:00:00 3244 days, 00:00:00 8.887671
You need this odd apply at the end because not yet full support for timedelta64[ns] scalars (e.g. like how we use Timestamps now for datetime64[ns], coming in 0.12)
Not sure if you still need it, but in Pandas 0.14 i usually use .astype('timedelta64[X]') method
http://pandas.pydata.org/pandas-docs/stable/timeseries.html (frequency conversion)
df = pd.DataFrame([ pd.Timestamp('20010101'), pd.Timestamp('20040605') ])
df.ix[0]-df.ix[1]
Returns:
0 -1251 days
dtype: timedelta64[ns]
(df.ix[0]-df.ix[1]).astype('timedelta64[Y]')
Returns:
0 -4
dtype: float64
Hope that will help
Let's specify that you have a pandas series named time_difference which has type
numpy.timedelta64[ns]
One way of extracting just the day (or whatever desired attribute) is the following:
just_day = time_difference.apply(lambda x: pd.tslib.Timedelta(x).days)
This function is used because the numpy.timedelta64 object does not have a 'days' attribute.
To convert any type of data into days just use pd.Timedelta().days:
pd.Timedelta(1985, unit='Y').days
84494
I have a pandas dataframe like the following
Year Month Day Securtiy Trade Value NewDate
2011 1 10 AAPL Buy 1500 0
My question is, how can I merge the columns Year, Month, Day into column NewDate
so that the newDate column looks like the following
2011-1-10
The best way is to parse it when reading as csv:
In [1]: df = pd.read_csv('foo.csv', sep='\s+', parse_dates=[['Year', 'Month', 'Day']])
In [2]: df
Out[2]:
Year_Month_Day Securtiy Trade Value NewDate
0 2011-01-10 00:00:00 AAPL Buy 1500 0
You can do this without the header, by defining column names while reading:
pd.read_csv(input_file, header=['Year', 'Month', 'Day', 'Security','Trade', 'Value' ], parse_dates=[['Year', 'Month', 'Day']])
If it's already in your DataFrame, you could use an apply:
In [11]: df['Date'] = df.apply(lambda s: pd.Timestamp('%s-%s-%s' % (s['Year'], s['Month'], s['Day'])), 1)
In [12]: df
Out[12]:
Year Month Day Securtiy Trade Value NewDate Date
0 2011 1 10 AAPL Buy 1500 0 2011-01-10 00:00:00
df['Year'] + '-' + df['Month'] + '-' + df['Date']
You can create a new Timestamp as follows:
df['newDate'] = df.apply(lambda x: pd.Timestamp('{0}-{1}-{2}'
.format(x.Year, x.Month, x.Day),
axix=1)
>>> df
Year Month Day Securtiy Trade Value NewDate newDate
0 2011 1 10 AAPL Buy 1500 0 2011-01-10