I need to find the median month value between two dates in a date frame. I am simplifying the case by showing four examples.
import pandas as pd
import numpy as np
import datetime
df=pd.DataFrame([["1/31/2016","3/1/2016"],
["6/15/2016","7/14/2016"],
["7/14/2016","8/15/2016"],
["8/7/2016","9/6/2016"]], columns=['FromDate','ToDate'])
df['Month'] = df.ToDate.dt.month-df.FromDate.dt.month
I am trying to append a column but I am not getting the desired result.
I need to see these values: [2,6,7,8].
You can calculate the average date explicitly by adding half the timedelta between 2 dates to the earlier date. Then just extract the month:
# convert to datetime if necessary
df[df.columns] = df[df.columns].apply(pd.to_datetime)
# calculate mean date, then extract month
df['Month'] = (df['FromDate'] + (df['ToDate'] - df['FromDate']) / 2).dt.month
print(df)
FromDate ToDate Month
0 2016-01-31 2016-03-01 2
1 2016-06-15 2016-07-14 6
2 2016-07-14 2016-08-15 7
3 2016-08-07 2016-09-06 8
You need to convert the string to datetime before using dt.month.
This line calculates the average month number :
df['Month'] = (pd.to_datetime(df['ToDate']).dt.month +
pd.to_datetime(df['FromDate']).dt.month)//2
print(df)
FromDate ToDate Month
0 1/31/2016 3/1/2016 2
1 6/15/2016 7/14/2016 6
2 7/14/2016 8/15/2016 7
3 8/7/2016 9/6/2016 8
This only works with both dates in the same year.
jpp's solution is fine but will in some cases give the wrong answer:
["1/1/2016","3/1/2016"] one would expect 2 because February is between January and March, but jpp's will give 1 corresponding to January.
Related
Initially, my dataframe had a Month column containing numbers representing the months.
Month
1
2
3
4
I typed df["Month"] = pd.to_datetime(df["Month"]) and I get this...
Month
970-01-01 00:00:00.0000000001
1970-01-01 00:00:00.000000002
1970-01-01 00:00:00.000000003
1970-01-01 00:00:00.000000004
I would like to just retain just the dates and not the time. Any solutions?
get the date from the column using df['Month'].dt.date
Use format='%m' in to_datetime:
df["Month"] = pd.to_datetime(df["Month"], format='%m')
print (df)
Month
0 1900-01-01
1 1900-02-01
2 1900-03-01
3 1900-04-01
My DataFrame looks like this:
id
date
value
1
2021-07-16
100
2
2021-09-15
20
1
2021-04-10
50
1
2021-08-27
30
2
2021-07-22
15
2
2021-07-22
25
1
2021-06-30
40
3
2021-10-11
150
2
2021-08-03
15
1
2021-07-02
90
I want to groupby the id, and return the difference of total value in a 90-days period.
Specifically, I want the values of last 90 days based on today, and based on 30 days ago.
For example, considering today is 2021-10-13, I would like to get:
the sum of all values per id between 2021-10-13 and 2021-07-15
the sum of all values per id between 2021-09-13 and 2021-06-15
And finally, subtract them to get the variation.
I've already managed to calculate it, by creating separated temporary dataframes containing only the dates in those periods of 90 days, grouping by id, and then merging these temp dataframes into a final one.
But I guess it should be an easier or simpler way to do it. Appreciate any help!
Btw, sorry if the explanation was a little messy.
If I understood correctly, you need something like this:
import pandas as pd
import datetime
## Calculation of the dates that we are gonna need.
today = datetime.datetime.now()
delta = datetime.timedelta(days = 120)
# Date of the 120 days ago
hundredTwentyDaysAgo = today - delta
delta = datetime.timedelta(days = 90)
# Date of the 90 days ago
ninetyDaysAgo = today - delta
delta = datetime.timedelta(days = 30)
# Date of the 30 days ago
thirtyDaysAgo = today - delta
## Initializing an example df.
df = pd.DataFrame({"id":[1,2,1,1,2,2,1,3,2,1],
"date": ["2021-07-16", "2021-09-15", "2021-04-10", "2021-08-27", "2021-07-22", "2021-07-22", "2021-06-30", "2021-10-11", "2021-08-03", "2021-07-02"],
"value": [100,20,50,30,15,25,40,150,15,90]})
## Casting date column
df['date'] = pd.to_datetime(df['date']).dt.date
grouped = df.groupby('id')
# Sum of last 90 days per id
ninetySum = grouped.apply(lambda x: x[x['date'] >= ninetyDaysAgo.date()]['value'].sum())
# Sum of last 90 days, starting from 30 days ago per id
hundredTwentySum = grouped.apply(lambda x: x[(x['date'] >= hundredTwentyDaysAgo.date()) & (x['date'] <= thirtyDaysAgo.date())]['value'].sum())
The output is
ninetySum - hundredTwentySum
id
1 -130
2 20
3 150
dtype: int64
You can double check to make sure these are the numbers you wanted by printing ninetySum and hundredTwentySum variables.
I have a df
date
2021-03-12
2021-03-17
...
2022-05-21
2022-08-17
I am trying to add a column year_week, but my year week starts at 2021-06-28, which is the first day of July.
I tried:
df['date'] = pd.to_datetime(df['date'])
df['year_week'] = (df['date'] - timedelta(days=datetime(2021, 6, 24).timetuple()
.tm_yday)).dt.isocalendar().week
I played around with the timedelta days values so that the 2021-06-28 has a value of 1.
But then I got problems with previous & dates exceeding my start date + 1 year:
2021-03-12 has a value of 38
2022-08-17 has a value of 8
So it looks like the valid period is from 2021-06-28 + 1 year.
date year_week
2021-03-12 38 # LY38
2021-03-17 39 # LY39
2021-06-28 1 # correct
...
2022-05-21 47 # correct
2022-08-17 8 # NY8
Is there a way to get around this? As I am aggregating the data by year week I get incorrect results due to the past & upcoming dates. I would want to have negative dates for the days before 2021-06-28 or LY38 denoting that its the year week of the last year, accordingly year weeks of 52+ or NY8 denoting that this is the 8th week of the next year?
Here is a way, I added two dates more than a year away. You need the isocalendar from the difference between the date column and the dayofyear of your specific date. Then you can select the different scenario depending on the year of your specific date. use np.select for the different result format.
#dummy dataframe
df = pd.DataFrame(
{'date': ['2020-03-12', '2021-03-12', '2021-03-17', '2021-06-28',
'2022-05-21', '2022-08-17', '2023-08-17']
}
)
# define start date
d = pd.to_datetime('2021-6-24')
# remove the nomber of day of year from each date
s = (pd.to_datetime(df['date']) - pd.Timedelta(days=d.day_of_year)
).dt.isocalendar()
# get the difference in year
m = (s['year'].astype('int32') - d.year)
# all condition of result depending on year difference
conds = [m.eq(0), m.eq(-1), m.eq(1), m.lt(-1), m.gt(1)]
choices = ['', 'LY','NY',(m+1).astype(str)+'LY', '+'+(m-1).astype(str)+'NY']
# create the column
df['res'] = np.select(conds, choices) + s['week'].astype(str)
print(df)
date res
0 2020-03-12 -1LY38
1 2021-03-12 LY38
2 2021-03-17 LY39
3 2021-06-28 1
4 2022-05-21 47
5 2022-08-17 NY8
6 2023-08-17 +1NY8
I think
pandas period_range can be of some help
pd.Series(pd.period_range("6/28/2017", freq="W", periods=Number of weeks you want))
I have a table that has a column Months_since_Start_fin_year and a Date column. I need to add the number of months in the first column to the date in the second column.
DateTable['Date']=DateTable['First_month']+DateTable['Months_since_Start_fin_year'].astype("timedelta64[M]")
This works OK for month 0, but month 1 already has a different time and for month 2 onwards has the wrong date.
Image of output table where early months have the correct date but month 2 where I would expect June 1st actually shows May 31st
It must be adding incomplete months, but I'm not sure how to fix it?
I have also tried
DateTable['Date']=DateTable['First_month']+relativedelta(months=DateTable['Months_since_Start_fin_year'])
but I get a type error that says
TypeError: cannot convert the series to <class 'int'>
My Months_since_Start_fin_year is type int32 and my First_month variable is datetime64[ns]
The problem with adding months as an offset to a date is that not all months are equally long (28-31 days). So you need pd.DateOffset which handles that ambiguity for you. .astype("timedelta64[M]") on the other hand only gives you the average days per month within a year (30 days 10:29:06).
Ex:
import pandas as pd
# a synthetic example since you didn't provide a mre
df = pd.DataFrame({'start_date': 7*['2017-04-01'],
'month_offset': range(7)})
# make sure we have datetime dtype
df['start_date'] = pd.to_datetime(df['start_date'])
# add month offset
df['new_date'] = df.apply(lambda row: row['start_date'] +
pd.DateOffset(months=row['month_offset']),
axis=1)
which would give you e.g.
df
start_date month_offset new_date
0 2017-04-01 0 2017-04-01
1 2017-04-01 1 2017-05-01
2 2017-04-01 2 2017-06-01
3 2017-04-01 3 2017-07-01
4 2017-04-01 4 2017-08-01
5 2017-04-01 5 2017-09-01
6 2017-04-01 6 2017-10-01
You can find similar examples here on SO, e.g. Add months to a date in Pandas. I only modified the answer there by using an apply to be able to take the months offset from one of the DataFrame's columns.
Want to calculate the difference of days between pandas date series -
0 2013-02-16
1 2013-01-29
2 2013-02-21
3 2013-02-22
4 2013-03-01
5 2013-03-14
6 2013-03-18
7 2013-03-21
and today's date.
I tried but could not come up with logical solution.
Please help me with the code. Actually I am new to python and there are lot of syntactical errors happening while applying any function.
You could do something like
# generate time data
data = pd.to_datetime(pd.Series(["2018-09-1", "2019-01-25", "2018-10-10"]))
pd.to_datetime("now") > data
returns:
0 False
1 True
2 False
you could then use that to select the data
data[pd.to_datetime("now") > data]
Hope it helps.
Edit: I misread it but you can easily alter this example to calculate the difference:
data - pd.to_datetime("now")
returns:
0 -122 days +13:10:37.489823
1 24 days 13:10:37.489823
2 -83 days +13:10:37.489823
dtype: timedelta64[ns]
You can try as Follows:
>>> from datetime import datetime
>>> df
col1
0 2013-02-16
1 2013-01-29
2 2013-02-21
3 2013-02-22
4 2013-03-01
5 2013-03-14
6 2013-03-18
7 2013-03-21
Make Sure to convert the column names to_datetime:
>>> df['col1'] = pd.to_datetime(df['col1'], infer_datetime_format=True)
set the current datetime in order to Further get the diffrence:
>>> curr_time = pd.to_datetime("now")
Now get the Difference as follows:
>>> df['col1'] - curr_time
0 -2145 days +07:48:48.736939
1 -2163 days +07:48:48.736939
2 -2140 days +07:48:48.736939
3 -2139 days +07:48:48.736939
4 -2132 days +07:48:48.736939
5 -2119 days +07:48:48.736939
6 -2115 days +07:48:48.736939
7 -2112 days +07:48:48.736939
Name: col1, dtype: timedelta64[ns]
With numpy you can solve it like difference-two-dates-days-weeks-months-years-pandas-python-2
. bottom line
df['diff_days'] = df['First dates column'] - df['Second Date column']
# for days use 'D' for weeks use 'W', for month use 'M' and for years use 'Y'
df['diff_days']=df['diff_days']/np.timedelta64(1,'D')
print(df)
if you want days as int and not as float use
df['diff_days']=df['diff_days']//np.timedelta64(1,'D')
From the pandas docs under Converting To Timestamps you will find:
"Converting to Timestamps To convert a Series or list-like object of date-like objects e.g. strings, epochs, or a mixture, you can use the to_datetime function"
I haven't used pandas before but this suggests your pandas date series (a list-like object) is iterable and each element of this series is an instance of a class which has a to_datetime function.
Assuming my assumptions are correct, the following function would take such a list and return a list of timedeltas' (a datetime object representing the difference between two date time objects).
from datetime import datetime
def convert(pandas_series):
# get the current date
now = datetime.now()
# Use a list comprehension and the pandas to_datetime method to calculate timedeltas.
return [now - pandas_element.to_datetime() for pandas_series]
# assuming 'some_pandas_series' is a list-like pandas series object
list_of_timedeltas = convert(some_pandas_series)