Develop Reference

How to show to average sales for each year within ten years for a specific city in Pandas?

What would be the correct way to show what was the average sales volume in Carlisle city for each year between
2010-2020?
Here is an abbreviated form of the large data frame showing only the columns and rows relevant to the question:
import pandas as pd
df = pd.DataFrame({'Date': ['01/09/2009','01/10/2009','01/11/2009','01/12/2009','01/01/2010','01/02/2010','01/03/2010','01/04/2010','01/05/2010','01/06/2010','01/07/2010','01/08/2010','01/09/2010','01/10/2010','01/11/2010','01/12/2010','01/01/2011','01/02/2011'],
'RegionName': ['Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle','Carlisle'],
'SalesVolume': [118,137,122,132,83,81,105,114,110,106,137,130,129,121,129,100,84,62]})
This is what I've tried:
import pandas as pd
from matplotlib import pyplot as plt
df = pd.read_csv ('C:/Users/user/AppData/Local/Programs/Python/Python39/Scripts/uk_hpi_dataset_2021_01.csv')
df.Date = pd.to_datetime(df.Date)
df['Year'] = pd.to_datetime(df['Date']).apply(lambda x:
'{year}'.format(year=x.year).zfill(2))
carlisle_vol = df[df['RegionName'].str.contains('Carlisle')]
carlisle_vol.groupby('Year')['SalesVolume'].mean()
print(sales_vol)
When I try to run this code, it doesn't filter the 'Date' column to only calculate the average SalesVolume for the years beginning in '01/01/2010' and ending at '01/12/2020'. For some reason, it also prints out every other column is well. Can anyone please help me to answer this question correctly?
This is the result I've got

>>> df.loc[(df["Date"].dt.year.between(2010, 2020))
& (df["RegionName"] == "Carlisle")] \
.groupby([pd.Grouper(key="Date", freq="Y")])["SalesVolume"].mean()
Date
2010-01-01 112.083333
2011-01-01 73.000000
Freq: A-DEC, Name: SalesVolume, dtype: float64
For further:
The only difference between the answer of #nocibambi is the groupby parameter and particularly the freq argument of pd.Grouper. Imagine your accounting year starts the 1st september.
Sales each 3 months:
>>> df
Date Sales
0 2010-09-01 1 # 1st group: mean=2.5
1 2010-12-01 2
2 2011-03-01 3
3 2011-06-01 4
4 2011-09-01 5 # 2nd group: mean=6.5
5 2011-12-01 6
6 2012-03-01 7
7 2012-06-01 8
>>> df.groupby(pd.Grouper(key="Date", freq="AS-SEP")).mean()
Sales
Date
2010-09-01 2.5
2011-09-01 6.5
Check the documentation to know all values of freq aliases and anchoring suffix

You can access year with the datetime accessor:
df[
(df["RegionName"] == "Carlisle")
& (df["Date"].dt.year >= 2010)
& (df["Date"].dt.year <= 2020)
].groupby(df.Date.dt.year)["SalesVolume"].mean()
>>>
Date
2010 112.083333
2011 73.000000
Name: SalesVolume, dtype: float64

Difference between timestamps as days [duplicate]

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

Pandas excel import changes the Date format

Im learning python (3.6 with anaconda) for my studies.
Im using pandas to import a xls file with 2 columns : Date (dd-mm-yyyy) and price.
But pandas changes the date format :
xls_file = pd.read_excel('myfile.xls')
print(xls_file.iloc[0, 0])
Im getting :
2010-01-04 00:00:00
instead of :
04-01-2010 or at least : 2010-01-04
I dont know why hh:mm:ss is added, I get the same result for each row from the Date column. I tried also different things using to_datetime but it didnt fix it.
Any idea ?
Thanks

What you need is to define the format that the datetime values get printed. There might be a more elegant way to do it but something like that will work:
In [11]: df
Out[11]:
id date
0 1 2017-09-12
1 2 2017-10-20
# Specifying the format
In [16]: print(pd.datetime.strftime(df.iloc[0,1], "%Y-%m-%d"))
2017-09-12
If you want to store the date as string in your specific format then you can also do something like:
In [17]: df["datestr"] = pd.datetime.strftime(df.iloc[0,1], "%Y-%m-%d")
In [18]: df
Out[18]:
id date datestr
0 1 2017-09-12 2017-09-12
1 2 2017-10-20 2017-09-12
In [19]: df.dtypes
Out[19]:
id int64
date datetime64[ns]
datestr object
dtype: object

Pandas: Difference of of two datetime64 objects yields NaT rather than correct timedelta value

This question "gets asked a lot" - but after looking carefully at the other answers I haven't found a solution that works in my case. It's a shame this is still such a sticking point.
I have a pandas dataframe with a column datetime and I simply want to calculate the time range covered by the data, in seconds (say).
from datetime import datetime
# You can create fake datetime entries any way you like, e.g.
df = pd.DataFrame({'datetime': pd.date_range('10/1/2001 10:00:00', \
periods=3, freq='10H'),'B':[4,5,6]})
# (a) This yields NaT:
timespan_a=df['datetime'][-1:]-df['datetime'][:1]
print timespan_a
# 0 NaT
# 2 NaT
# Name: datetime, dtype: timedelta64[ns]
# (b) This does work - but why?
timespan_b=df['datetime'][-1:].values.astype("timedelta64")-\
df['datetime'][:1].values.astype("timedelta64")
print timespan_b
# [72000000000000]
Why doesn't (a) work?
Why is (b) required rather? (it also gives a one-element numpy array rather than a timedelta object)
My pandas is at version 0.20.3, which rules out an earlier known bug.
Is this a dynamic-range issue?

There is problem different indexes, so one item Series cannot align and get NaT.
Solution is convert first or second values to numpy array by values:
timespan_a = df['datetime'][-1:]-df['datetime'][:1].values
print (timespan_a)
2 20:00:00
Name: datetime, dtype: timedelta64[ns]
Or set both index values to same:
a = df['datetime'][-1:]
b = df['datetime'][:1]
print (a)
2 2001-10-02 06:00:00
Name: datetime, dtype: datetime64[ns]
a.index = b.index
print (a)
0 2001-10-02 06:00:00
Name: datetime, dtype: datetime64[ns]
print (b)
0 2001-10-01 10:00:00
Name: datetime, dtype: datetime64[ns]
timespan_a = a - b
print (timespan_a)
0 20:00:00
Name: datetime, dtype: timedelta64[ns]
If want working with scalars:
a = df.loc[df.index[-1], 'datetime']
b = df.loc[0, 'datetime']
print (a)
2001-10-02 06:00:00
print (b)
2001-10-01 10:00:00
timespan_a = a - b
print (timespan_a)
0 days 20:00:00
Another solution, thank you Anton vBR:
timespan_a = df.get_value(len(df)-1,'datetime')- df.get_value(0,'datetime')
print (timespan_a)
0 days 20:00:00

pick month start and end data in python

I have stock data downloaded from yahoo finance. I want to pickup data in the row corresponding to monthly start and month end. I am trying to do it with python pandas data frame. But I am not getting correct method to get the starting & ending of the month. will be great full if somebody can help me in solving this.
Please note that if 1st of the month is holiday and there is no data for that, I need to pick up 2nd day's data. Same rule applies to last of the month also. Thanks in advance.
Example data is
2016-01-05,222.80,222.80,217.00,217.75,15074800,217.75
2016-01-04,226.95,226.95,220.05,220.70,14092000,220.70
2015-12-31,225.95,226.55,224.00,224.45,11558300,224.45
2015-12-30,229.00,229.70,224.85,225.80,11702800,225.80
2015-12-29,228.85,229.95,227.50,228.20,7263200,228.20
2015-12-28,229.05,229.95,228.00,228.90,8756800,228.90
........
........
2015-12-04,240.00,242.15,238.05,241.10,11115100,241.10
2015-12-03,244.15,244.50,240.40,241.10,7155600,241.10
2015-12-02,250.55,250.65,243.75,244.60,10881700,244.60
2015-11-30,249.65,253.00,245.00,250.20,12865400,250.20
2015-11-27,243.00,250.50,242.80,249.70,15149900,249.70
2015-11-26,241.95,244.90,241.00,242.50,13629800,242.50

First, you should convert your date column to datetime format, then group by month, then sort groupby Series by date and take the first/last from it using head/tail methods, like so:
In [37]: df
Out[37]:
0 1 2 3 4 5 6
0 2016-01-05 222.80 222.80 217.00 217.75 15074800 217.75
1 2016-01-04 226.95 226.95 220.05 220.70 14092000 220.70
2 2015-12-31 225.95 226.55 224.00 224.45 11558300 224.45
3 2015-12-30 229.00 229.70 224.85 225.80 11702800 225.80
4 2015-12-29 228.85 229.95 227.50 228.20 7263200 228.20
5 2015-12-28 229.05 229.95 228.00 228.90 8756800 228.90
In [25]: import datetime
In [29]: df[0] = df[0].apply(lambda x: datetime.datetime.strptime(x, '%Y-%m-%d')
)
In [36]: df.groupby(df[0].apply(lambda x: x.month)).apply(lambda x: x.sort_value
s(0).head(1))
Out[36]:
0 1 2 3 4 5 6
0
1 1 2016-01-04 226.95 226.95 220.05 220.7 14092000 220.7
12 5 2015-12-28 229.05 229.95 228.00 228.9 8756800 228.9
In [38]: df.groupby(df[0].apply(lambda x: x.month)).apply(lambda x: x.sort_value
s(0).tail(1))
Out[38]:
0 1 2 3 4 5 6
0
1 0 2016-01-05 222.80 222.80 217.0 217.75 15074800 217.75
12 2 2015-12-31 225.95 226.55 224.0 224.45 11558300 224.45
You can merge the result dataframes, using pd.concat()

For the first / last day of each month, you can use .resample() with 'BMS' and 'BM' for Business Month (Start) like so (using pandas 0.18 syntax):
df.resample('BMS').first()
df.resample('BM').last()
This assumes that your data have a DateTimeIndex as usual when downloaded from yahoo using pandas_datareader:
from datetime import datetime
from pandas_datareader.data import DataReader
df = DataReader('FB', 'yahoo', datetime(2015, 1, 1), datetime(2015, 3, 31))['Open']
df.head()
Date
2015-01-02 78.580002
2015-01-05 77.980003
2015-01-06 77.230003
2015-01-07 76.760002
2015-01-08 76.739998
Name: Open, dtype: float64
df.tail()
Date
2015-03-25 85.500000
2015-03-26 82.720001
2015-03-27 83.379997
2015-03-30 83.809998
2015-03-31 82.900002
Name: Open, dtype: float64
do:
df.resample('BMS').first()
Date
2015-01-01 78.580002
2015-02-02 76.110001
2015-03-02 79.000000
Freq: BMS, Name: Open, dtype: float64
and
df.resample('BM').last()
to get:
Date
2015-01-30 78.000000
2015-02-27 80.680000
2015-03-31 82.900002
Freq: BM, Name: Open, dtype: float64

Assuming you have downloaded data from Yahoo:
> import pandas.io.data as web
> import datetime
> start = datetime.datetime(2016,1,1)
> end = datetime.datetime(2016,5,1)
> df = web.DataReader("AAPL", "yahoo", start, end)
You simply pick the month end and start rows with:
df[df.index.is_month_end]
df[df.index.is_month_start]
If you want to access a specific row, like the first row of the first starting day of the selected starting days, you simply do:
df[df.index.is_month_start].ix[0]