Pandas group hourly data into daily sums with date index - python

I am working on a code that takes hourly data for a month and groups it into 24 hour sums. My problem is that I would like the index to read the date/year and I am just getting an index of 1-30.
The code I am using is
df = df.iloc[:,16:27].groupby([lambda x: x.day]).sum()
example of output I am getting
DateTime data
1 1772.031568
2 19884.42243
3 28696.72159
4 24906.20355
5 9059.120325
example of output I would like
DateTime data
1/1/2017 1772.031568
1/2/2017 19884.42243
1/3/2017 28696.72159
1/4/2017 24906.20355
1/5/2017 9059.120325


This is an old question, but I don't think the accepted solution is the best in this particular case. What you want to accomplish is to down sample time series data, and Pandas has built-in functionality for this called resample(). For your example you will do:
df = df.iloc[:,16:27].resample('D').sum()
or if the datetime column is not the index
df = df.iloc[:,16:27].resample('D', on='datetime_column_name').sum()
There are (at least) 2 benefits from doing it this way as opposed to accepted answer:
Resample can up sample and down sample, groupby() can only down sample
No lambdas, list comprehensions or date formatting functions required.
For more information and examples, see documentation here: resample()


If your index is a datetime, you can build a combined groupby clause:
df = df.iloc[:,16:27].groupby([lambda x: "{}/{}/{}".format(x.day, x.month, x.year)]).sum()
or even better:
df = df.iloc[:,16:27].groupby([lambda x: x.strftime("%d%m%Y")]).sum()


if your index was not datetime object.
import pandas as pd
df = pd.DataFrame({'data': [1772.031568, 19884.42243,28696.72159, 24906.20355,9059.120325]},index=[1,2,3,4,5])
print df.head()
rng = pd.date_range('1/1/2017',periods =len(df.index), freq='D')
df.set_index(rng,inplace=True)
print df.head()
will result in
data
1 1772.031568
2 19884.422430
3 28696.721590
4 24906.203550
5 9059.120325
data
2017-01-01 1772.031568
2017-01-02 19884.422430
2017-01-03 28696.721590
2017-01-04 24906.203550
2017-01-05 9059.120325


First you need to create an index on your datetime column to expose functions that break the datetime into smaller pieces efficiently (like the year and month of the datetime).
Next, you need to group by the year, month and day of the index if you want to apply an aggregate method (like sum()) to each day of the year, and retain separate aggregations for each day.
The reset_index() and rename() functions allow us to rename our group_by categories to their original names. This "flattens" out our data, making the category an actual column on the resulting dataframe.
import pandas as pd
date_index = pd.DatetimeIndex(df.created_at)
# 'df.created_at' is the datetime column in your dataframe
counted = df.group_by([date_index.year, date_index.month, date_index.day])\
.agg({'column_to_sum': 'sum'})\
.reset_index()\
.rename(columns={'level_1': 'year',
'level_2': 'month',
'level_3': 'day'})
# Resulting dataframe has columns "column_to_sum", "year", "month", "day" available


You can exploit panda's DatetimeIndex:
working_df=df.iloc[:, 16:27]
result = working_df.groupby(pd.DatetimeIndex(working_df.DateTime)).date).sum()
This if you DateTime column is actually DateTime (and be careful of the timezone).
In this way you will have valid datetime in the index, so that you can easily do other manipulations.

Related

Select Pandas dataframe rows based on 'hour' datetime

I have a pandas dataframe 'df' with a column 'DateTimes' of type datetime.time.
The entries of that column are hours of a single day:
00:00:00
.
.
.
23:59:00
Seconds are skipped, it counts by minutes.
How can I choose rows by hour, for example the rows between 00:00:00 and 00:01:00?
If I try this:
df.between_time('00:00:00', '00:00:10')
I get an error that index must be a DateTimeIndex.
I set the index as such with:
df=df.set_index(keys='DateTime')
but I get the same error.
I can't seem to get 'loc' to work either. Any suggestions?

Here a working example of what you are trying to do:
times = pd.date_range('3/6/2012 00:00', periods=100, freq='S', tz='UTC')
df = pd.DataFrame(np.random.randint(10, size=(100,1)), index=times)
df.between_time('00:00:00', '00:00:30')
Note the index has to be of type DatetimeIndex.
I understand you have a column with your dates/times. The problem probably is that your column is not of this type, so you have to convert it first, before setting it as index:
# Method A
df.set_index(pd.to_datetime(df['column_name'], drop=True)
# Method B
df.index = pd.to_datetime(df['column_name'])
df = df.drop('col', axis=1)
(The drop is only necessary if you want to remove the original column after setting it as index)

Check out these links:
convert column to date type: Convert DataFrame column type from string to datetime
filter dataframe on dates: Filtering Pandas DataFrames on dates
Hope this helps

map with named function vs identical lambda function providing different responses, pandas

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

Convert Column to Date Format (Pandas Dataframe)

I have a pandas dataframe as follows:
Symbol Date
A 02/20/2015
A 01/15/2016
A 08/21/2015
I want to sort it by Date, but the column is just an object.
I tried to make the column a date object, but I ran into an issue where that format is not the format needed. The format needed is 2015-02-20, etc.
So now I'm trying to figure out how to have numpy convert the 'American' dates into the ISO standard, so that I can make them date objects, so that I can sort by them.
How would I convert these american dates into ISO standard, or is there a more straight forward method I'm missing within pandas?

You can use pd.to_datetime() to convert to a datetime object. It takes a format parameter, but in your case I don't think you need it.
>>> import pandas as pd
>>> df = pd.DataFrame( {'Symbol':['A','A','A'] ,
'Date':['02/20/2015','01/15/2016','08/21/2015']})
>>> df
Date Symbol
0 02/20/2015 A
1 01/15/2016 A
2 08/21/2015 A
>>> df['Date'] =pd.to_datetime(df.Date)
>>> df.sort('Date') # This now sorts in date order
Date Symbol
0 2015-02-20 A
2 2015-08-21 A
1 2016-01-15 A
For future search, you can change the sort statement:
>>> df.sort_values(by='Date') # This now sorts in date order
Date Symbol
0 2015-02-20 A
2 2015-08-21 A
1 2016-01-15 A

sort method has been deprecated and replaced with sort_values. After converting to datetime object using df['Date']=pd.to_datetime(df['Date'])
df.sort_values(by=['Date'])
Note: to sort in-place and/or in a descending order (the most recent first):
df.sort_values(by=['Date'], inplace=True, ascending=False)

#JAB's answer is fast and concise. But it changes the DataFrame you are trying to sort, which you may or may not want.
(Note: You almost certainly will want it, because your date columns should be dates, not strings!)
In the unlikely event that you don't want to change the dates into dates, you can also do it a different way.
First, get the index from your sorted Date column:
In [25]: pd.to_datetime(df.Date).order().index
Out[25]: Int64Index([0, 2, 1], dtype='int64')
Then use it to index your original DataFrame, leaving it untouched:
In [26]: df.ix[pd.to_datetime(df.Date).order().index]
Out[26]:
Date Symbol
0 2015-02-20 A
2 2015-08-21 A
1 2016-01-15 A
Magic!
Note: for Pandas versions 0.20.0 and later, use loc instead of ix, which is now deprecated.

Since pandas >= 1.0.0 we have the key argument in DataFrame.sort_values. This way we can sort the dataframe by specifying a key and without adjusting the original dataframe:
df.sort_values(by="Date", key=pd.to_datetime)
Symbol Date
0 A 02/20/2015
2 A 08/21/2015
1 A 01/15/2016

The data containing the date column can be read by using the below code:
data = pd.csv(file_path,parse_dates=[date_column])
Once the data is read by using the above line of code, the column containing the information about the date can be accessed using pd.date_time() like:
pd.date_time(data[date_column], format = '%d/%m/%y')
to change the format of date as per the requirement.

data['Date'] = data['Date'].apply(pd.to_datetime) # non-null datetime64[ns]

Pivoting out Datetimes and then calling an operation in Pandas/Python

I've seen several articles about using datetime and dateutil to convert into datetime objects.
However, I can't seem to figure out how to convert a column into a datetime object so I can pivot out that columns and perform operations against it.
I have a dataframe as such:
Col1 Col 2
a 1/1/2013
a 1/12/2013
b 1/5/2013
b 4/3/2013 ....etc
What I want is :
pivott = pivot_table( df, rows ='Col1', values='Col2', and then I want to get the range of dates for each value in Col1)
I am not sure how to correctly approach this. Even after using
df['Col2']= pd.to_datetime(df['Col2'])
I couldn't do operations against the dates since they are strings...
Any advise?

Use datetime.strptime
import pandas as pd
from datetime import datetime
df = pd.read_csv('somedata.csv')
convertdatetime = lambda d: datetime.strptime(d,'%d/%m/%Y')
converted = df['DATE_TIME_IN_STRING'].apply(convertdatetime)
converted[:10] # you should be getting dtype: datetime64[ns]

Filtering Pandas DataFrames on dates

I have a Pandas DataFrame with a 'date' column. Now I need to filter out all rows in the DataFrame that have dates outside of the next two months. Essentially, I only need to retain the rows that are within the next two months.
What is the best way to achieve this?

If date column is the index, then use .loc for label based indexing or .iloc for positional indexing.
For example:
df.loc['2014-01-01':'2014-02-01']
See details here http://pandas.pydata.org/pandas-docs/stable/dsintro.html#indexing-selection
If the column is not the index you have two choices:
Make it the index (either temporarily or permanently if it's time-series data)
df[(df['date'] > '2013-01-01') & (df['date'] < '2013-02-01')]
See here for the general explanation
Note: .ix is deprecated.

Previous answer is not correct in my experience, you can't pass it a simple string, needs to be a datetime object. So:
import datetime
df.loc[datetime.date(year=2014,month=1,day=1):datetime.date(year=2014,month=2,day=1)]

And if your dates are standardized by importing datetime package, you can simply use:
df[(df['date']>datetime.date(2016,1,1)) & (df['date']<datetime.date(2016,3,1))]
For standarding your date string using datetime package, you can use this function:
import datetime
datetime.datetime.strptime

If you have already converted the string to a date format using pd.to_datetime you can just use:
df = df[(df['Date'] > "2018-01-01") & (df['Date'] < "2019-07-01")]

The shortest way to filter your dataframe by date:
Lets suppose your date column is type of datetime64[ns]
# filter by single day
df_filtered = df[df['date'].dt.strftime('%Y-%m-%d') == '2014-01-01']
# filter by single month
df_filtered = df[df['date'].dt.strftime('%Y-%m') == '2014-01']
# filter by single year
df_filtered = df[df['date'].dt.strftime('%Y') == '2014']

If your datetime column have the Pandas datetime type (e.g. datetime64[ns]), for proper filtering you need the pd.Timestamp object, for example:
from datetime import date
import pandas as pd
value_to_check = pd.Timestamp(date.today().year, 1, 1)
filter_mask = df['date_column'] < value_to_check
filtered_df = df[filter_mask]

If the dates are in the index then simply:
df['20160101':'20160301']

You can use pd.Timestamp to perform a query and a local reference
import pandas as pd
import numpy as np
df = pd.DataFrame()
ts = pd.Timestamp
df['date'] = np.array(np.arange(10) + datetime.now().timestamp(), dtype='M8[s]')
print(df)
print(df.query('date > #ts("20190515T071320")')
with the output
date
0 2019-05-15 07:13:16
1 2019-05-15 07:13:17
2 2019-05-15 07:13:18
3 2019-05-15 07:13:19
4 2019-05-15 07:13:20
5 2019-05-15 07:13:21
6 2019-05-15 07:13:22
7 2019-05-15 07:13:23
8 2019-05-15 07:13:24
9 2019-05-15 07:13:25
date
5 2019-05-15 07:13:21
6 2019-05-15 07:13:22
7 2019-05-15 07:13:23
8 2019-05-15 07:13:24
9 2019-05-15 07:13:25
Have a look at the pandas documentation for DataFrame.query, specifically the mention about the local variabile referenced udsing # prefix. In this case we reference pd.Timestamp using the local alias ts to be able to supply a timestamp string

So when loading the csv data file, we'll need to set the date column as index now as below, in order to filter data based on a range of dates. This was not needed for the now deprecated method: pd.DataFrame.from_csv().
If you just want to show the data for two months from Jan to Feb, e.g. 2020-01-01 to 2020-02-29, you can do so:
import pandas as pd
mydata = pd.read_csv('mydata.csv',index_col='date') # or its index number, e.g. index_col=[0]
mydata['2020-01-01':'2020-02-29'] # will pull all the columns
#if just need one column, e.g. Cost, can be done:
mydata['2020-01-01':'2020-02-29','Cost']
This has been tested working for Python 3.7. Hope you will find this useful.

I'm not allowed to write any comments yet, so I'll write an answer, if somebody will read all of them and reach this one.
If the index of the dataset is a datetime and you want to filter that just by (for example) months, you can do following:
df.loc[df.index.month == 3]
That will filter the dataset for you by March.

How about using pyjanitor
It has cool features.
After pip install pyjanitor
import janitor
df_filtered = df.filter_date(your_date_column_name, start_date, end_date)

You could just select the time range by doing: df.loc['start_date':'end_date']

In pandas version 1.1.3 I encountered a situation where the python datetime based index was in descending order. In this case
df.loc['2021-08-01':'2021-08-31']
returned empty. Whereas
df.loc['2021-08-31':'2021-08-01']
returned the expected data.

Another solution if you would like to use the .query() method.
It allows you to use write readable code like .query(f"{start} < MyDate < {end}") on the trade off, that .query() parses strings and the columns values must be in pandas date format (so that it is also understandable for .query())
df = pd.DataFrame({
'MyValue': [1,2,3],
'MyDate': pd.to_datetime(['2021-01-01','2021-01-02','2021-01-03'])
})
start = datetime.date(2021,1,1).strftime('%Y%m%d')
end = datetime.date(2021,1,3).strftime('%Y%m%d')
df.query(f"{start} < MyDate < {end}")
(following the comment from #Phillip Cloud, answer from #Retozi)

import the pandas library
import pandas as pd
STEP 1: convert the date column into a string using the pd.to_datetime() method
df['date']=pd.to_datetime(df["date"],unit='s')
STEP 2: perform the filtering in any predetermined manner ( i.e 2 months)
df = df[(df["date"] >"2022-03-01" & df["date"] < "2022-05-03")]
STEP 3 : Check the output
print(df)

# 60 days from today
after_60d = pd.to_datetime('today').date() + datetime.timedelta(days=60)
# filter date col less than 60 days date
df[df['date_col'] < after_60d]

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