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Pandas pd.cut() - binning datetime column / series

Attempting to do a bin using pd.cut() but it is fairly elaborate-

A collegue sends me multiple files with report dates such as:

 '03-16-2017 to 03-22-2017'
 '03-23-2017 to 03-29-2017'
 '03-30-2017 to 04-05-2017'

They are all combined into a single dataframe and given a column name, df['Filedate'] so that every record in the file has the correct filedate.

The last day is a cutoff point, so I created a new column df['Filedate_bin'] which converts the last day to 3/22/2017, 3/29/2017, 4/05/2017 as a string.

Then I created a list: Filedate_bin_list= df.Filedate_bin.unique(). As a result I have a unique list of string cutoff dates that I would like to use as bins.

Importing different data into dataframe, there is a column of transaction dates: 3/28/2017, 3/29/2017, 3/30/2017, 4/1/2017, 4/2/2017, etc. Assigning them to a bin is difficult, it tried:

df['bin'] = pd.cut(df.Processed_date, Filedate_bin_list)

Received TypeError: unsupported operand type for -: 'str' and 'str'

Went back and tried converting the Filedate_bin to datetime, format='%m/%d/%Y' and get

TypeError: Cannot cast ufunc less input from dtype('<m8[ns]') to dtype ('<m8') with casting rule 'same_kind'.

Is there a better way to bin my processed_date(s) to text bins?

Am trying to tie in my processed dates 3/27/2017 to '03-23-2017 to 03-29-2017'

like image 205
Arthur D. Howland Avatar asked Apr 19 '17 16:04

Arthur D. Howland


1 Answers

UPDATE: starting from Pandas v0.20.1 (May 5, 2017) pd.cut and pd.qcut support datetime64 and timedelta64 dtypes (GH14714, GH14798).

Thanks @lighthouse65 for checking this!


Old answer:

Consider this approach:

df = pd.DataFrame(pd.date_range('2000-01-02', freq='1D', periods=15), columns=['Date'])

bins_dt = pd.date_range('2000-01-01', freq='3D', periods=6)
bins_str = bins_dt.astype(str).values

labels = ['({}, {}]'.format(bins_str[i-1], bins_str[i]) for i in range(1, len(bins_str))]

df['cat'] = pd.cut(df.Date.astype(np.int64)//10**9,
                   bins=bins_dt.astype(np.int64)//10**9,
                   labels=labels)

Result:

In [59]: df
Out[59]:
         Date                       cat
0  2000-01-02  (2000-01-01, 2000-01-04]
1  2000-01-03  (2000-01-01, 2000-01-04]
2  2000-01-04  (2000-01-01, 2000-01-04]
3  2000-01-05  (2000-01-04, 2000-01-07]
4  2000-01-06  (2000-01-04, 2000-01-07]
5  2000-01-07  (2000-01-04, 2000-01-07]
6  2000-01-08  (2000-01-07, 2000-01-10]
7  2000-01-09  (2000-01-07, 2000-01-10]
8  2000-01-10  (2000-01-07, 2000-01-10]
9  2000-01-11  (2000-01-10, 2000-01-13]
10 2000-01-12  (2000-01-10, 2000-01-13]
11 2000-01-13  (2000-01-10, 2000-01-13]
12 2000-01-14  (2000-01-13, 2000-01-16]
13 2000-01-15  (2000-01-13, 2000-01-16]
14 2000-01-16  (2000-01-13, 2000-01-16]

In [60]: df.dtypes
Out[60]:
Date    datetime64[ns]
cat           category
dtype: object

Explanation:

df.Date.astype(np.int64)//10**9 - converts datetime values into UNIX epoch (timestamp - # of seconds since 1970-01-01 00:00:00):

In [65]: df.Date.astype(np.int64)//10**9
Out[65]:
0     946771200
1     946857600
2     946944000
3     947030400
4     947116800
5     947203200
6     947289600
7     947376000
8     947462400
9     947548800
10    947635200
11    947721600
12    947808000
13    947894400
14    947980800
Name: Date, dtype: int64

the same will applyied to bins:

In [66]: bins_dt.astype(np.int64)//10**9
Out[66]: Int64Index([946684800, 946944000, 947203200, 947462400, 947721600, 947980800], dtype='int64')

labels:

In [67]: labels
Out[67]:
['(2000-01-01, 2000-01-04]',
 '(2000-01-04, 2000-01-07]',
 '(2000-01-07, 2000-01-10]',
 '(2000-01-10, 2000-01-13]',
 '(2000-01-13, 2000-01-16]']
like image 77
MaxU - stop WAR against UA Avatar answered Sep 23 '22 08:09

MaxU - stop WAR against UA