Pandas: Remove rows whose date does not follow specified format

Question

In my dataset, I have a date column in which the data is of mixed format.

• Some rows only have YYYY (e.g. 2009)
• Others have MMM-YY (e.g. Jan-08)
• Yet others have a full date MM/DD/YYYY (e.g. 01/15/2006)

How might I remove the entries that do not follow the format MM/DD/YYYY? I'm not quite sure where to start here.

Below, I have provided the .head() of the data as a dict.

{'Collection Date': {0: '2001',
  1: '2002',
  2: '2006',
  3: '2/19/2006',
  4: '2/28/2006'},
 'Complete Genome': {0: 'No', 1: 'No', 2: 'No', 3: 'No', 4: 'No'},
 'Country': {0: 'Egypt', 1: 'Egypt', 2: 'Egypt', 3: 'Egypt', 4: 'Egypt'},
 'Flu Season': {0: '-N/A-', 1: '-N/A-', 2: '-N/A-', 3: '-N/A-', 4: '-N/A-'},
 'Host Species': {0: 'IRD:Human',
  1: 'IRD:Human',
  2: 'IRD:Bird/Avian',
  3: 'IRD:Chicken/Avian',
  4: 'IRD:Avian'},
 'Protein Name': {0: 'NA', 1: 'NA', 2: 'NA', 3: 'NA', 4: 'HA'},
 'Segment': {0: 6, 1: 6, 2: 6, 3: 6, 4: 4},
 'Segment Length': {0: 1428, 1: 1449, 2: 1441, 3: 1363, 4: 1707},
 'Sequence Accession': {0: 'AJ457944',
  1: 'AJ457943',
  2: 'GU050304',
  3: 'GQ184251',
  4: 'KF178948'},
 'State/Province': {0: '-N/A-',
  1: '-N/A-',
  2: '-N/A-',
  3: '-N/A-',
  4: '-N/A-'},
 'Strain Name': {0: '(A/Egypt/84/2001(H1N2))',
  1: '(A/Egypt/96/2002(H1N2))',
  2: 'A/avian/Egypt/920431/2006(H9N2)',
  3: 'A/chicken/Egypt/06207-NLQP/2006(H5N1)',
  4: 'A/chicken/Egypt/0626/2006'},
 'Subtype': {0: 'H1N2', 1: 'H1N2', 2: 'H9N2', 3: 'H5N1', 4: 'H5N1'}}

Answer 1

Rather than applying Regular expression to match just MM/DD/YYYY, if your dates are only ever YYYY or MMM/YYYY or MM/DD/YYYY then you can exploit the fact that MM/DD/YYYY is a string of length 10:

In [8]:

import pandas as pd

pd.set_option('display.notebook_repr_html', False)
df = pd.DataFrame({'date':['01/03/1987', '2003', 'Jan-08', '31/01/2010']})
df
Out[8]:
         date
0  01/03/1987
1        2003
2      Jan-08
3  31/01/2010

[4 rows x 1 columns]
In [9]:

df.ix[df.date.str.len() !=10]
Out[9]:
     date
1    2003
2  Jan-08

[2 rows x 1 columns]

You can then just use to_datetime:

In [16]:

df1 = df.ix[df.date.str.len() !=10]
df1
Out[16]:
     date
1    2003
2  Jan-08

[2 rows x 1 columns]
In [17]:

df1.date = pd.to_datetime(df1.date)
df1
Out[17]:
                 date
1 2003-01-01 00:00:00
2 2014-01-08 00:00:00

[2 rows x 1 columns]

Just for completeness if you wanted to filter using regexp:

df.ix[~df.date.str.contains('(\d{2})[/](\d{2})[/](\d{4})')]

would work, note the ~ which is a negation

Answer 2

You can use pd.to_datetime with the option errors='coerce' to convert invalid dates to NaT and then filter out NaTs with dropna()

Example:

>>> df = pd.DataFrame({'date':['01/03/1987', '2003', 'Jan-08', '31/01/2010', '2/13/2016'],'value':range(5)})
>>> df
         date  value
0  01/03/1987      0
1        2003      1
2      Jan-08      2
3  31/01/2010      3
4   2/13/2016      4

Format: DD/MM/YYYY

>>> pd.to_datetime(df['date'], format='%d/%m/%Y', errors='coerce')
0   1987-03-01
1          NaT
2          NaT
3   2010-01-31
4          NaT
>>> df['date'] = pd.to_datetime(df['date'], format='%d/%m/%Y', errors='coerce')
>>> df.dropna()
        date  value
0 1987-03-01      0
3 2010-01-31      3