Populating column in pandas dataframe using a function call

Question

How can I use .loc with .str.match() to update column values but with a function call? The code I'm trying is like;

df.loc[df['Col1'].str.match(r'\d\d/\d\d/\d\d\d\d', na=False), 'Col2'] = _my_func(df['Col1'])

a simple regex pattern to find date format, and then _myfunc();

def _my_func(data)
    for row in data.iteritems():
        day = int(row[1][:2])
        month = int(row[1][3:5])
        year = int(row[1][6:])
        fecha = datetime.datetime(year, month, day, 0, 0, 0)
        diff =  fecha - datetime.datetime.now()
        if diff.days > 0:
            return 'Yes'
        elif diff.days < 0:
            return 'No'

Is this a correct way to return values from the function into the dataframe?

Also if I insert a print('test') into the _my_func just before either return, it only prints test one time, instead of a print for each row in the data passed to the function, does anyone know why? Thank you.

Your function takes a pd.Series but returns just one yes or no answer. This would be why your print test only occurs once. For this to work using this method (i.e. instead of using df['Col1'].apply(_my_func)) you'd need to maintain all the 'yes' and 'no' results into an iterable (list or pd.Series) — Dillon
– Dillon, Commented Oct 1, 2018 at 11:59

Nordle · Accepted Answer · 2018-10-01 12:05:28Z

2

You can try it using apply() function.

For example:

df['loc1'] = df['loc1'].apply(_my_func)

Then it would take each row of the dataframe and pass it as input to the function _my_func.

edited Oct 1, 2018 at 12:05

Nordle

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answered Oct 1, 2018 at 12:03

Bhaskar

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Comments

Nordle · Accepted Answer · 2018-10-01 14:53:59Z

1

Following my comment:

def _my_func(x):
    day = int(x[:2])
    month = int(x[3:5])
    year = int(x[6:])
    fecha = datetime.datetime(year, month, day, 0, 0, 0)
    diff = fecha - datetime.datetime.now()
    if diff.days > 0:
        return 'Yes'
    elif diff.days < 0:
        return 'No'

Followed by:

df.loc[df['Col1'].str.match(r'\d\d/\d\d/\d\d\d\d', na=False), 'Col2'] = df['Col1'].apply(_my_func)

edited Oct 1, 2018 at 14:53

Nordle

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answered Oct 1, 2018 at 12:02

Dillon

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1 Comment

Nordle Over a year ago

Exactly what I needed, I edited out the first index in your answer because it wasn't needed after making the apply changes, thanks!

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