Passing Column(s) to Function with Multiple Condtitions - Python Pandas

Question

I created a function that assigns customers to a "bucket" based on their annual purchase history. The function operates as intended when I pass individual values in (curryear, last year). How do I pass all values from two sepearate columns in curryear, lastyear?

When I try the following I receive

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

My code:

#FUNCTION FOR CATEGORIZING ANNUAL CUSTOMER PURCHASE BEHAVIOR
def bucket(curryear, lastyear):
    if ((lastyear > 0) & (curryear <= 0)):
        return 'Attrition'
    elif ((lastyear > curryear) & (curryear > 0)):
        return 'Organic Attrition'
    elif ((lastyear <= 0) & (curryear > 0)):
        return 'New Sales'
    elif ((curryear > lastyear) & (lastyear > 0)):
        return 'Organic Growth'
    elif ((lastyear == 0) & (curryear == 0)):
        return 'None'
    else:
        return 'Flat'

bucket(df['2019'],df['2018'])

Here is a sample of the data I am using: Sample Data

mabergerx · Accepted Answer · 2020-07-06 14:26:36Z

1

The error pretty much says the exact reason for the error you get (testing stuff like >0 for a whole column is ambiguous, as you could mean to check if every value is above 0 or just a single value from the column). You could apply the function you wrote on the individual values row-wise like this:

def bucket(curryear, lastyear):
    if ((lastyear > 0) & (curryear <= 0)):
        return 'Attrition'
    elif ((lastyear > curryear) & (curryear > 0)):
        return 'Organic Attrition'
    elif ((lastyear <= 0) & (curryear > 0)):
        return 'New Sales'
    elif ((curryear > lastyear) & (lastyear > 0)):
        return 'Organic Growth'
    elif ((lastyear == 0) & (curryear == 0)):
        return 'None'
    else:
        return 'Flat'

df["bucket"] = df.apply(lambda x: bucket(x["2019"], x["2018"]), axis=1)

answered Jul 6, 2020 at 14:26

mabergerx

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YeonwooSung · Accepted Answer · 2020-07-06 14:27:07Z

0

Basically, pandas has a function called "apply", which supports the lambda function.

df['bucket'] = df.apply(lambda x: bucket(x.2019, x.2018), axis=1)

answered Jul 6, 2020 at 14:27

YeonwooSung

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Quang Hoang · Accepted Answer · 2020-07-06 14:28:47Z

0

Rewrite your function so that it can be parallelized over the columns:

def bucket(curryear, lastyear):
    ly_pos, cy_pos = lastyear > 0, curryear > 0
    out = np.select( (ly_pos & (~cy_pos), (lastyear > curryear) & cy_pos,
                      (~ly_pos) & cy_pos, (curryear>lastyear)&ly_pos,
                      (lastyear==0) & (curryear==0)
                     ),
                     ('Attritrion', 'Organic Attrition',
                      'New Sales', 'Organic Growth', 'None'),
                    'Flat'
                   )
    return out

bucket(df['2019'], df['2018'])

answered Jul 6, 2020 at 14:28

Quang Hoang

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a0nic · Accepted Answer · 2020-07-06 14:57:44Z

0

@mabergerx This worked - big thanks. This is how I executed it which I realize is probably inefficient.

df['2013 Bucket'] = df.apply(lambda x: bucket(x["2013"], x["2012"]), axis=1)
df['2014 Bucket'] = df.apply(lambda x: bucket(x["2014"], x["2013"]), axis=1)
df['2015 Bucket'] = df.apply(lambda x: bucket(x["2015"], x["2014"]), axis=1)
df['2016 Bucket'] = df.apply(lambda x: bucket(x["2016"], x["2015"]), axis=1)
df['2017 Bucket'] = df.apply(lambda x: bucket(x["2017"], x["2016"]), axis=1)
df['2018 Bucket'] = df.apply(lambda x: bucket(x["2018"], x["2017"]), axis=1)
df['2019 Bucket'] = df.apply(lambda x: bucket(x["2019"], x["2018"]), axis=1)

answered Jul 6, 2020 at 14:57

a0nic

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