How to pass large number arguments to **kwargs

Question

I have following dataframe :

   p     p_loc  s            s0  s0_loc  s1  s1_loc  s2 s2_loc  s3   s3_loc
0  ABCD  M      AB,AC,AD     AB  A       AC   M      AD  Q      NaN  U
1  PQRS  N      PQ,PR,PS,QR  PQ  B       PR   N      PS  R      QR   V

I'm splitting column s into columns s0,s1,s2,s3 using , as delimeter. I want to pass values from s0,s1,s2,s3,s4,s5,s0_loc,s1_loc,s2_loc,s3_loc to following function which accepts following arguments:

def func(p,p_loc,**args):
    print(p)
    print(p_loc)
    print(kwargs)

I'm passing arguments as follows:

df['new'] = df.apply(lambda x: func(df['p'],df['p_loc'],df['s0'],df['s1'],df['s2'],df['s3'],df['s0_loc'],df['s1_loc'],df['s2_loc'],df['s3_loc']))

I want to know is there any simpler way to pass these arguments(Say, suppose my split of column s results in more than 15-20 columns.. How can I pass these many arguments in a simpler way)...???

Thanks in advance...

Answer 1

Create a set of positional arguments like:

>>> not_kwargs = {'p', 'p_loc', 's'}

Next filter out kwargs like:

>>> data = df.to_dict()
>>> kwargs = {key:data[key] for key in data.keys() ^ not_kwargs}

If you are using python-2.7 , ^ only operates on set .

>>> kwargs = {key:data[key] for key in set(data.keys()) ^ not_kwargs}

Call function:

>>> func(data['p'], data['p_loc'], **kwargs)

Answer 2

Not sure if this is what you want, but you can put the columns you require in a seperate list and generate the arguments when you need them.

def func(p, p_loc, *args):
    print(p)
    print(p_loc)
    print(args)

list_of_cols = ['s0', 's1', 's2', 's3', 's0_loc', 's1_loc', 's2_loc','s3_loc']
df['new'] = df.apply(lambda x: func(df['p'], df['p_loc'], *[df[col] for col in list_of_cols]))

Answer 3

Is this what you're looking for?

You can pass a dict into kwargs like this:

d = {
    'p': df['p'],
    'p_loc': df['p_loc'],
    's0': df['s0'],
    's1': df['s1'],
    ...
    }

def func(**kwargs):
    print(kwargs)

func(**d)