passing a variable into apply() in pandas

user_2908485 0
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I am having trouble getting the syntax right for applying a function to a dataframe. I am trying to create a new column in a dataframe by joining the strings in two other columns, passing in a separator. I get the error

TypeError: ("apply_join() missing 1 required positional argument: 'sep'", 'occurred at index cases')

If I add sep to the apply_join() function call, that also fails:

  File "unite.py", line 37, in unite
    tibble_extra = df[cols].apply(apply_join, sep)
NameError: name 'sep' is not defined

import pandas as pd
from io import StringIO

tibble3_csv = """country,year,cases,population
Afghanistan,1999,745,19987071
Afghanistan,2000,2666,20595360
Brazil,1999,37737,172006362
Brazil,2000,80488,174504898
China,1999,212258,1272915272
China,2000,213766,1280428583"""
with StringIO(tibble3_csv) as fp:
    tibble3 = pd.read_csv(fp)
print(tibble3)

def str_join_elements(x, sep=""):
    assert type(sep) is str
    return sep.join((str(xi) for xi in x))

def unite(df, cols, new_var, combine=str_join_elements):
    def apply_join(x, sep):
        joinstr = str_join(x, sep)
        return pd.Series({new_var[i]:s for i, s in enumerate(joinstr)})
  
    fixed_vars = df.columns.difference(cols)
    tibble = df[fixed_vars].copy()
    tibble_extra = df[cols].apply(apply_join)
  
    return pd.concat([tibble, tibble_extra], axis=1) 
table3_again = unite(tibble3, ['cases', 'population'], 'rate', combine=lambda x: str_join_elements(x, "/"))
print(table3_again)
pandas apply python
2 Answers

Use lambda when you have multiple parameters i.e 

df[cols].apply(lambda x: apply_join(x,sep),axis=1)

Or pass parameters with the help of args parameter i.e 

 df[cols].apply(apply_join,args=[sep],axis=1)
user_48006520

You just add it into the apply statement:

tibble_extra = df[cols].apply(apply_join, sep=...)

Also, you should specify the axis. It may work without it, but its a good habit to prevent errors:

tibble_extra = df[cols].apply(apply_join, sep=..., axis=1(columns) or 0(rows|default))
user_65111940
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