Syntax error in lambda with if expression

Question

basic_salary['dem_education_level_numeric']=basic_salary['dem_education_level'].apply(lambda x : 3 if (x == 'high'))

I am getting below error:

SyntaxError: invalid syntax (<ipython-input-20-80adac538241>, line 1)
  File "<ipython-input-20-80adac538241>", line 1
    basic_salary['dem_education_level_numeric']=basic_salary['dem_education_level'].apply(lambda x : 3 if (x == 'high'))
                                                                                                                       ^
SyntaxError: invalid syntax

Please help

Possible duplicate of Does Python have a ternary conditional operator? — Christian König
– Christian König, Commented May 16, 2017 at 5:11
Side note: apply() is deprecated since 2.3. It has been supplemented with direct call syntax: (lambda x:<whatever>)(value). — ivan_pozdeev
– ivan_pozdeev, Commented May 16, 2017 at 5:26

Taku · Accepted Answer · 2017-05-16 05:16:49Z

4

If you're going to have a if inside a lambda statement, an else must also be defined. Every lambda statement must return something, even if it's None. Change your lambda statement to this, by returning None if x isn't "high":

lambda x : 3 if x == 'high' else None # paralysis isn't necessary

edited May 16, 2017 at 5:16

answered May 16, 2017 at 5:11

Taku

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A.J. Uppal · Accepted Answer · 2017-05-16 05:13:05Z

1

Lambda statements need to return something, and without an else clause the return may be ambiguous. Change to:

basic_salary['dem_education_level_numeric']=basic_salary['dem_education_level'].apply(lambda x : 3 if x == 'high' else 0)

answered May 16, 2017 at 5:13

A.J. Uppal

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