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?](http://stackoverflow.com/questions/394809/does-python-have-a-ternary-conditional-operator) — Christian König, May 16 '17 at 05:11
Side note: `apply()` is deprecated since 2.3. It has been supplemented with direct call syntax: `(lambda x:)(value)`. — ivan_pozdeev, May 16 '17 at 05:26

Taku · Answer 1 · 2017-05-16T05:16:49.463

3

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 '17 at 05:16

answered May 16 '17 at 05:11

Taku

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score 1 · Answer 2 · answered May 16 '17 at 05:13

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 '17 at 05:13

A.J. Uppal

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Syntax error in lambda with if expression

2 Answers2