Python // Lambda: invalid syntax

Question

I'm trying to iterate on a dataframe. I want to replace a few characters by anothers, unless the item I'm iterating is null/nan/NaN/etc.

For that I'm trying to use this line below:

lista['ultima_receita'] = lista['ultima_receita'].apply(lambda rstr: float(rstr.replace('.','').replace(',','.')[3:]) if pd.isnull(rstr) == False)

However it keeps getting me a invalid synthax error:

    lista['ultima_receita'] = lista['ultima_receita'].apply(lambda rstr: float(rstr.replace('.','').replace(',','.')[3:]) if pd.isnull(rstr) == False)
                                                                                                                                                     ^
SyntaxError: invalid syntax

I've tried everything I could and didn't find out the reason why the synthax is wrong. Can someone help?

Cireo · Accepted Answer · 2017-06-23 00:32:32Z

3

Try making your question simpler. The lambda is indeed the issue, so this has nothing to do with pandas here.

>>> lambda rstr: float(rstr.replace('.','').replace(',','.')[3:]) if pd.isnull(rstr) == False
SyntaxError: invalid syntax  # At the end of 'False' above

or even simpler:

>>> lambda x: "foo" if "bar" == False
SyntaxError: invalid syntax

This is because python needs an else for the A if B else C construction. If you want to do a conditional modification you could make this else rstr, or use other pandas / numpy logic to do different logic.

>>> func = lambda x: "foo" if "bar" in x else x
>>> func("isobaric"), func("agnostic")
('foo', 'agnostic')

answered Jun 23, 2017 at 0:32

Cireo

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