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I have code here where I'm searching through a series on a pandas dataframe.

                    df[(df['SlyWeekofFiscalYear']==wk) &
                       (df['IB']==bnd) &
                       (df['slyfiscalyear']==yr)]
                    ['Wholesale'].sum()

It's in a function that passes a kwarg bnd=None. Is there a way to disregard the 2nd line of code if bnd=None?

Currently I have a long if statement but I'd like to tidy up the code if possible.

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3 Answers 3

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You can change the second line to ((bnd is None) | (df['IB'] == bnd)). If bnd is None, this produces an all true Series, and since you have & operators, it will have no effect on the result.

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5 Comments

Will the second boolean vector be evaluated if the first condition is true?
@IanS Normally, unlike or or and. | and & have no short circuits. But can't find the docs at this moment.
That's what I suspected. Your word is enough for me :)
So my solution could be marginally faster!
@IanS Hmm, I am suspicious how faster it could be though. | is vectorized and usually pretty fast.
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Have you tried replacing the second line with some thing that's always true if bnd=None?

Somelike:

((df['IB']==bnd) | (bnd is None))
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Perfect - I used your suggestion. So simple but yet it escaped me!
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You could try a ternary statement:

(df['IB'] == bnd if bnd is not None else True)

The scalar value True should be broadcast properly to a vector of the right length.

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