> What you are hoping for here is known as "short circuit" or "lazy" evaluation.
In SciPy, we have the private utility function `_lazywhere`[1] for this. Cheers, Lucas [1] https://github.com/scipy/scipy/blob/f44326023dc51758495491fc9f06858fd38358a0/scipy/_lib/_util.py#L88-L156 > On 26 Apr 2024, at 00:18, rosko37 <[email protected]> wrote: > > What you are hoping for here is known as "short circuit" or "lazy" > evaluation. Namely, this would work if np.where(cond, x, y) only evaluates x > if cond is true and only evaluates y if cond is false. In this case, not only > can it handle situations where one of the true/false cases "breaks" one of > the possible return values, it would also potentially save a lot of work if > one of x of y is expensive to evaluate. This would clearly be the case if > there were an explicit branch in the code, i.e. something of the form > > for row in score_df: > if number[row] = 0: > return None > else: > return 100 - ((100*rank_column[row] - 50)/number[row]) > > It COULD also be true for a vectorized operation IF it is implemented that > way. However, numpy.where is NOT implemented like this. Its parameters are > ALL numpy arrays themselves, and are each precomputed separately. Only then > is the Boolean array that represents the outcomes of the conditional > "combined" with the other two arrays x and y to produce the result. > > What you want is to do something along the lines of fancy indexing, where you > evaluate the condition on the array to get an explicit Boolean mask, and then > use this to select a "slice" (not a real contiguous slice, but a subset) of > rows of the dataframe to pass to the expression that may break for zero > values. > > On Thu, Apr 25, 2024 at 5:19 AM 840362492--- via NumPy-Discussion > <[email protected] <mailto:[email protected]>> wrote: >> 0 >> >> In my code, I use the following calculation for a column in the dataframe: >> np.where(df_score['number'] ! = 0, 100 - ((100 * df_score[rank_column] >> -50)/df_score['number']), None),I have used df_score['number']! = 0, but the >> code is still wrong, ZeroDivisionError: float division by zero, even if I >> put df_score['number']! = 0 changed to df_score['number'] > 0, why? >> >> pandas version:1.1.5 numpy version:1.24.4 >> >> Here are my numbers： 12.00000 12.00000 12.00000 12.00000 12.00000 0.00000 >> 0.00000 0.00000 0.00000 0.00000 12.00000 12.00000 12.00000 >> >> I want to know why it went wrong and what should be done to fix it? Thank >> you for your help >> _______________________________________________ >> NumPy-Discussion mailing list -- [email protected] >> <mailto:[email protected]> >> To unsubscribe send an email to [email protected] >> <mailto:[email protected]> >> https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ >> Member address: [email protected] <mailto:[email protected]> > _______________________________________________ > NumPy-Discussion mailing list -- [email protected] > <mailto:[email protected]> > To unsubscribe send an email to [email protected] > <mailto:[email protected]> > https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ > Member address: [email protected] <mailto:[email protected]>
_______________________________________________ NumPy-Discussion mailing list -- [email protected] To unsubscribe send an email to [email protected] https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ Member address: [email protected]
