, I get an invalid syntax espana['R_score'] = espana["Recency"].apply(lambda x : (1 if x < q["Recency"][0.30]) elif (2 if x < q["Recency"][0.55]) else 3) All I'm trying to do is have a column such that for some column ("Recency") before it, it returns a number (either 3,2,or 1) for each row depending on whether the number is below the .30 percentile, above the .55 percentile and anything above that is given a 1 What's wrong with code? Also do I need to explicitly state in the second case that I also need to be above the the .30 percentile or will this take care of itself? Note, for clarification,  q["Recency"][0.30] just picks off the "Recency row column corresponding to the index .30 of a table I created (using the quantile function)

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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When I type the following, I get an invalid syntax espana['R_score'] = espana["Recency"].apply(lambda x : (1 if x < q["Recency"][0.30]) elif (2 if x < q["Recency"][0.55]) else 3) All I'm trying to do is have a column such that for some column ("Recency") before it, it returns a number (either 3,2,or 1) for each row depending on whether the number is below the .30 percentile, above the .55 percentile and anything above that is given a 1 What's wrong with code? Also do I need to explicitly state in the second case that I also need to be above the the .30 percentile or will this take care of itself? Note, for clarification,  q["Recency"][0.30] just picks off the "Recency row column corresponding to the index .30 of a table I created (using the quantile function)
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