On 10/22/2014 12:40 PM, Chris Angelico wrote: > That's true when it's fundamentally arithmetic. But part of that > readability difference is the redundancy in the second one. What if it > weren't so redundant? > > 'Negative' if x < 0 else 'Low' if x < 10 else 'Mid' if x < 20 else 'High' > > You can't easily turn that into a dict lookup, nor indexing. It's > either a chained if/elif tree or nested if/else expressions, which > come to the same thing. No, you can't turn that into a dict lookup, but this is one of the canonical use cases for the bisect module:
>>> from bisect import bisect >>> breakpoints = [0, 10, 20] >>> labels = ['Negative', 'Low', 'Mid', 'High'] >>> values = [-5, 5, 15, 25] >>> [labels[bisect(breakpoints, value)] for value in values] ['Negative', 'Low', 'Mid', 'High'] It's also worth noting that using bisect is O(log(n)) instead of O(n), but if you're going to hit a point where the asymptotic behavior matters I'm sure you will have long since abandoned a manually-written if/elif chain. MMR... -- https://mail.python.org/mailman/listinfo/python-list
