Python 3: use functools.reduce: >>> from functools import reduce >>> reduce(lambda x, y: x*y, [1,2,3,4,5,6]) 720 Python 2: use reduce: >>> reduce(lambda x, y: x*y, [1,2,3,4,5,6]) 720 For compatible with 2 and 3 use pip install six, then: >>> from six.moves import reduce >>> reduce(lambda x, y: x*y, [1,2,3,4,5,6]) 720
Very interesting question, and clever trick. Let’s look at a simple example of getting a single byte manipulated. Using unsigned 8 bit for simplicity. Imagine your number is xxaxxbxx and you want ab000000. The solution consisted of two steps: a bit masking, followed by multiplication. The bit mask is a simple AND operation that turns … Read more
Short answer: Not likely. Long answer: Your compiler has an optimizer in it that knows how to multiply as quickly as your target processor architecture is capable. Your best bet is to tell the compiler your intent clearly (i.e. i*2 rather than i << 1) and let it decide what the fastest assembly/machine code sequence … Read more
You can also write: [e] * n You should note that if e is for example an empty list you get a list with n references to the same list, not n independent empty lists. Performance testing At first glance it seems that repeat is the fastest way to create a list with n identical … Read more