Python 3.x does have reduce, you just have to do a from functools import reduce. It also has “dict comprehensions”, which have exactly the syntax in your example. Python 2.7 and 3.x also have a Counter class which does exactly what you want: from collections import Counter cnt = Counter(“abracadabra”) In Python 2.6 or earlier, … Read more
Here it is the updated version of the algorithm based on the newer paper: var pow_2_32 = 0xFFFFFFFF + 1; function HyperLogLog(std_error) { function log2(x) { return Math.log(x) / Math.LN2; } function rank(hash, max) { var r = 1; while ((hash & 1) == 0 && r <= max) { ++r; hash >>>= 1; } … Read more
There’s always this method: n = 1; if ( i >= 100000000 ) { n += 8; i /= 100000000; } if ( i >= 10000 ) { n += 4; i /= 10000; } if ( i >= 100 ) { n += 2; i /= 100; } if ( i >= 10 ) … Read more
The Counter class in the collections module is purpose built to solve this type of problem: from collections import Counter words = “apple banana apple strawberry banana lemon” Counter(words.split()) # Counter({‘apple’: 2, ‘banana’: 2, ‘strawberry’: 1, ‘lemon’: 1})
I am surprised no-one has mentioned the simplest solution,max() with the key list.count: max(lst,key=lst.count) Example: >>> lst = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67] >>> max(lst,key=lst.count) 4 This works in Python 3 or 2, but note that it only returns the most frequent item and … Read more
Like many problems to do with strings, this can be done in a simple way with a regex. >>> word = ‘Llanfairpwllgwyn|gyllgogerychwyrndrobwllllantysiliogogogoch’ >>> import re >>> pattern = re.compile(r’ch|dd|ff|ng|ll|ph|rh|th|[^\W\d_]’, flags=re.IGNORECASE) >>> len(pattern.findall(word)) 51 The character class [^\W\d_] (from here) matches word-characters that are not digits or underscores, i.e. letters, including those with diacritics.
For arbitrary-length integers, bin(n).count(“1”) is the fastest I could find in pure Python. I tried adapting Óscar’s and Adam’s solutions to process the integer in 64-bit and 32-bit chunks, respectively. Both were at least ten times slower than bin(n).count(“1”) (the 32-bit version took about half again as much time). On the other hand, gmpy popcount() … Read more
True is equal to 1. >>> sum([True, True, False, False, False, True]) 3