There is currently no API for the hashtags feature on Facebook edit: there was however a public posts search function which will return some public posts with a certain hashtag if you use that hashtag as the search string in API version 1.0 – there is no equivalent in version 2.0 onwards It ignores the … Read more
Here’s another option: s=”some string” numbers = sum(c.isdigit() for c in s) letters = sum(c.isalpha() for c in s) spaces = sum(c.isspace() for c in s) others = len(s) – numbers – letters – spaces
Two things come to my mind: std::count_if and then compare the result to 1. To avoid traversing the whole container in case eg the first two elements already match the predicate I would use two calls looking for matching elements. Something along the line of auto it = std::find_if(begin,end,predicate); if (it == end) return false; … Read more
SELECT foo, count(bar) FROM mytable GROUP BY bar ORDER BY count(bar) DESC; The group by statement tells aggregate functions to group the result set by a column. In this case “group by bar” says count the number of fields in the bar column, grouped by the different “types” of bar. A better explanation is here: … Read more
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
You can use array_count_values function $array = array(‘apple’, ‘orange’, ‘pear’, ‘banana’, ‘apple’, ‘pear’, ‘kiwi’, ‘kiwi’, ‘kiwi’); print_r(array_count_values($array)); will output Array ( [apple] => 2 [orange] => 1 [pear] => 2 etc… )
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