One way to do is to use std::string find function: #include <string> #include <iostream> int main() { int occurrences = 0; std::string::size_type pos = 0; std::string s = “FooBarFooBarFoo”; std::string target = “Foo”; while ((pos = s.find(target, pos )) != std::string::npos) { ++ occurrences; pos += target.length(); } std::cout << occurrences << std::endl; }
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 is a nice tutorial, it is what you need. (Source: coursesweb.net/php-mysql) Register and show online users and visitors Count Online users and visitors using a MySQL table In this tutorial you can learn how to register, to count, and display in your webpage the number of online users and visitors. The principle is this: … Read more
PHP has no support for a SQL where sort of thing, especially not with an array of arrays. But you can do the counting your own while you iterate over the data: $count = array(); foreach($fruit as $one) { @$count[$one[‘color’]]++; } printf(“You have %d green fruit(s).\n”, $count[‘green’]); The alternative is to write yourself some little … Read more
As mentioned in THIS ANSWER using operator.countOf() is the way to go but you can also use a generator within sum() function as following: sum(value == 0 for value in D.values()) # Or the following which is more optimized sum(1 for v in D.values() if v == 0) Or as a slightly more optimized and … Read more
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
Using data.table, which has the function rleid(): require(data.table) # v1.9.5+ rleid(x) # [1] 1 2 3 4 5 5 6 7 7 8 8 8 9
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
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})
In Python 2.7 (or newer), you can use collections.Counter: import collections a = [1,1,1,1,2,2,2,2,3,3,4,5,5] counter=collections.Counter(a) print(counter) # Counter({1: 4, 2: 4, 3: 2, 5: 2, 4: 1}) print(counter.values()) # [4, 4, 2, 1, 2] print(counter.keys()) # [1, 2, 3, 4, 5] print(counter.most_common(3)) # [(1, 4), (2, 4), (3, 2)] print(dict(counter)) # {1: 4, 2: 4, … Read more