Here is a good procedure for resetting any sequence to 0 from Oracle guru Tom Kyte. Great discussion on the pros and cons in the links below too. [email protected]> create or replace procedure reset_seq( p_seq_name in varchar2 ) is l_val number; begin execute immediate ‘select ‘ || p_seq_name || ‘.nextval from dual’ INTO l_val; execute … Read more
Understanding slicing
You need to keep track of two variables, the current count (count) of the consecutive values and the maximum count (max_count) so far. When you observe a different value, you reset count and update max_count and continue the loop. def get_longest_seq(l, val): count = 0 max_count = 0 for e in l: if e == … Read more
#include <stdio.h> int seq(void){ static int v = 5, d = 1; int ret = v; v += d; if(v == 5 || v == 7) d = -d; return ret; } int main(void) { for(int i = 0; i < 20; ++i) printf(“%d “, seq()); return 0; }
from nltk import ngrams from collections import Counter def simple(seq, n): return Counter(ngrams(seq, n)).most_common(1)[0][1] / float(len(seq) – n + 1)
When using combinations() you are getting all pairs, even the non-adjacent ones. You can create a function that will return the adjacent pairs. I’ve tried the following code and it worked, maybe it can give you some insight: import os import re from collections import Counter def pairs(text): ans = re.findall(r'[A-Za-z]+’, text) return (tuple(ans[i:i+2]) for … Read more