Subtract objects in one NSArray from another array
Something like this? NSMutableArray *array = [NSMutableArray arrayWithArray:wants]; [array removeObjectsInArray:needs];
Something like this? NSMutableArray *array = [NSMutableArray arrayWithArray:wants]; [array removeObjectsInArray:needs];
There are really two separate issues. The first is keeping the elements of the array in proper order so that there are no “holes” after removing an element. The second is actually resizing the array itself. Arrays in C are allocated as a fixed number of contiguous elements. There is no way to actually remove … Read more
You should be able to use subqueries for that: SELECT (SELECT COUNT(*) FROM … WHERE …) – (SELECT COUNT(*) FROM … WHERE …) AS Difference Just tested it: Difference ———– 45 (1 row(s) affected)
I can reproduce this on my Q6600 (Python 2.6.2); increasing the range to 100000000: (‘+=’, 11.370000000000001) (‘-=’, 10.769999999999998) First, some observations: This is 5% for a trivial operation. That’s significant. The speed of the native addition and subtraction opcodes is irrelevant. It’s in the noise floor, completely dwarfed by the bytecode evaluation. That’s talking about … Read more
<?php echo $newdate = date(“m-Y”, strtotime(“-1 months”)); output 07-2016
You could do this. adjusted_datetime = (datetime_from_form.to_time – n.hours).to_datetime
Solution Use pd.concat followed by drop_duplicates(keep=False) pd.concat([df1, df2, df2]).drop_duplicates(keep=False) It looks like a b 1 3 4 Explanation pd.concat adds the two DataFrames together by appending one right after the other. if there is any overlap, it will be captured by the drop_duplicates method. However, drop_duplicates by default leaves the first observation and removes every … Read more
set.difference, set.union… can take any iterable as the second arg while both need to be sets to use -, there is no difference in the output. Operation Equivalent Result s.difference(t) s – t new set with elements in s but not in t With .difference you can do things like: s1 = set([1,2,3]) print(s1.difference(*[[3],[4],[5]])) {1, … Read more
Okay so you can do that in two steps, taken from @zoechi (a big contributor to Flutter): Define the base time, let us say: var date = new DateTime(2018, 1, 13); Now, you want the new date: var newDate = new DateTime(date.year, date.month – 1, date.day); And you will get 2017-12-13