There’s no built-in JavaScript function to do this, but you can write your own fairly easily: function pad(n) { return (n < 10) ? (“0” + n) : n; } EDIT: Meanwhile there is a native JS function that does that. See String#padStart console.log(String(5).padStart(2, ‘0’));
In python 3.6, the fstring or “formatted string literal” mechanism was introduced. f”{a:02}” is the equivalent of the .format format below, but a little bit more terse. python 3 before 3.6 prefers a somewhat more verbose formatting system: “{0:0=2d}”.format(a) You can take shortcuts here, the above is probably the most verbose variant. The full documentation … Read more
Strings & VARCHAR. Do not try storing phone numbers as actual numbers. it will ruin the formatting, remove preceding 0s and other undesirable things. You may, if you choose to, restrict user inputs to just numeric values but even in that case, keep your backing persisted data as characters/strings and not numbers. Be aware of … Read more
Both lines you posted are fine, but you can do it purely in integers, and it will be the most efficient: def sum_digits(n): s = 0 while n: s += n % 10 n //= 10 return s or with divmod: def sum_digits2(n): s = 0 while n: n, remainder = divmod(n, 10) s += … Read more
Python’s string library has isdigit() method: string[0].isdigit()
The reason it is only a suggestion is that you could quite easily write a print function that ignored the options value. The built-in printing and formatting functions do use the options value as a default. As to the second question, since R uses finite precision arithmetic, your answers aren’t accurate beyond 15 or 16 … Read more
Well, the most efficient way, presuming you know the size of the integer, would be a lookup. Should be faster than the much shorter logarithm based approach. If you don’t care about counting the ‘-‘, remove the + 1. #include <climits> // generic solution template <class T> int numDigits(T number) { int digits = 0; … Read more
length is a property, not a method. You can’t call it, hence you don’t need parenthesis (): function getlength(number) { return number.toString().length; } UPDATE: As discussed in the comments, the above example won’t work for float numbers. To make it working we can either get rid of a period with String(number).replace(‘.’, ”).length, or count the … Read more
You can do it in O(n) (where n is the number of digits) like this: Starting from the right, you find the first pair-of-digits such that the left-digit is smaller than the right-digit. Let’s refer to the left-digit by “digit-x”. Find the smallest number larger than digit-x to the right of digit-x, and place it … Read more
how about let isnum = /^\d+$/.test(val);