Check to See If a Value Is Within a Range

How to elegantly check if a number is within a range?

There are a lot of options:

int x = 30;
if (Enumerable.Range(1,100).Contains(x))  //true

And indeed basic if more elegantly can be written with reversing order in the first check:

if (1 <= x && x <= 100)   //true

Also, check out this SO post for regex options.

Notes:

LINQ solution is strictly for style points - since Contains iterates over all items its complexity is O(range_size) and not O(1) normally expected from a range check.

More generic version for other ranges (notice that second argument is count, not end):
```
if (Enumerable.Range(start, end - start + 1).Contains(x)
```
There is temptation to write if solution without && like 1 <= x <= 100 - that look really elegant, but in C# leads to a syntax error "Operator '<=' cannot be applied to operands of type 'bool' and 'int'"

Check if a value is within a range of numbers

You're asking a question about numeric comparisons, so regular expressions really have nothing to do with the issue. You don't need "multiple if" statements to do it, either:

if (x >= 0.001 && x <= 0.009) {
  // something
}

You could write yourself a "between()" function:

function between(x, min, max) {
  return x >= min && x <= max;
}
// ...
if (between(x, 0.001, 0.009)) {
  // something
}

Check to see if a value is within a range?

Assuming that the values of ID are unique:

DT[, list(OK = 1 %in% seq(time.s, time.e)), by = ID]

giving;

   ID    OK
1:  1  TRUE
2:  2  TRUE
3:  3 FALSE
4:  4 FALSE

How do I determine if a value is within a range?

It is because you have declared i variable as int and you are taking input as string so when it is checking condition it is getting null value in i variable and not able to enter if block check below code

    #include <stdio.h>

    int main(){

      int i;
        printf("Value to check Interval \n");
        scanf("%d",&i);

      if (i>4 && i<6){
        printf("%d Value is in first interval\n", i);
      }
    }

try compiling your code without if condition i variable will return a null value

How to find out if a number is in a list of ranges?

Simply iterate over the list, take both values and create a range from those values and check if b in range(...), also use enumerate, start it from 1 and you will get in which consecutive range in the list the number is.

a = [[167772352, 167772415], [167772160, 167772223], [167772288, 167772351], [167772224, 167772255]]
b = 167772241

for index, (start, end) in enumerate(a, start=1):
    if b in range(start, end + 1):
        print(index)
        break

You can also use a list comprehension:

a = [[167772352, 167772415], [167772160, 167772223], [167772288, 167772351], [167772224, 167772255]]
b = 167772241

index = [b in range(start, end + 1) for start, end in a].index(True) + 1
print(index)

Also note the end + 1 used in both ranges, that is because the range doesn't include its end value so adding 1 means that the range is inclusive on both sides. Also both methods will get the index that starts from one, which is how you would usually count (1, 2, 3, 4, ...) as you have stated in your question that b should be in the fourth range (which means that you started counting from 1)

How to check if a number is within a range in shell

If you are using Bash, you are better off using the arithmetic expression, ((...)) for readability and flexibility:

if ((number >= 2 && number <= 5)); then
  # your code
fi

To read in a loop until a valid number is entered:

#!/bin/bash

while :; do
  read -p "Enter a number between 2 and 5: " number
  [[ $number =~ ^[0-9]+$ ]] || { echo "Enter a valid number"; continue; }
  if ((number >= 2 && number <= 5)); then
    echo "valid number"
    break
  else
    echo "number out of range, try again"
  fi
done

((number >= 2 && number <= 5)) can also be written as ((2 <= number <= 5)).

Determine whether integer is between two other integers

if 10000 <= number <= 30000:
    pass

For details, see the docs.