How to Check Whether All Elements of Array Are in Between Two Values

How to check whether all elements of array are in between two values

The docs for np.nadarray.all state:

Returns True if all elements evaluate to `True`.

The implication is that the method is generally used with Boolean arrays. If not used in this way, you may see all sorts of unintended consequences from that fact 0 is considered False and 1 True:

np.arange(10).all()  # False
np.ones(10).all()    # True
np.zeros(10).all()   # False

Comparison operators of a 1d array versus a number automatically produce Boolean arrays. You can also use the & operator to combine multiple comparisons element-wise. So you can rewrite your logic:

def foo(rotation, speed):
    avgs = rotation / speed
    return ((avgs >= 55) & (avgs <= 55.2)).all()

Notice there's no need for an if statement here. The result of the np.ndarray.all is a Boolean value, i.e. either True or False.

Furthermore, a more succinct way of writing your logic is possible via np.isclose and its atol ("absolute tolerance") argument:

def foo(rotation, speed):
    avgs = rotation / speed
    return np.isclose(A, 55.1, atol=0.1)

Check if array values are between two values of another array

You can use ufunc.outer like so:

steps = np.arange(0, 22.5, 2.5)
vect = np.random.normal(10, 5.0, 10)

np.less.outer(vect,steps[1:]) & np.greater_equal.outer(vect,steps[:-1])

Sample output:

array([[False, False, False, False, False, False,  True, False],
       [False, False, False, False,  True, False, False, False],
       [False,  True, False, False, False, False, False, False],
       [False, False, False,  True, False, False, False, False],
       [False, False, False,  True, False, False, False, False],
       [False, False, False, False,  True, False, False, False],
       [False, False, False, False, False, False, False,  True],
       [False, False, False,  True, False, False, False, False],
       [False, False,  True, False, False, False, False, False],
       [False, False, False,  True, False, False, False, False]])

Or as DataFrame

pd.DataFrame(np.less.outer(vect,steps[1:]) & np.greater_equal.outer(vect,steps[:-1]), columns=[f"{steps[i]:.1f}-{steps[i+1]:.1f}" for i in range(len(steps)-1)])

Output:

   0.0-2.5  2.5-5.0  5.0-7.5  7.5-10.0  10.0-12.5  12.5-15.0  15.0-17.5  17.5-20.0
0    False    False    False     False      False      False       True      False
1    False    False    False     False       True      False      False      False
2    False     True    False     False      False      False      False      False
3    False    False    False      True      False      False      False      False
4    False    False    False      True      False      False      False      False
5    False    False    False     False       True      False      False      False
6    False    False    False     False      False      False      False       True
7    False    False    False      True      False      False      False      False
8    False    False     True     False      False      False      False      False
9    False    False    False      True      False      False      False      False

How to check if value is between two values stored in an array

This will return the higher index, for the next larger value. I have assumed from your example that the array is already sorted

let largerIndex  = pressureLevels.firstIndex(where: { $0 > tempPressure}) 
let smallerIndex = largerIndex - 1

Note the edge cases, if there is no value larger than tempPressure then largerIndex will be nil and if all values in the array are larger then largerIndex will be 0

how to see if there is a value in a list that is between two values from another list

It really depends on what you want it to return. I wrote a code that will return the first pattern that it finds, but with some changes I'm sure it would not be difficult to return all combinations.

def get_between(a, b):
    a, b = sorted(a), sorted(b)
    
    for b_value in b:
        smaller = None
        greater = None
        for a_value in a:
            if b_value > a_value:
                smaller = a_value
            elif b_value < a_value:
                greater = a_value
            
            if smaller and greater:
                return f"{b_value} is between {smaller} and {greater}"
    
    return "There is no such combination"

a = [1, 4, 12]
b = [2, 13]
print(get_between(a, b))

The output on that case will be 2 is between 1 and 4, but you can adapt the return value to be whatever you want.

Easy way to test if each element in an numpy array lies between two values?

One solution would be:

import numpy as np
a = np.array([1, 2, 3, 4, 5])
(a > 1) & (a < 5)
# array([False,  True,  True,  True, False])

results array between two values - javascript

A slightly different approach, with a customized callback function for filtering.

function filterNumbers(min, max) {    return function (a) { return a >= min && a <= max; };}
var numbers = ['10', '15', '20', '25', '30', '35', '40', '45', '50'],    result = numbers.filter(filterNumbers(10, 30));
document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');

var array = [18, 23, 20, 17, 21, 18, 22, 19, 18, 20],    lower = 18,    upper = 20,    result = array.reduce(function (r, a) {        return r + (a >= lower && a <= upper);    }, 0);
console.log(result);

const allEqual = arr => arr.every( v => v === arr[0] )
allEqual( [1,1,1,1] )  // true

[1,1,1,1].every( (val, i, arr) => val === arr[0] )   // true