CODEWITHSUNDEEP
python
Naheem Olaniyan
Naheem Olaniyan

Reputation: 93

Faster way to find if a number is between two number in Python

Please is there a faster way to find out if a number is between two numbers. My current code is below. Thanks

lists = [2.3, 4, 3,5.5, 6.5, 7.5, 6, 8]
newlist = []
a = 2
b = 7
for i in lists:
    if min(a, b) < i < max(a, b):
        newlist.append(i)
print(newlist)

Upvotes: 0

Views: 1990

Answers (3)

Daweo
Daweo

Reputation: 36370

When looking for where your code is slow, profile built-in module is handy. Here we can use it following way

import cProfile as profile
def func():
    lists = [2.3, 4, 3,5.5, 6.5, 7.5, 6, 8]
    newlist = []
    a = 2
    b = 7
    for i in lists:
        if min(a, b) < i < max(a, b):
            newlist.append(i)
    print(newlist)
profile.run('func()')

output is

[2.3, 4, 3, 5.5, 6.5, 6]
         27 function calls in 0.000 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.000    0.000 <stdin>:1(func)
        1    0.000    0.000    0.000    0.000 <string>:1(<module>)
        1    0.000    0.000    0.000    0.000 {built-in method builtins.exec}
        8    0.000    0.000    0.000    0.000 {built-in method builtins.max}
        8    0.000    0.000    0.000    0.000 {built-in method builtins.min}
        1    0.000    0.000    0.000    0.000 {built-in method builtins.print}
        6    0.000    0.000    0.000    0.000 {method 'append' of 'list' objects}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

As you might deduce max and min was used 8 times each but in this case once for min and once for max would be enough. Sample data is too tiny to say anything useful about time of execution of components. If you wish you might use more data (longer lists) and look for results.

Upvotes: 0

j1-lee
j1-lee

Reputation: 13929

Try the following:

a, b = min(a, b), max(a, b)
newlist = [x for x in lists if a < x < b]

With 100000 iterations, I found it 3 times faster than the original code. Using list comprehension instead of if helps a little, but most improvements come from pre-defining max and min before list comprehension (or if);

  • 0.1944 sec.: list comprehension + min & max predefined
  • 0.2672 sec.: if + min & max predefined
  • 0.5600 sec.: original (if + min & max at each iteration)

Upvotes: 2

Exotic Cut 5276
Exotic Cut 5276

Reputation: 255

lists = [2.3, 4, 3,5.5, 6.5, 7.5, 6, 8]
newlist = []
a = 2
b = 7
minimum = min(a, b)
maximum = max(a, b)
for i in lists:
    if minimum < i < maximum:
        newlist.append(i)
print(newlist)

This will make things faster as we are not computing minimum and maximum everytime when the loop runs and conditions are checked.

Upvotes: 2

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