How to find whether a number belongs to a particular range in Python?

Question

Suppose I want to check if x belongs to range 0 to 0.5. How can I do it?

Can I use the range function for that?

score 79 · Answer 1 · answered Mar 06 '09 at 08:53

79

No, you can't do that. range() expects integer arguments. If you want to know if x is inside this range try some form of this:

print 0.0 <= x <= 0.5

Be careful with your upper limit. If you use range() it is excluded (range(0, 5) does not include 5!)

answered Mar 06 '09 at 08:53

1

-1: No reference to the documentation and no example of what range really does. +1: Being polite in the face of an absurd question. – S.Lott Mar 06 '09 at 11:25
17

"Range" is also an english word, not only a function name ;) – Josef Kufner Jul 11 '14 at 22:47

vartec · Answer 2 · 2009-03-06T09:24:58.757

26

print 'yes' if 0 < x < 0.5 else 'no'

range() is for generating arrays of consecutive integers

edited Mar 06 '09 at 09:24

answered Mar 06 '09 at 08:51

vartec

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thank you..i got range() concept is there any built-in method ? – user46646 Mar 06 '09 at 09:05
built-in method for writing 0 < x < 0.5? – vartec Mar 06 '09 at 09:11
how's writing 0 – SilentGhost Mar 06 '09 at 09:12
i want to know if there is any method – user46646 Mar 06 '09 at 09:16
26

Using built-in methods: (0.0).__lt__(x).__and__((5.0).__gt__(x)) – Mar 06 '09 at 09:21
@vartec: Add `print` before 'yes' to match the question exactly. – jfs Mar 06 '09 at 09:24
@Sebastian: done, however, I'm not sure anymore what the question is exactly ;-) – vartec Mar 06 '09 at 09:25

score 8 · Answer 3 · answered Mar 06 '09 at 08:54

8

>>> s = 1.1
>>> 0<= s <=0.2
False
>>> 0<= s <=1.2
True

answered Mar 06 '09 at 08:54

SilentGhost

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score 8 · Answer 4 · edited Feb 14 '18 at 23:54

8

To check whether some number n is in the inclusive range denoted by the two number a and b you do either

if   a <= n <= b:
    print "yes"
else:
    print "no"

use the replace >= and <= with > and < to check whether n is in the exclusive range denoted by a and b (i.e. a and b are not themselves members of the range).

Range will produce an arithmetic progression defined by the two (or three) arguments converted to integers. See the documentation. This is not what you want I guess.

edited Feb 14 '18 at 23:54

chris

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answered Mar 06 '09 at 09:12

VoidPointer

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1

your alternative version is just outright wrong – SilentGhost Mar 06 '09 at 09:20
2

wrong as in concept-wrong, not typo-wrong – SilentGhost Mar 06 '09 at 09:27
Well, it works for integers. For floats it *can* work, depending on values of b and n. – Abgan Mar 06 '09 at 12:39
It doesn't work for integers. Consider a = b = n = 1. – chris Feb 14 '18 at 23:41

score 6 · Answer 5 · answered Aug 14 '09 at 14:06

6

I would use the numpy library, which would allow you to do this for a list of numbers as well:

from numpy import array
a = array([1, 2, 3, 4, 5, 6,])
a[a < 2]

answered Aug 14 '09 at 14:06

dalloliogm

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for within a range (0.0 to 5.0) do: a[numpy.logical_and(0.0 – ecoe Oct 21 '13 at 19:11
This is slower and more complex than the highest-rated answer. – chris Feb 14 '18 at 23:43

score 6 · Answer 6 · answered Mar 11 '15 at 14:39

6

if num in range(min, max):
  """do stuff..."""
else:
  """do other stuff..."""

answered Mar 11 '15 at 14:39

b..

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This won't work for the OP's example of `float`s. See @vartec's answer. – Sanjay Manohar Aug 22 '15 at 15:50
Likely, this is a very slow implementation. – chris Feb 14 '18 at 23:42
1

Actually, it looks like it's implemented efficiently in CPython as `min <= num < max` (with special checks for `stride != 1`). You can check out the implementation in `range_contains` and `range_contains_long` in `cpython/Objects/rangeobject.c` – Connor Brinton Nov 14 '19 at 16:30

score 3 · Answer 7 · answered Mar 06 '09 at 11:40

3

Old faithful:

if n >= a and n <= b:

And it doesn't look like Perl (joke)

answered Mar 06 '09 at 11:40

Ali Afshar

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9

This is not Pythonic. – Asclepius Feb 22 '14 at 02:27

How to find whether a number belongs to a particular range in Python?

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