CODEWITHSUNDEEP
pythonif-statementrange
B9M8
B9M8

Reputation: 47

How to use a float in an "if x in range" statement

I'm trying to write an if statement that takes a float as a range.

x = 8.2

if x in range(0, 4.4):
    print('one')
if x in range(4.5, 8):
    print('two')
if x in range(8.1, 9.9):
    print('three')
if x > 10:
    print('four')

I have also tried this to no avail:

if x > 0 <= 4.4

Upvotes: 2

Views: 2717

Answers (4)

user20523129
user20523129

Reputation:

For floating points you don't need to use range instead you can use comparisons operators like:

num = 4
if (num > 3.3) print("less than four")
elif (num > 3.7) print("less than four")
elif (num > 3. 9) print("less than four")
else print("Equal to four")

Upvotes: 0

Mike
Mike

Reputation: 1164

Use

if 0 < x <= 4.4:

where x is in the middle. It's equivalent to

if 0 < x and x <= 4.4:

range is not suitable for that task.

Upvotes: 3

B9M8
B9M8

Reputation: 47

x = 8.3

if 0 <= x <= 4.4:
    print('one')
if 4.5 <= x <= 8:
    print('two')
if 8.1 <= x <= 9.9:
    print('three')
if x > 10:
    print('four')

Upvotes: 1

Barmar
Barmar

Reputation: 781068

You don't need range(). Just use comparisons, and use elif so that the ranges are exclusive.

if x < 4.5:
    print('one')
elif x < 8:
    print('two')
elif x < 10:
    print('three')
else:
    print('four')

This also solves the problem that you had gaps between your ranges.

Upvotes: 4

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