Is there any range() function in python for float numbers for example
a=0.6
if a in range(0,1):
a=3
How can i implement this?
If you're looking to check whether a is in between two numbers it's better to use:
0 <=a<=1
Otherwise , if you do need a list of say 0 to 1 in 0.1 jumps you can use this code to generate it:
lst = map(lambda x: x/10.0, range(11))
Similar to the question linked by Begueradj but slightly different (note, floats are not the same as decimals):
import decimal
def drange(start, stop, step=decimal.Decimal('1')):
while start < stop:
yield start
start += step
print(list(drange(
decimal.Decimal('1.25'),
decimal.Decimal('2.34'),
decimal.Decimal('0.1'),
)))
Output:
[Decimal('1.25'), Decimal('1.35'), Decimal('1.45'), Decimal('1.55'),
Decimal('1.65'), Decimal('1.75'), Decimal('1.85'), Decimal('1.95'),
Decimal('2.05'), Decimal('2.15'), Decimal('2.25')]
If I'm reading correctly, you want to test if a number is between two other numbers, so use:
a = 0.6
if 0 <= a < 1: # change to `<= 1` to be inclusive
a = 3
You don't need to generate a range and do membership testing - unless you have a discrete set of values that your a
should match - the builtin range
in Python 3.x can do efficient lookups for int
s as it can optimise membership testing. If you have a large amount of discrete values in a large range, then you'd be better of doing it mathematically anyway.
assuming you have numpy installed do:
>>>
import numpy
>>>
print np.arange(0,1,0.1)array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9])
if you don't have Numpy installed, definitely go get it.