CODEWITHSUNDEEP
pythonnormalization
Akshat Bhargava
Akshat Bhargava

Reputation: 61

loop to normalize range (0,10) in to (0,1)

What I need is to divide/spread 0 to 1. according to single number which is more than 2. like number 5 so 0 to 5 will be divided like this

0.00  
0.25
0.50
0.75
1.00

5 values in a list

and my other question is what to do to get a sequence like this where middle number is 1 and first and last number is 0 , if number is 10.

0.00
0.25
0.50
0.75
1.00
1.00
0.75
0.50
0.25
0.00

Upvotes: 3

Views: 572

Answers (3)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476659

The upper bound of the range(..) is exclusive (meaning it is not enumerated), so you need to add one step to the range(..) function:

for i in range(0,11):
     b = i*(1.0/10)
     print b

That being said, if you want to create such array, you can use numpy.arange(..):

>>> import numpy as np
>>> np.arange(0, 1.1, 0.1)
array([0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1. ])

This thus allows you to specify floats for the offset, end, and step parameter.

As for your second question, you can itertools.chain iterables together, like:

from itertools import chain

for i in chain(range(0, 11), range(10, -1, -1)):
    print(i/10.0)

Here we thus have one range(..) that iterates from 0 to 10 (both inclusive), and one that iterates from 10, to 0 (both inclusive).

Upvotes: 5

Nouman
Nouman

Reputation: 7303

range 0 to 10 will give you numbers from 0 to 9. Here is some practical to explain:

>>> list(range(0,10))  
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

>>> list(range(0,11))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

>>> 
>>> list(range(0,1))
[0]

>>>

Upvotes: 2

Robson
Robson

Reputation: 823

You should use range(0,11) to get all the numbers from 0 to 10.

Upvotes: 2

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