How do I create a list with numbers between two values?

Question

How do I create a list of numbers between two values? For example, a list between 11 and 16:

[11, 12, 13, 14, 15, 16]

Answer 1

Use range. In Python 2, it returns a list directly:

>>> range(11, 17)
[11, 12, 13, 14, 15, 16]

In Python 3, range is an iterator. To convert it to a list:

>>> list(range(11, 17))
[11, 12, 13, 14, 15, 16]

Note: The second number in range(start, stop) is exclusive. So, stop = 16+1 = 17.

---

To increment by steps of 0.5, consider using numpy's arange() and .tolist():

>>> import numpy as np
>>> np.arange(11, 17, 0.5).tolist()

[11.0, 11.5, 12.0, 12.5, 13.0, 13.5,
 14.0, 14.5, 15.0, 15.5, 16.0, 16.5]

See: How do I use a decimal step value for range()?

Answer 2

You seem to be looking for range():

>>> x1=11
>>> x2=16
>>> range(x1, x2+1)
[11, 12, 13, 14, 15, 16]
>>> list1 = range(x1, x2+1)
>>> list1
[11, 12, 13, 14, 15, 16]

For incrementing by 0.5 instead of 1, say:

>>> list2 = [x*0.5 for x in range(2*x1, 2*x2+1)]
>>> list2
[11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5, 16.0]

Answer 3

Try:

range(x1, x2+1)  

That is a list in Python 2.x and behaves mostly like a list in Python 3.x. If you are running Python 3 and need a list that you can modify, then use:

list(range(x1, x2+1))

Answer 4

Use list comprehension in python. Since you want 16 in the list too.. Use x2+1. Range function excludes the higher limit in the function.

list=[x for x in range(x1, x2+1)]

Answer 5

If you are looking for range like function which works for float type, then here is a very good article.

def frange(start, stop, step=1.0):
    ''' "range()" like function which accept float type''' 
    i = start
    while i < stop:
        yield i
        i += step
# Generate one element at a time.
# Preferred when you don't need all generated elements at the same time. 
# This will save memory.
for i in frange(1.0, 2.0, 0.5):
    print i   # Use generated element.
# Generate all elements at once.
# Preferred when generated list ought to be small.
print list(frange(1.0, 10.0, 0.5))    

Output:

1.0
1.5
[1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, 9.5]

Answer 6

assuming you want to have a range between x to y

range(x,y+1)

>>> range(11,17)
[11, 12, 13, 14, 15, 16]
>>>

use list for 3.x support

Answer 7

I got here because I wanted to create a range between -10 and 10 in increments of 0.1 using list comprehension. Instead of doing an overly complicated function like most of the answers above I just did this

simple_range = [ x*0.1 for x in range(-100, 100) ]

By changing the range count to 100 I now get my range of -10 through 10 by using the standard range function. So if you need it by 0.2 then just do range(-200, 200) and so on etc

Answer 8

In python you can do this very eaisly

start=0
end=10
arr=list(range(start,end+1))
output: arr=[0,1,2,3,4,5,6,7,8,9,10]

or you can create a recursive function that returns an array upto a given number:

ar=[]
def diff(start,end):
    if start==end:
        d.append(end)
        return ar
    else:
        ar.append(end)
        return diff(start-1,end) 

output: ar=[10,9,8,7,6,5,4,3,2,1,0]

Answer 9

While @Jared's answer for incrementing works for 0.5 step size, it fails for other step sizes due to rounding issues:

np.arange(11, 17, 0.1).tolist()
# [11.0,11.1,11.2,11.299999999999999, ...   16.79999999999998, 16.899999999999977]

Instead I needed something like this myself, working not just for 0.5:

# Example 11->16 step 0.5
s = 11
e = 16
step = 0.5
my_list = [round(num, 2) for num in np.linspace(s,e,(e-s)*int(1/step)+1).tolist()]
# [11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5, 16.0]

# Example 0->1 step 0.1
s = 0
e = 1
step = 0.1
my_list = [round(num, 2) for num in np.linspace(s,e,(e-s)*int(1/step)+1).tolist()]
# [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]

Answer 10

The most elegant way to do this is by using the range function however if you want to re-create this logic you can do something like this :

def custom_range(*args):
    s = slice(*args)
    start, stop, step = s.start, s.stop, s.step
    if 0 == step:
        raise ValueError("range() arg 3 must not be zero")
    i = start
    while i < stop if step > 0 else i > stop:
        yield i
        i += step

>>> [x for x in custom_range(10, 3, -1)]

This produces the output:

[10, 9, 8, 7, 6, 5, 4]

As expressed before by @Jared, the best way is to use the range or numpy.arrange however I find the code interesting to be shared.

Answer 11

Every answer above assumes range is of positive numbers only. Here is the solution to return list of consecutive numbers where arguments can be any (positive or negative), with the possibility to set optional step value (default = 1).

def any_number_range(a,b,s=1):
""" Generate consecutive values list between two numbers with optional step (default=1)."""
if (a == b):
    return a
else:
    mx = max(a,b)
    mn = min(a,b)
    result = []
    # inclusive upper limit. If not needed, delete '+1' in the line below
    while(mn < mx + 1):
        # if step is positive we go from min to max
        if s > 0:
            result.append(mn)
            mn += s
        # if step is negative we go from max to min
        if s < 0:
            result.append(mx)
            mx += s
    return result

For instance, standard command list(range(1,-3)) returns empty list [], while this function will return [-3,-2,-1,0,1]

Updated: now step may be negative. Thanks @Michael for his comment.

Answer 12

@YTZ's answer worked great in my case. I had to generate a list from 0 to 10000 with a step of 0.01 and simply adding 0.01 at each iteration did not work due to rounding issues.

Therefore, I used @YTZ's advice and wrote the following function:

import numpy as np


def generate_floating_numbers_in_range(start: int, end: int, step: float):
    """
    Generate a list of floating numbers within a specified range.

    :param start: range start
    :param end: range end
    :param step: range step
    :return:
    """
    numbers = np.linspace(start, end,(end-start)*int(1/step)+1).tolist()
    return [round(num, 2) for num in numbers]

Answer 13

If you're open to numpy, there are a few functions in it that can generate numbers between 11 and 16:

1. numbers between 11 and 16 with step=0.5: np.arange(11, 16.1, 0.5)
2. n evenly-spaced numbers between 11 and 16: np.linspace(11, 16, n)
3. n random integers between 11 and 16: np.random.randint(11, 17, n)
4. n random numbers between 11 and 16: np.random.rand(n)*(16-11) + 11
5. n numbers between 11 and 16 that are the log of evenly-spaced numbers (results in a concave curve): np.log(np.linspace(np.e**11, np.e**16, n))
6. n numbers between 11 and 16 that are the normalization of evenly-spaced numbers on log scale (results in a convex curve): (np.logspace(11, 16, n, base=np.e) - np.e**11)/(np.e**16 - np.e**11)*(16-11) + 11

The numbers generated by rules (1), (2), (5), (6) are plotted as follows (for n=50). As you can see, similar to the built-in range(), the output size of np.arange() is controlled by its step; however, the output size of np.linspace(), np.logspace() and np.random functions are directly controlled by the third positional argument.