# Python – Create list with numbers between 2 values?

## The Question :

418 people think this question is useful

How would I create a list with values between two values I put in? For example, the following list is generated for values from 11 to 16:

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



740 people think this answer is useful

Use range. In Python 2.x it returns a list so all you need is:

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



In Python 3.x range is a iterator. So, you need to convert it to a list:

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



Note: The second number is exclusive. So, here it needs to be 16+1 = 17

EDIT:

To respond to the question about incrementing by 0.5, the easiest option would probably be to use 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]



26 people think this answer is useful

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]



9 people think this answer is useful

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))



5 people think this answer is useful

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]



4 people think this answer is useful

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)]

4 people think this answer is useful

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

3 people think this answer is useful

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]

1 people think this answer is useful

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.

1 people think this answer is useful

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.

1 people think this answer is useful

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]



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