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

## Question:

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

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

## Answers:

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]
```

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)]
```

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]
```

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

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))
```

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]
```

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.

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.

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]

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]
```

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

@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]
```