How can I generate a list of consecutive numbers?
Question:
Say if you had a number input 8 in python and you wanted to generate a list of consecutive numbers up to 8 like
[0, 1, 2, 3, 4, 5, 6, 7, 8]
How could you do this?
Answers:
In Python 3, you can use the builtin range function like this
>>> list(range(9))
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Note 1: Python 3.x’s range function, returns a range object. If you want a list you need to explicitly convert that to a list, with the list function like I have shown in the answer.
Note 2: We pass number 9 to range function because, range function will generate numbers till the given number but not including the number. So, we give the actual number + 1.
Note 3: There is a small difference in functionality of range in Python 2 and 3. You can read more about that in this answer.
Using Python’s built in range function:
Python 2
input = 8
output = range(input + 1)
print output
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Python 3
input = 8
output = list(range(input + 1))
print(output)
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Just to give you another example, although range(value) is by far the best way to do this, this might help you later on something else.
list = []
calc = 0
while int(calc) < 9:
list.append(calc)
calc = int(calc) + 1
print list
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Depending on how you want the result, you can also print each number in a for loop:
def numbers():
for i in range(int(input('How far do you wanna go? '))+1):
print(i)
So if the user input was 7 for example:
How far do you wanna go? 7
0
1
2
3
4
5
6
7
You can also delete the ‘+1’ in the for loop and place it on the print statement, which will change it to starting at 1 instead of 0.
Note :- Certainly in python-3x you need to use Range function It works to generate numbers on demand, standard method to use Range function to make a list of consecutive numbers is
x=list(range(10))
#"list"_will_make_all_numbers_generated_by_range_in_a_list
#number_in_range_(10)_is_an_option_you_can_change_as_you_want
print (x)
#Output_is_ [0,1,2,3,4,5,6,7,8,9]
Also if you want to make an function to generate a list of consecutive numbers by using Range function watch this code !
def consecutive_numbers(n) :
list=[i for i in range(n)]
return (list)
print(consecutive_numbers(10))
Good Luck!
You can use itertools.count() to generate unbounded sequences. (itertools is in the Python standard library). Docs here:
https://docs.python.org/3/library/itertools.html#itertools.count
You can use list comprehensions for this problem as it will solve it in only two lines.
n = int(input("Enter the range of the list:n"))
l1 = [i for i in range(n)] #Creates list of numbers in the range 0 to n
print(l1)
Here is a way to generate n consecutive numbers in equal intervals between them starting from 0 to 100 using numpy:
import numpy as np
myList = np.linspace(0, 100, n)
Say if you had a number input 8 in python and you wanted to generate a list of consecutive numbers up to 8 like
[0, 1, 2, 3, 4, 5, 6, 7, 8]
How could you do this?
In Python 3, you can use the builtin range function like this
>>> list(range(9))
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Note 1: Python 3.x’s range function, returns a range object. If you want a list you need to explicitly convert that to a list, with the list function like I have shown in the answer.
Note 2: We pass number 9 to range function because, range function will generate numbers till the given number but not including the number. So, we give the actual number + 1.
Note 3: There is a small difference in functionality of range in Python 2 and 3. You can read more about that in this answer.
Using Python’s built in range function:
Python 2
input = 8
output = range(input + 1)
print output
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Python 3
input = 8
output = list(range(input + 1))
print(output)
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Just to give you another example, although range(value) is by far the best way to do this, this might help you later on something else.
list = []
calc = 0
while int(calc) < 9:
list.append(calc)
calc = int(calc) + 1
print list
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Depending on how you want the result, you can also print each number in a for loop:
def numbers():
for i in range(int(input('How far do you wanna go? '))+1):
print(i)
So if the user input was 7 for example:
How far do you wanna go? 7
0
1
2
3
4
5
6
7
You can also delete the ‘+1’ in the for loop and place it on the print statement, which will change it to starting at 1 instead of 0.
Note :- Certainly in python-3x you need to use Range function It works to generate numbers on demand, standard method to use Range function to make a list of consecutive numbers is
x=list(range(10))
#"list"_will_make_all_numbers_generated_by_range_in_a_list
#number_in_range_(10)_is_an_option_you_can_change_as_you_want
print (x)
#Output_is_ [0,1,2,3,4,5,6,7,8,9]
Also if you want to make an function to generate a list of consecutive numbers by using Range function watch this code !
def consecutive_numbers(n) :
list=[i for i in range(n)]
return (list)
print(consecutive_numbers(10))
Good Luck!
You can use itertools.count() to generate unbounded sequences. (itertools is in the Python standard library). Docs here:
https://docs.python.org/3/library/itertools.html#itertools.count
You can use list comprehensions for this problem as it will solve it in only two lines.
n = int(input("Enter the range of the list:n"))
l1 = [i for i in range(n)] #Creates list of numbers in the range 0 to n
print(l1)
Here is a way to generate n consecutive numbers in equal intervals between them starting from 0 to 100 using numpy:
import numpy as np
myList = np.linspace(0, 100, n)