CODEWITHSUNDEEP
pythonpython-3.xlist
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Reputation: 1006

How to create a list of numbers following the natural number sequence according to their length?

I want the list of numbers as:

n = int(input("Enter count")

if n=1 , lst =["1"]
if n=2, lst=["1","2"]
if n=3, lst =["1","2","3"]

The list should contain the elements as strings, not integers.

Any idea how to do this?

Upvotes: 1

Views: 775

Answers (4)

XMehdi01
XMehdi01

Reputation: 1

You could use for-loop, and each iteration append i+1 to a list, Since for-loop start from index 0:

n = int(input("n:")) #5
l = []
for i in range(n):
    l.append(str(i+1))
print(l) #['1', '2', '3', '4', '5']

With list comprehension:

print([str(i+1) for i in range(n)])

Upvotes: 0

Akar Periwal
Akar Periwal

Reputation: 1

You can use map and range..

n = int(input('Enter count:')
lst = list(map(str, list(range(1,n+1))))

Upvotes: 0

MD Mushfirat Mohaimin
MD Mushfirat Mohaimin

Reputation: 2066

This code is efficient

n = int(input("Enter count:"))
lst = [str(x) for x in range(1,n+1)]

Upvotes: 0

Andreas
Andreas

Reputation: 9197

You can use range to create the numbers and since you want strings, you can afterwards map them to strings:

n = int(input("Enter count")
list(map(str, range(1, n+1)))

Upvotes: 1

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