How to print in Python 3.10 a series of lists [1], [1, 2], [1, 2, 3], [1, 2, 3, ..., n] in one liner-CodePudding

I have a list [1, 2, 3, 4, ..., n]. In a one-liner list comprehension, how can I print the progression [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], ..., [1, 2, 3, 4, ..., n-1], where n represents an arbitrary natural number?

CodePudding user response：

Now if you want this principle to operate on any starting list (not necessarily [1,2,...]), you can do this:

rawlist=['a','b','c','d']

newlist=[rawlist[:i] for i in range(1,len(rawlist) 1)]

CodePudding user response：

Try this:

n = 10
res = [list(range(1,i)) for i in range(2,n 1)]
print(res)

Output:

[
 [1], # range(1, 2) 
 [1, 2], # range(1, 3)
 [1, 2, 3],
 [1, 2, 3, 4],
 [1, 2, 3, 4, 5],
 [1, 2, 3, 4, 5, 6],
 [1, 2, 3, 4, 5, 6, 7],
 [1, 2, 3, 4, 5, 6, 7, 8],
 [1, 2, 3, 4, 5, 6, 7, 8, 9] # range(1, 10)
]

CodePudding user response：

How about a for loop in a for loop?

n = 10

print([[x for x in range(1,i 1)] for i in range(1,n 1)])

This prints an array of the numbers 1 to i for each value of i in the range 1 to n.

CodePudding user response：

This should give you every list from [1] to [1,...n-1]

[print([x for x in range(1, y 1)]) for y in range(1, n)]

CodePudding user response：

Starting with our [1, ..., n] list, where n in this case is 5:

>>> a_list = list(range(1, 6))
>>> a_list
[1, 2, 3, 4, 5]

we can just iterate over the list and create another list for each number, with the last one ending in n-1:

>>> print(*[list(range(1, x)) for x in a_list[1:]], sep=", ")
[1], [1, 2], [1, 2, 3], [1, 2, 3, 4]

Or, doing it based on n directly rather than a_list:

>>> n = 5
>>> print(*[list(range(1, x)) for x in range(2, n 1)], sep=", ")
[1], [1, 2], [1, 2, 3], [1, 2, 3, 4]

Note that using a list comprehension is not necessary; you can replace the outer [] in both examples with () (i.e. a parenthesized generator expression rather than a list comprehension) and it will print exactly the same thing.