repeat a sequence of numbers N times and incrementally increase the values in the sequence Python

Question

I have a number sequence as below

sequence = (0,0,1,1,1,1)

I want the number sequence to repeat a specified number of times but incrementally increase the values within the sequence

so if n= 3 then the sequence would go 1,1,2,2,2,2,3,3,4,4,4,4,5,5,6,6,6,6

I can make a sequence like below but the range function is not quite right in this instance

n = 3
CompleteSequence =  [j + k for j in range(0, n, 2) for k in sequence]
CompleteSequence 
[0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3]

I have tried itertools

import itertools
sequence = (0,0,1,1,1,1)
n=3
list1 = list(itertools.repeat(sequence,n))
list1

[(0, 0, 1, 1, 1, 1), (0, 0, 1, 1, 1, 1), (0, 0, 1, 1, 1, 1)]

How can I go I achieve the pattern I need? a combination of range and itertools?

Asked By: Spooked

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Source

Answer 1

Use range(start, stop, step) – i.e. with step=2 to only iterate over the odd values:

>>> [[i] * 2 + [i + 1] * 4 for i in range(1, 6, 2)]
[[1, 1, 2, 2, 2, 2], [3, 3, 4, 4, 4, 4], [5, 5, 6, 6, 6, 6]]

Full example:

# they are the same
sequence = (0, 0, 1, 1, 1, 1)
assert [0] * 2 + [1] * 4 == list(sequence)

n = 3

result = [[i] * 2 + [i + 1] * 4 for i in range(1, n * 2, 2)]
print(result)

# in the case you need to unpack values into one list
result = []
for i in range(1, n * 2, 2):
    result += [i] * 2 + [i + 1] * 4
print(result)

Result:

[[1, 1, 2, 2, 2, 2], [3, 3, 4, 4, 4, 4], [5, 5, 6, 6, 6, 6]]
[1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6]

Answered By: rv.kvetch

Answer 2

You can achieve this pattern with `range()` and an `increment` variable:

sequence = (0,0,1,1,1,1)
n = 3

CompleteSequence = []
increment = 1
for i in range(1, n+1):
    for j in sequence:
        CompleteSequence.append(j + increment)
    increment += 2

CompleteSequence now holds:

[1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6]

Answered By: Marcelo Paco

Answer 3

s = [0,0,1,1,1,1]
n = 3
r = [
  i + offset
    for offset in range(1, n*2+1, 2)
    for i in s
]

Demo : https://trinket.io/python3/3b1d5c0261

Answered By: MatBailie

Answer 4

You can also use:

from itertools import repeat

sequence = (0, 0, 1, 1, 1, 1)
n = 3
list1 = [i*2+j+1 for i, s in enumerate(repeat(sequence, n)) for j in s]
print(list1)

# Output
[1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6]

Answered By: Corralien

Answer 5

Your "incrementally increase" and "the incrementing pattern is what I am mainly trying to achieve" sound like:

from itertools import accumulate

incrementing_pattern = [1, 0, 1, 0, 0, 0]
n = 3

sequence = accumulate(n * incrementing_pattern)

print(*sequence)

Output (Attempt This Online!):

1 1 2 2 2 2 3 3 4 4 4 4 5 5 6 6 6 6

Answered By: Kelly Bundy

repeat a sequence of numbers N times and incrementally increase the values in the sequence Python

Question:

Answers:

You can achieve this pattern with `range()` and an `increment` variable:

repeat a sequence of numbers N times and incrementally increase the values in the sequence Python

Question:

Answers:

You can achieve this pattern with range() and an increment variable:

You can achieve this pattern with `range()` and an `increment` variable: