CODEWITHSUNDEEP
python
MacR6
MacR6

Reputation: 115

take a list of numbers (1 element) and break it into chunks with python?

I would like to take a var containing numbers and create a list of 2 digit numbers.

For instance:

x = 123456

I want to create a list of 2 digit chunks

y = [12,34,56]

I can't seem to figure this one out.

Upvotes: 0

Views: 171

Answers (4)

kostbash
kostbash

Reputation: 206

If x is string:

x = '1234563'

a = [x[i * 2 : (i + 1) * 2] for i in range(len(x) // 2)]

If x is int:

x = 1234563

l = len(str(x))
a = [(x // (10 ** (i - 2))) % 100 for i in range(l, 2, -2)]

Upvotes: 0

Toni V
Toni V

Reputation: 41

def trunk(numbers, chunkSize):
    new_list = []
    nums = str(numbers)
    for x in xrange(0, len(nums), chunkSize):
        new_list.append(int(nums[x:chunkSize+x]))
    return new_list

>>> x = 123456
>>> trunk(x, 2)
[12, 34, 56]
>>> x = 12345
>>> trunk(x, 2)
[12, 34, 5]

Upvotes: 0

ndpu
ndpu

Reputation: 22571

>>> x = 123456
>>> [int(str(x)[i:i+2]) for i in range(0, len((str(x)), 2)]
[12, 34, 56]

Upvotes: 0

Hyperboreus
Hyperboreus

Reputation: 32439

Use modulo and floor division.

def chunks(n):
    if n < 0: raise Exception ("Don't")
    while n:
        yield n % 100
        n //= 100

a = [c for c in chunks (123456)][::-1]
print(a)

Also PS: For input 12345 the output is [1, 23, 45].

And PPS: Is this for FFT multiplication?

Upvotes: 3

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