Right split a string into groups of 3

Question

What is the most Pythonic way to right split into groups of threes? I've seen this answer https://stackoverflow.com/a/2801117/1461607 but I need it to be right aligned. Preferably a simple efficient one-liner without imports.

• '123456789' = ['123','456','789']
• '12345678' = ['12','345','678']
• '1234567' = ['1','234','567']

Answer 1

Another way, not sure about efficiency (it'd be better if they were already numbers instead of strings), but is another way of doing it in 2.7+.

for i in map(int, ['123456789', '12345678', '1234567']):
    print i, '->', format(i, ',').split(',')

#123456789 -> ['123', '456', '789']
#12345678 -> ['12', '345', '678']
#1234567 -> ['1', '234', '567']

Answer 2

simple (iterated from the answer in your link):

[int(a[::-1][i:i+3][::-1]) for i in range(0, len(a), 3)][::-1]

Explanation : a[::-1] is the reverse list of a We will compose the inversion with the slicing.

Step one : reverse the list

 a           =   a[::-1]
'123456789' - > '987654321'

Step Two : Slice in parts of three's

 a[i]           =   a[i:i+3]
 '987654321'    ->  '987','654','321'

Step Three : invert the list again to present the digits in increasing order

 a[i]           =  int(a[i][::-1])
 '987','654','321' -> 789, 654, 123

Final Step : invert the whole list

 a           =   a[::-1]
 789, 456, 123 -> 123, 456, 789

Bonus : Functional synthetic sugar

It's easier to debug when you have proper names for functions

invert = lambda a: a[::-1]
slice  = lambda array, step : [ int( invert( array[i:i+step]) ) for i in range(len(array),step)  ]

answer = lambda x: invert ( slice ( invert (x) , 3 ) )
answer('123456789')
#>> [123,456,789]