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.
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']
format(num, ',') formats a string into conventional , delimited blocks of 3, so that splitting on , gives you the required result.format(int('1234567'), ',').split(',') if you wanted...
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.
a = a[::-1]
'123456789' - > '987654321'
a[i] = a[i:i+3]
'987654321' -> '987','654','321'
a[i] = int(a[i][::-1])
'987','654','321' -> 789, 654, 123
a = a[::-1]
789, 456, 123 -> 123, 456, 789
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]
ValueError: invalid literal for int() with base 10: '' for a = '123456789'.This is the best I came up with:
[a[max(i-3,0):i] for i in range(len(a), 0, -3)][::-1]
and another one, which works without inverting the list, but is slightly more ugly:
[a[max(0,i):i+3] for i in range((len(a)-1)%3-2, len(a), 3)]
The shortest isn't always the most Pythonic.
def by3(s):
out = []
while len(s):
out.insert(0, s[-3:])
s = s[:-3]
return out
>>> by3(a)
['12', '345', '678']
A four liner:
splitted_number = []
while number:
number, r = number[:-3], number[-3:]
splitted_number.insert(0, r)
My 2cents if someone find this usefull: textwrap is built-in module that do the thing.
from textwrap import wrap
a = '123456789'
b = wrap(a, 3)
>>> b ['123', '456', '789']
Expanding @Manu answer,
from textwrap import wrap
a = '123456789'
b = wrap(a, 3)
print(b)
newlist = []
for i in b:
newlist.append(int(i))
print(newlist)
['123', '456', '789']
[123, 456, 789]
[Program finished]
or
n = 1234567890
v = []
while n > 0:
n, r = divmod(n, 1000)
v.insert(0, r)
print(v)
[1, 234, 567, 890]
[Program finished]
