Another way of generating a list of special elements in python

Question

listA = ['000','001','010','011', '100','101','111','110']

To get the above list I usually use the following code. My aim is to get all the possible combinations of 3 numbers where each number can either be 0 or 1.

listA = list(range(0,222))
listA = [str(x).zfill(3) for x in listA]

 listA = [x for x in listA if not x[-3] in [str(z) for z in range(2,10,1)]]
 listA = [x for x in listA if not x[-2] in [str(z) for z in range(2,10,1)]]
 listA = [x for x in listA if not x[-1] in [str(z) for z in range(2,10,1)]]

This only works if list(range(0,222)) is relatively small, for large range like (0,222222222222), listA = list(range(0,222222222222)) returns Memory error. Is there another way to do this?

This is the Cartesian product of a list (`['0', '1']`) with itself (three times in total), with each result joined up. Please see the linked duplicate. — Karl Knechtel, Mar 08 '23 at 17:26

score 0 · Accepted Answer · answered Apr 17 '22 at 12:59

0

You could try using itertools.product:

>>> from itertools import product
>>> a = [''.join(p) for p in product('01', repeat=3)]
>>> a
['000', '001', '010', '011', '100', '101', '110', '111']

answered Apr 17 '22 at 12:59

Sash Sinha

18,743
3
23
40

dawg · Answer 2 · 2022-04-17T14:05:03.180

Isn't this just binary strings from 0 to some end number?

You could do:

def zn1(up):
    w=len(f'{up:b}')
    return [f'{x:0{w}b}' for x in range(up+1)]

Or, if memory is an issue, use a generator:

def zn1(up):
    w=len(f'{up:b}')
    for n in range(up+1):
        yield f'{x:0{w}b}'

The first for example:

>>> zn1(7)
['000', '001', '010', '011', '100', '101', '110', '111']

>>> zn1(2000)
['00000000000', '00000000001', '00000000010', '00000000011', '00000000100', 
...
'11111001100', '11111001101', '11111001110', '11111001111', '11111010000']

Another way of generating a list of special elements in python

2 Answers2