What's the fastest way to convert a list of booleans into a binary string in python?
e.g. boolList2BinString([True, True, False]) = '0b110'.
Also, how would I convert that binary string into the binary literal? Would this take more time than just converting from the boolean list to the binary literal immediatley? How would one do this?
e.g. boolList2Bin([True, True, False]) = 0b110.
Thanks!
Regarding your first question, you can use a list comprehension* and a conditional expression:
>>> def boolList2BinString(lst):
... return '0b' + ''.join(['1' if x else '0' for x in lst])
...
>>> boolList2BinString([True, True, False])
'0b110'
>>>
Regarding your second, you cannot "convert that binary string into the binary literal". As their name suggests, literals must be literally typed out:
>>> x = 0b110
>>>
Perhaps you meant that you want the quotes removed from the output? If so, use print:
>>> def boolList2BinString(lst):
... return '0b' + ''.join(['1' if x else '0' for x in lst])
...
>>> boolList2BinString([True, True, False])
'0b110'
>>> print(boolList2BinString([True, True, False]))
0b110
>>>
*Note: I purposefully chose to use a list comprehension with str.join instead of a generator expression because the former is generally faster.
join can handle generators, so you do not need to create a list first.if/else you can use '01'[x] because bools can be used as indexes.
str.join. But the link I gave in my answer shows that it is generally faster to use one. Also, if you use timeit.timeit, you will see that the conditional expression is slightly faster than '01'[x].Convert the list to a decent binary (will be a long int):
number = reduce(lambda a, b: (a<<1) + int(b), [ True, True, False ])
And then, if you really need a "binary string", as you put it, use
bin(number)
to generate that string.
EDIT
You also can use this code:
number = sum(int(bit) << position
for (position, bit) in
enumerate(reversed([True, True, False])))
The mechanism behind it is the same as before.
data = [True, True, False]
print bin(int("".join(str(int(item)) for item in data), 2))
# 0b110
values = [True,False,False,True]
bin(sum(int(v)*2**i for i,v in enumerate(values[::-1]) ))
actually
In [7]: %timeit bin(sum(int(v)*2**i for i,v in enumerate(values[::-1]) ))
10000 loops, best of 3: 108 us per loop
In [8]: %timeit '0b' + ''.join(['1' if x else '0' for x in values])
100000 loops, best of 3: 5.25 us per loop
In [9]: %timeit bin(int("".join(str(int(item)) for item in values), 2))
10000 loops, best of 3: 29.5 us per loop
In [10]: %timeit bin(reduce(lambda a, b: (a<<1) + int(b), values))
10000 loops, best of 3: 31.3 us per loop
my solution is the slowest :( ...
L = [True, False, True]
''.join(map(str, map(int, L))) # '101'.
Py_True-- is probably fastest. /s Seriously though, how serious is the "fastest" requirement? Did you add that just because or have you tried something and it was too slow (if so, please add details!).