How can I check that a list has one and only one truthy value?

Question

In python, I have a list that should have one and only one truthy value (that is, bool(value) is True). Is there a clever way to check for this

Answer 1

@JonClements` solution extended for at most N True values:

# Extend any() to n true values
def _NTrue(i, n=1):
    for x in xrange(n):
        if any(i): # False for empty
            continue
        else:
            return False
    return True

def NTrue(iterable, n=1):
    i = iter(iterable)
    return any(i) and not _NTrue(i, n)

edit: better version

def test(iterable, n=1): 
    i = iter(iterable) 
    return sum(any(i) for x in xrange(n+1)) <= n 

edit2: include at least m True's and at most n True's

def test(iterable, n=1, m=1): 
    i = iter(iterable) 
    return  m <= sum(any(i) for x in xrange(n+1)) <= n

Answer 2

It depends if you are just looking for the value True or are also looking for other values that would evaluate to True logically (like 11 or "hello"). If the former:

def only1(l):
    return l.count(True) == 1

If the latter:

def only1(l):
    return sum(bool(e) for e in l) == 1

since this would do both the counting and the conversion in a single iteration without having to build a new list.

Answer 3

You can do:

x = [bool(i) for i in x]
return x.count(True) == 1

Or

x = map(bool, x)
return x.count(True) == 1

Building on @JoranBeasley's method:

sum(map(bool, x)) == 1

Answer 4

For completeness' sake and to demonstrate advanced use of Python's control flow for for loop iteration, one can avoid the extra accounting in the accepted answer, making this slightly faster.:

def one_bool_true(iterable):
    it = iter(iterable)
    for i in it:
        if i:
            break
    else:            #no break, didn't find a true element
        return False
    for i in it:     # continue consuming iterator where left off
        if i: 
            return False
    return True      # didn't find a second true.

The above's simple control flow makes use of Python's sophisticated feature of loops: the else. The semantics are that if you finish iterating over the iterator that you are consuming without break-ing out of it, you then enter the else block.

Here's the accepted answer, which uses a bit more accounting.

def only1(l):
    true_found = False
    for v in l:
        if v:
            # a True was found!
            if true_found:
                # found too many True's
                return False 
            else:
                # found the first True
                true_found = True
    # found zero or one True value
    return true_found

to time these:

import timeit
>>> min(timeit.repeat(lambda: one_bool_true([0]*100 + [1, 1])))
13.992251592921093
>>> min(timeit.repeat(lambda: one_bool_true([1, 1] + [0]*100)))
2.208037032979064
>>> min(timeit.repeat(lambda: only1([0]*100 + [1, 1])))
14.213872335107908
>>> min(timeit.repeat(lambda: only1([1, 1] + [0]*100)))
2.2482982632641324
>>> 2.2482/2.2080
1.0182065217391305
>>> 14.2138/13.9922
1.0158373951201385

So we see that the accepted answer takes a bit longer (slightly more than one and a half of a percent).

Naturally, using the built-in any, written in C, is much faster (see Jon Clement's answer for the implementation - this is the short form):

>>> min(timeit.repeat(lambda: single_true([0]*100 + [1, 1])))
2.7257133318785236
>>> min(timeit.repeat(lambda: single_true([1, 1] + [0]*100)))
2.012824866380015

Answer 5

If there is only one True, then the length of the Trues should be one:

def only_1(l): return 1 == len(filter(None, l))