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? Right now, I am just iterating across the list and manually checking:

def only1(l)
    true_found = False
    for v in l:
        if v and not true_found:
            true_found=True
        elif v and true_found:
             return False #"Too Many Trues"
    return true_found

This seems inelegant and not very pythonic. Is there a cleverer way to do this?

Answer 1

John Clements answer is really elegant but the first thing I would think would be:

if sum(map(bool, iterable)) == 1: ...

This takes advantage of the fact that True == 1 and False == 0 in python and will work for any type of values in the iterable.

Answer 2

One that doesn't require imports:

def single_true(iterable):
    i = iter(iterable)
    return any(i) and not any(i)

Alternatively, perhaps a more readable version:

def single_true(iterable):
    iterator = iter(iterable)

    # consume from "i" until first true or it's exhausted
    has_true = any(iterator) 

    # carry on consuming until another true value / exhausted
    has_another_true = any(iterator) 

    # True if exactly one true found
    return has_true and not has_another_true

This:

• Looks to make sure i has any true value
• Keeps looking from that point in the iterable to make sure there is no other true value

Answer 3

What about:

len([v for v in l if type(v) == bool and v])

If you only want to count boolean True values.

Answer 4

Here's something that ought to work for anything truthy, though it has no short-circuit. I found it while looking for a clean way to forbid mutually-exclusive arguments:

if sum(1 for item in somelist if item) != 1:
    raise ValueError("or whatever...")

Answer 5

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 6

I wanted to earn the necromancer badge, so I generalized the Jon Clements' excellent answer, preserving the benefits of short-circuiting logic and fast predicate checking with any and all.

Thus here is:

N(trues) = n

def n_trues(iterable, n=1):
    i = iter(iterable)
    return all(any(i) for j in range(n)) and not any(i)

N(trues) <= n:

def up_to_n_trues(iterable, n=1):
    i = iter(iterable)
    all(any(i) for j in range(n))
    return not any(i)

N(trues) >= n:

def at_least_n_trues(iterable, n=1):
    i = iter(iterable)
    return all(any(i) for j in range(n))

m <= N(trues) <= n

def m_to_n_trues(iterable, m=1, n=1):
    i = iter(iterable)
    assert m <= n
    return at_least_n_trues(i, m) and up_to_n_trues(i, n - m)

Answer 7

The most verbose solution is not always the most unelegant solution. Therefore I add just a minor modification (in order to save some redundant boolean evaluations):

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

---

Here are some timings for comparison:

# file: test.py
from itertools import ifilter, islice

def OP(l):
    true_found = False
    for v in l:
        if v and not true_found:
            true_found=True
        elif v and true_found:
             return False #"Too Many Trues"
    return true_found

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

def FJ(l):
    return len(list(islice(ifilter(None, l), 2))) == 1

def JonClements(iterable):
    i = iter(iterable)
    return any(i) and not any(i)

def moooeeeep(l):
    true_found = False
    for v in l:
        if v:
            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

My output:

$ python -mtimeit -s 'import test; l=[True]*100000' 'test.OP(l)' 
1000000 loops, best of 3: 0.523 usec per loop
$ python -mtimeit -s 'import test; l=[True]*100000' 'test.DavidRobinson(l)' 
1000 loops, best of 3: 516 usec per loop
$ python -mtimeit -s 'import test; l=[True]*100000' 'test.FJ(l)' 
100000 loops, best of 3: 2.31 usec per loop
$ python -mtimeit -s 'import test; l=[True]*100000' 'test.JonClements(l)' 
1000000 loops, best of 3: 0.446 usec per loop
$ python -mtimeit -s 'import test; l=[True]*100000' 'test.moooeeeep(l)' 
1000000 loops, best of 3: 0.449 usec per loop

As can be seen, the OP solution is significantly better than most other solutions posted here. As expected, the best ones are those with short circuit behavior, especially that solution posted by Jon Clements. At least for the case of two early True values in a long list.

Here the same for no True value at all:

$ python -mtimeit -s 'import test; l=[False]*100000' 'test.OP(l)' 
100 loops, best of 3: 4.26 msec per loop
$ python -mtimeit -s 'import test; l=[False]*100000' 'test.DavidRobinson(l)' 
100 loops, best of 3: 2.09 msec per loop
$ python -mtimeit -s 'import test; l=[False]*100000' 'test.FJ(l)' 
1000 loops, best of 3: 725 usec per loop
$ python -mtimeit -s 'import test; l=[False]*100000' 'test.JonClements(l)' 
1000 loops, best of 3: 617 usec per loop
$ python -mtimeit -s 'import test; l=[False]*100000' 'test.moooeeeep(l)' 
100 loops, best of 3: 1.85 msec per loop

I did not check the statistical significance, but interestingly, this time the approaches suggested by F.J. and especially that one by Jon Clements again appear to be clearly superior.

Answer 8

def only1(l)
    sum(map(lambda x: 1 if x else 0, l)) == 1

Explanation: The map function maps a list to another list, doing True => 1 and False => 0. We now have a list of 0s and 1s instead of True or False. Now we simply sum this list and if it is 1, there was only one True value.

Answer 9

This seems to work and should be able to handle any iterable, not justlists. It short-circuits whenever possible to maximize efficiency. Works in both Python 2 and 3.

def only1(iterable):
    for i, x in enumerate(iterable):  # check each item in iterable
        if x: break                   # truthy value found
    else:
        return False                  # no truthy value found
    for x in iterable[i+1:]:          # one was found, see if there are any more
        if x: return False            #   found another...
    return True                       # only a single truthy value found

testcases = [  # [[iterable, expected result], ... ]
    [[                          ], False],
    [[False, False, False, False], False],
    [[True,  False, False, False], True],
    [[False, True,  False, False], True],
    [[False, False, False, True],  True],
    [[True,  False, True,  False], False],
    [[True,  True,  True,  True],  False],
]

for i, testcase in enumerate(testcases):
    correct = only1(testcase[0]) == testcase[1]
    print('only1(testcase[{}]): {}{}'.format(i, only1(testcase[0]),
                                             '' if correct else
                                             ', error given '+str(testcase[0])))

Output:

only1(testcase[0]): False
only1(testcase[1]): False
only1(testcase[2]): True
only1(testcase[3]): True
only1(testcase[4]): True
only1(testcase[5]): False
only1(testcase[6]): False

Answer 10

import collections

def only_n(l, testval=True, n=1):
    counts = collections.Counter(l)
    return counts[testval] == n

Linear time. Uses the built-in Counter class, which is what you should be using to check counts.

Re-reading your question, it looks like you actually want to check that there is only one truthy value, rather than one True value. Try this:

import collections

def only_n(l, testval=True, coerce=bool, n=1):
    counts = collections.Counter((coerce(x) for x in l))
    return counts[testval] == n

While you can get better best case performance, nothing has better worst-case performance. This is also short and easy to read.

Here's a version optimised for best-case performance:

import collections
import itertools

def only_n(l, testval=True, coerce=bool, n=1):
    counts = collections.Counter()
    def iterate_and_count():
        for x in itertools.imap(coerce,l):
            yield x
            if x == testval and counts[testval] > n:
               break
    counts.update(iterate_and_count())
    return counts[testval] == n

The worst case performance has a high k (as in O(kn+c)), but it is completely general.

Here's an ideone to experiment with performance: http://ideone.com/ZRrv2m

Answer 11

@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 12

Is this what you're looking for?

sum(l) == 1

Answer 13

if sum([bool(x) for x in list]) == 1

(Assuming all your values are booleanish.)

This would probably be faster just summing it

sum(list) == 1   

although it may cause some problems depending on the data types in your list.

Answer 14

A one-line answer that retains the short-circuiting behavior:

from itertools import ifilter, islice

def only1(l):
    return len(list(islice(ifilter(None, l), 2))) == 1

This will be significantly faster than the other alternatives here for very large iterables that have two or more true values relatively early.

ifilter(None, itr) gives an iterable that will only yield truthy elements (x is truthy if bool(x) returns True). islice(itr, 2) gives an iterable that will only yield the first two elements of itr. By converting this to a list and checking that the length is equal to one we can verify that exactly one truthy element exists without needing to check any additional elements after we have found two.

Here are some timing comparisons:

• Setup code:

  In [1]: from itertools import islice, ifilter
  
  In [2]: def fj(l): return len(list(islice(ifilter(None, l), 2))) == 1
  
  In [3]: def david(l): return sum(bool(e) for e in l) == 1
  

• Exhibiting short-circuit behavior:

  In [4]: l = range(1000000)
  
  In [5]: %timeit fj(l)
  1000000 loops, best of 3: 1.77 us per loop
  
  In [6]: %timeit david(l)
  1 loops, best of 3: 194 ms per loop
  

• Large list where short-circuiting does not occur:

  In [7]: l = [0] * 1000000
  
  In [8]: %timeit fj(l)
  100 loops, best of 3: 10.2 ms per loop
  
  In [9]: %timeit david(l)
  1 loops, best of 3: 189 ms per loop
  

• Small list:

  In [10]: l = [0]
  
  In [11]: %timeit fj(l)
  1000000 loops, best of 3: 1.77 us per loop
  
  In [12]: %timeit david(l)
  1000000 loops, best of 3: 990 ns per loop
  

So the sum() approach is faster for very small lists, but as the input list gets larger my version is faster even when short-circuiting is not possible. When short-circuiting is possible on a large input, the performance difference is clear.

Answer 15

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))

Answer 16

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 17

>>> l = [0, 0, 1, 0, 0]
>>> has_one_true = len([ d for d in l if d ]) == 1
>>> has_one_true
True

Answer 18

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