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

Each Answer to this Q is separated by one/two green lines.

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?

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

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.

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.

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 : from itertools import islice, ifilter

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

In : def david(l): return sum(bool(e) for e in l) == 1
``````
• Exhibiting short-circuit behavior:

``````In : l = range(1000000)

In : %timeit fj(l)
1000000 loops, best of 3: 1.77 us per loop

In : %timeit david(l)
1 loops, best of 3: 194 ms per loop
``````
• Large list where short-circuiting does not occur:

``````In : l =  * 1000000

In : %timeit fj(l)
100 loops, best of 3: 10.2 ms per loop

In : %timeit david(l)
1 loops, best of 3: 189 ms per loop
``````
• Small list:

``````In : l = 

In : %timeit fj(l)
1000000 loops, best of 3: 1.77 us per loop

In : %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.

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)
``````

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

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
``````

``````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.

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

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

This seems to work and should be able to handle any iterable, not just`list`s. 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) == testcase
print('only1(testcase[{}]): {}{}'.format(i, only1(testcase),
'' if correct else
', error given '+str(testcase)))
``````

Output:

``````only1(testcase): False
only1(testcase): False
only1(testcase): True
only1(testcase): True
only1(testcase): True
only1(testcase): False
only1(testcase): False
``````

@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
``````

``````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.

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(*100 + [1, 1])))
13.992251592921093
>>> min(timeit.repeat(lambda: one_bool_true([1, 1] + *100)))
2.208037032979064
>>> min(timeit.repeat(lambda: only1(*100 + [1, 1])))
14.213872335107908
>>> min(timeit.repeat(lambda: only1([1, 1] + *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(*100 + [1, 1])))
2.7257133318785236
>>> min(timeit.repeat(lambda: single_true([1, 1] + *100)))
2.012824866380015
``````

Is this what you’re looking for?

``````sum(l) == 1
``````

``````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

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...")
``````

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