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What standard algorithm can determine if exactly one of a container satisfies a predicate?


What is the best algorithm for an overridden System.Object.GetHashCode?What does the C++ standard state the size of int, long type to be?What is the difference between a generative and a discriminative algorithm?What exactly is nullptr?Algorithm to reduce satisfiability javaC++11 introduced a standardized memory model. What does it mean? And how is it going to affect C++ programming?Image Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionWhat is the optimal algorithm for the game 2048?Get iterator from a container uniformly random from iterators that satisfy a predicateis_partitioned behavior when no elements satisfy predicate






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7















I need an STL algorithm that takes a predicate and a collection and returns true if one and only one member of the collection satisfies the predicate, otherwise returns false.



How would I do this using STL algorithms?



E.g., to replace the following with STL algorithm code to express the same return value.



int count = 0;

for( auto itr = c.begin(); itr != c.end(); ++itr ) 
 if ( predicate( *itr ) ) 
 if ( ++count > 1 ) 
 break;
 
 


return 1 == count;





c++ algorithm c++11 counting c++-standard-library







share|improve this question



















  • 3





    count_if handles the algorithm part. You'll still need to check ==1

    – Kenny Ostrom
    8 hours ago











  • Are you looking for std::any_of?

    – Jesper Juhl
    8 hours ago






  • 2





    "How would I do this using STL algorithms?" - How about studying the available algorithms and then pick the one that satisfies your need? And if none does; write your own. Research is a beautiful thing.

    – Jesper Juhl
    8 hours ago






  • 1





    Is the range sorted?

    – Galik
    8 hours ago






  • 4





    @JesperJuhl std::any_of() returns whether AT LEAST 1 element satisfies the predicate. There may be more than 1. std::any_of() does not return whether EXACTLY 1 element satisfies the predicate, which is what the OP wants.

    – Remy Lebeau
    8 hours ago


















7















I need an STL algorithm that takes a predicate and a collection and returns true if one and only one member of the collection satisfies the predicate, otherwise returns false.



How would I do this using STL algorithms?



E.g., to replace the following with STL algorithm code to express the same return value.



int count = 0;

for( auto itr = c.begin(); itr != c.end(); ++itr ) 
 if ( predicate( *itr ) ) 
 if ( ++count > 1 ) 
 break;
 
 


return 1 == count;





c++ algorithm c++11 counting c++-standard-library







share|improve this question



















  • 3





    count_if handles the algorithm part. You'll still need to check ==1

    – Kenny Ostrom
    8 hours ago











  • Are you looking for std::any_of?

    – Jesper Juhl
    8 hours ago






  • 2





    "How would I do this using STL algorithms?" - How about studying the available algorithms and then pick the one that satisfies your need? And if none does; write your own. Research is a beautiful thing.

    – Jesper Juhl
    8 hours ago






  • 1





    Is the range sorted?

    – Galik
    8 hours ago






  • 4





    @JesperJuhl std::any_of() returns whether AT LEAST 1 element satisfies the predicate. There may be more than 1. std::any_of() does not return whether EXACTLY 1 element satisfies the predicate, which is what the OP wants.

    – Remy Lebeau
    8 hours ago














7












7








7


1






I need an STL algorithm that takes a predicate and a collection and returns true if one and only one member of the collection satisfies the predicate, otherwise returns false.



How would I do this using STL algorithms?



E.g., to replace the following with STL algorithm code to express the same return value.



int count = 0;

for( auto itr = c.begin(); itr != c.end(); ++itr ) 
 if ( predicate( *itr ) ) 
 if ( ++count > 1 ) 
 break;
 
 


return 1 == count;





c++ algorithm c++11 counting c++-standard-library







share|improve this question
















I need an STL algorithm that takes a predicate and a collection and returns true if one and only one member of the collection satisfies the predicate, otherwise returns false.



How would I do this using STL algorithms?



E.g., to replace the following with STL algorithm code to express the same return value.



int count = 0;

for( auto itr = c.begin(); itr != c.end(); ++itr ) 
 if ( predicate( *itr ) ) 
 if ( ++count > 1 ) 
 break;
 
 


return 1 == count;





c++ algorithm c++11 counting c++-standard-library




c++ algorithm c++11 counting c++-standard-library






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 8 hours ago









JeJo

5,91831028




5,91831028










asked 9 hours ago









WilliamKFWilliamKF

15.6k50149248




15.6k50149248







  • 3





    count_if handles the algorithm part. You'll still need to check ==1

    – Kenny Ostrom
    8 hours ago











  • Are you looking for std::any_of?

    – Jesper Juhl
    8 hours ago






  • 2





    "How would I do this using STL algorithms?" - How about studying the available algorithms and then pick the one that satisfies your need? And if none does; write your own. Research is a beautiful thing.

    – Jesper Juhl
    8 hours ago






  • 1





    Is the range sorted?

    – Galik
    8 hours ago






  • 4





    @JesperJuhl std::any_of() returns whether AT LEAST 1 element satisfies the predicate. There may be more than 1. std::any_of() does not return whether EXACTLY 1 element satisfies the predicate, which is what the OP wants.

    – Remy Lebeau
    8 hours ago













  • 3





    count_if handles the algorithm part. You'll still need to check ==1

    – Kenny Ostrom
    8 hours ago











  • Are you looking for std::any_of?

    – Jesper Juhl
    8 hours ago






  • 2





    "How would I do this using STL algorithms?" - How about studying the available algorithms and then pick the one that satisfies your need? And if none does; write your own. Research is a beautiful thing.

    – Jesper Juhl
    8 hours ago






  • 1





    Is the range sorted?

    – Galik
    8 hours ago






  • 4





    @JesperJuhl std::any_of() returns whether AT LEAST 1 element satisfies the predicate. There may be more than 1. std::any_of() does not return whether EXACTLY 1 element satisfies the predicate, which is what the OP wants.

    – Remy Lebeau
    8 hours ago








3




3





count_if handles the algorithm part. You'll still need to check ==1

– Kenny Ostrom
8 hours ago





count_if handles the algorithm part. You'll still need to check ==1

– Kenny Ostrom
8 hours ago













Are you looking for std::any_of?

– Jesper Juhl
8 hours ago





Are you looking for std::any_of?

– Jesper Juhl
8 hours ago




2




2





"How would I do this using STL algorithms?" - How about studying the available algorithms and then pick the one that satisfies your need? And if none does; write your own. Research is a beautiful thing.

– Jesper Juhl
8 hours ago





"How would I do this using STL algorithms?" - How about studying the available algorithms and then pick the one that satisfies your need? And if none does; write your own. Research is a beautiful thing.

– Jesper Juhl
8 hours ago




1




1





Is the range sorted?

– Galik
8 hours ago





Is the range sorted?

– Galik
8 hours ago




4




4





@JesperJuhl std::any_of() returns whether AT LEAST 1 element satisfies the predicate. There may be more than 1. std::any_of() does not return whether EXACTLY 1 element satisfies the predicate, which is what the OP wants.

– Remy Lebeau
8 hours ago






@JesperJuhl std::any_of() returns whether AT LEAST 1 element satisfies the predicate. There may be more than 1. std::any_of() does not return whether EXACTLY 1 element satisfies the predicate, which is what the OP wants.

– Remy Lebeau
8 hours ago













2 Answers
2






active

oldest

votes


















17














Two things come to my mind:



std::count_if and then compare the result to 1.



To avoid traversing the whole container in case eg the first two elements already match the predicate I would use two calls looking for matching elements. Something along the line of



auto it = std::find_if(begin,end,predicate);
if (it == end) return false;
++it;
return std::none_of(it,end,predicate);


Or if you prefer it more compact:



auto it = std::find_if(begin,end,predicate); 
return (it != end) && std::none_of(std::next(it),end,predicate);


Credits goes to Remy Lebeau for compacting, Deduplicator for debracketing and Blastfurnance for realizing that we can also use none_of the std algorithms.






share|improve this answer




















  • 3





    I like the short-circuiting of the second option.

    – NathanOliver
    8 hours ago






  • 2





    I would shorten it further to this: auto it = std::find_if(begin,end,predicate); return ((it != end) && (std::find_if(std::next(it),end,predicate) == end));

    – Remy Lebeau
    8 hours ago






  • 1





    @formerlyknownas_463035818 I didn't think of short-circuiting. Nice picks! +1.

    – JeJo
    8 hours ago






  • 1





    I only dislike the outer extra-parentheses of the returned expression.

    – Deduplicator
    8 hours ago






  • 3





    The second std::find_if could be replaced by std::none_of which returns a bool directly without having to compare with end. It also expresses the intent of "no more matches".

    – Blastfurnace
    7 hours ago



















8














You can use std::count_if to count and return if it is one.



For example:



#include <iostream>
#include <algorithm> // std::count_if
#include <vector>

template<typename Type, typename Pred>
bool isOnlyOne(const std::vector<Type>& vec, Pred pred)

 return std::count_if(vec.cbegin(), vec.cend(), pred) == 1;


int main()

 std::vector<int> vec2, 4, 3;
 const auto pred = [](const int ele) return ele & 1; ;
 std::cout << std::boolalpha << isOnlyOne(vec, pred);
 return 0;



output:



true





share|improve this answer

























  • Yes, but that misses the key requirement of not counting past 2.

    – WilliamKF
    8 hours ago






  • 2





    @WilliamKF I didn't get your point!. Could you explain!

    – JeJo
    8 hours ago






  • 1





    I mean once count reaches 2, it is unnecessary to continue computing the predicate.

    – WilliamKF
    8 hours ago











  • @WilliamKF Shamefully agree with that. The other answer shows a better alternative. In case, you are interested to convert your shown algorithm to a better version by packing them into a templated function, here is an example:wandbox.org/permlink/zej1d4L0J7RoS5PH which will short circuit as in your code.

    – JeJo
    7 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









17














Two things come to my mind:



std::count_if and then compare the result to 1.



To avoid traversing the whole container in case eg the first two elements already match the predicate I would use two calls looking for matching elements. Something along the line of



auto it = std::find_if(begin,end,predicate);
if (it == end) return false;
++it;
return std::none_of(it,end,predicate);


Or if you prefer it more compact:



auto it = std::find_if(begin,end,predicate); 
return (it != end) && std::none_of(std::next(it),end,predicate);


Credits goes to Remy Lebeau for compacting, Deduplicator for debracketing and Blastfurnance for realizing that we can also use none_of the std algorithms.






share|improve this answer




















  • 3





    I like the short-circuiting of the second option.

    – NathanOliver
    8 hours ago






  • 2





    I would shorten it further to this: auto it = std::find_if(begin,end,predicate); return ((it != end) && (std::find_if(std::next(it),end,predicate) == end));

    – Remy Lebeau
    8 hours ago






  • 1





    @formerlyknownas_463035818 I didn't think of short-circuiting. Nice picks! +1.

    – JeJo
    8 hours ago






  • 1





    I only dislike the outer extra-parentheses of the returned expression.

    – Deduplicator
    8 hours ago






  • 3





    The second std::find_if could be replaced by std::none_of which returns a bool directly without having to compare with end. It also expresses the intent of "no more matches".

    – Blastfurnace
    7 hours ago
















17














Two things come to my mind:



std::count_if and then compare the result to 1.



To avoid traversing the whole container in case eg the first two elements already match the predicate I would use two calls looking for matching elements. Something along the line of



auto it = std::find_if(begin,end,predicate);
if (it == end) return false;
++it;
return std::none_of(it,end,predicate);


Or if you prefer it more compact:



auto it = std::find_if(begin,end,predicate); 
return (it != end) && std::none_of(std::next(it),end,predicate);


Credits goes to Remy Lebeau for compacting, Deduplicator for debracketing and Blastfurnance for realizing that we can also use none_of the std algorithms.






share|improve this answer




















  • 3





    I like the short-circuiting of the second option.

    – NathanOliver
    8 hours ago






  • 2





    I would shorten it further to this: auto it = std::find_if(begin,end,predicate); return ((it != end) && (std::find_if(std::next(it),end,predicate) == end));

    – Remy Lebeau
    8 hours ago






  • 1





    @formerlyknownas_463035818 I didn't think of short-circuiting. Nice picks! +1.

    – JeJo
    8 hours ago






  • 1





    I only dislike the outer extra-parentheses of the returned expression.

    – Deduplicator
    8 hours ago






  • 3





    The second std::find_if could be replaced by std::none_of which returns a bool directly without having to compare with end. It also expresses the intent of "no more matches".

    – Blastfurnace
    7 hours ago














17












17








17







Two things come to my mind:



std::count_if and then compare the result to 1.



To avoid traversing the whole container in case eg the first two elements already match the predicate I would use two calls looking for matching elements. Something along the line of



auto it = std::find_if(begin,end,predicate);
if (it == end) return false;
++it;
return std::none_of(it,end,predicate);


Or if you prefer it more compact:



auto it = std::find_if(begin,end,predicate); 
return (it != end) && std::none_of(std::next(it),end,predicate);


Credits goes to Remy Lebeau for compacting, Deduplicator for debracketing and Blastfurnance for realizing that we can also use none_of the std algorithms.






share|improve this answer















Two things come to my mind:



std::count_if and then compare the result to 1.



To avoid traversing the whole container in case eg the first two elements already match the predicate I would use two calls looking for matching elements. Something along the line of



auto it = std::find_if(begin,end,predicate);
if (it == end) return false;
++it;
return std::none_of(it,end,predicate);


Or if you prefer it more compact:



auto it = std::find_if(begin,end,predicate); 
return (it != end) && std::none_of(std::next(it),end,predicate);


Credits goes to Remy Lebeau for compacting, Deduplicator for debracketing and Blastfurnance for realizing that we can also use none_of the std algorithms.







share|improve this answer














share|improve this answer



share|improve this answer








edited 7 hours ago









Blastfurnace

14.4k54259




14.4k54259










answered 8 hours ago









formerlyknownas_463035818formerlyknownas_463035818

21.4k43075




21.4k43075







  • 3





    I like the short-circuiting of the second option.

    – NathanOliver
    8 hours ago






  • 2





    I would shorten it further to this: auto it = std::find_if(begin,end,predicate); return ((it != end) && (std::find_if(std::next(it),end,predicate) == end));

    – Remy Lebeau
    8 hours ago






  • 1





    @formerlyknownas_463035818 I didn't think of short-circuiting. Nice picks! +1.

    – JeJo
    8 hours ago






  • 1





    I only dislike the outer extra-parentheses of the returned expression.

    – Deduplicator
    8 hours ago






  • 3





    The second std::find_if could be replaced by std::none_of which returns a bool directly without having to compare with end. It also expresses the intent of "no more matches".

    – Blastfurnace
    7 hours ago













  • 3





    I like the short-circuiting of the second option.

    – NathanOliver
    8 hours ago






  • 2





    I would shorten it further to this: auto it = std::find_if(begin,end,predicate); return ((it != end) && (std::find_if(std::next(it),end,predicate) == end));

    – Remy Lebeau
    8 hours ago






  • 1





    @formerlyknownas_463035818 I didn't think of short-circuiting. Nice picks! +1.

    – JeJo
    8 hours ago






  • 1





    I only dislike the outer extra-parentheses of the returned expression.

    – Deduplicator
    8 hours ago






  • 3





    The second std::find_if could be replaced by std::none_of which returns a bool directly without having to compare with end. It also expresses the intent of "no more matches".

    – Blastfurnace
    7 hours ago








3




3





I like the short-circuiting of the second option.

– NathanOliver
8 hours ago





I like the short-circuiting of the second option.

– NathanOliver
8 hours ago




2




2





I would shorten it further to this: auto it = std::find_if(begin,end,predicate); return ((it != end) && (std::find_if(std::next(it),end,predicate) == end));

– Remy Lebeau
8 hours ago





I would shorten it further to this: auto it = std::find_if(begin,end,predicate); return ((it != end) && (std::find_if(std::next(it),end,predicate) == end));

– Remy Lebeau
8 hours ago




1




1





@formerlyknownas_463035818 I didn't think of short-circuiting. Nice picks! +1.

– JeJo
8 hours ago





@formerlyknownas_463035818 I didn't think of short-circuiting. Nice picks! +1.

– JeJo
8 hours ago




1




1





I only dislike the outer extra-parentheses of the returned expression.

– Deduplicator
8 hours ago





I only dislike the outer extra-parentheses of the returned expression.

– Deduplicator
8 hours ago




3




3





The second std::find_if could be replaced by std::none_of which returns a bool directly without having to compare with end. It also expresses the intent of "no more matches".

– Blastfurnace
7 hours ago






The second std::find_if could be replaced by std::none_of which returns a bool directly without having to compare with end. It also expresses the intent of "no more matches".

– Blastfurnace
7 hours ago














8














You can use std::count_if to count and return if it is one.



For example:



#include <iostream>
#include <algorithm> // std::count_if
#include <vector>

template<typename Type, typename Pred>
bool isOnlyOne(const std::vector<Type>& vec, Pred pred)

 return std::count_if(vec.cbegin(), vec.cend(), pred) == 1;


int main()

 std::vector<int> vec2, 4, 3;
 const auto pred = [](const int ele) return ele & 1; ;
 std::cout << std::boolalpha << isOnlyOne(vec, pred);
 return 0;



output:



true





share|improve this answer

























  • Yes, but that misses the key requirement of not counting past 2.

    – WilliamKF
    8 hours ago






  • 2





    @WilliamKF I didn't get your point!. Could you explain!

    – JeJo
    8 hours ago






  • 1





    I mean once count reaches 2, it is unnecessary to continue computing the predicate.

    – WilliamKF
    8 hours ago











  • @WilliamKF Shamefully agree with that. The other answer shows a better alternative. In case, you are interested to convert your shown algorithm to a better version by packing them into a templated function, here is an example:wandbox.org/permlink/zej1d4L0J7RoS5PH which will short circuit as in your code.

    – JeJo
    7 hours ago















8














You can use std::count_if to count and return if it is one.



For example:



#include <iostream>
#include <algorithm> // std::count_if
#include <vector>

template<typename Type, typename Pred>
bool isOnlyOne(const std::vector<Type>& vec, Pred pred)

 return std::count_if(vec.cbegin(), vec.cend(), pred) == 1;


int main()

 std::vector<int> vec2, 4, 3;
 const auto pred = [](const int ele) return ele & 1; ;
 std::cout << std::boolalpha << isOnlyOne(vec, pred);
 return 0;



output:



true





share|improve this answer

























  • Yes, but that misses the key requirement of not counting past 2.

    – WilliamKF
    8 hours ago






  • 2





    @WilliamKF I didn't get your point!. Could you explain!

    – JeJo
    8 hours ago






  • 1





    I mean once count reaches 2, it is unnecessary to continue computing the predicate.

    – WilliamKF
    8 hours ago











  • @WilliamKF Shamefully agree with that. The other answer shows a better alternative. In case, you are interested to convert your shown algorithm to a better version by packing them into a templated function, here is an example:wandbox.org/permlink/zej1d4L0J7RoS5PH which will short circuit as in your code.

    – JeJo
    7 hours ago













8












8








8







You can use std::count_if to count and return if it is one.



For example:



#include <iostream>
#include <algorithm> // std::count_if
#include <vector>

template<typename Type, typename Pred>
bool isOnlyOne(const std::vector<Type>& vec, Pred pred)

 return std::count_if(vec.cbegin(), vec.cend(), pred) == 1;


int main()

 std::vector<int> vec2, 4, 3;
 const auto pred = [](const int ele) return ele & 1; ;
 std::cout << std::boolalpha << isOnlyOne(vec, pred);
 return 0;



output:



true





share|improve this answer















You can use std::count_if to count and return if it is one.



For example:



#include <iostream>
#include <algorithm> // std::count_if
#include <vector>

template<typename Type, typename Pred>
bool isOnlyOne(const std::vector<Type>& vec, Pred pred)

 return std::count_if(vec.cbegin(), vec.cend(), pred) == 1;


int main()

 std::vector<int> vec2, 4, 3;
 const auto pred = [](const int ele) return ele & 1; ;
 std::cout << std::boolalpha << isOnlyOne(vec, pred);
 return 0;



output:



true






share|improve this answer














share|improve this answer



share|improve this answer








edited 8 hours ago

























answered 8 hours ago









JeJoJeJo

5,91831028




5,91831028












  • Yes, but that misses the key requirement of not counting past 2.

    – WilliamKF
    8 hours ago






  • 2





    @WilliamKF I didn't get your point!. Could you explain!

    – JeJo
    8 hours ago






  • 1





    I mean once count reaches 2, it is unnecessary to continue computing the predicate.

    – WilliamKF
    8 hours ago











  • @WilliamKF Shamefully agree with that. The other answer shows a better alternative. In case, you are interested to convert your shown algorithm to a better version by packing them into a templated function, here is an example:wandbox.org/permlink/zej1d4L0J7RoS5PH which will short circuit as in your code.

    – JeJo
    7 hours ago

















  • Yes, but that misses the key requirement of not counting past 2.

    – WilliamKF
    8 hours ago






  • 2





    @WilliamKF I didn't get your point!. Could you explain!

    – JeJo
    8 hours ago






  • 1





    I mean once count reaches 2, it is unnecessary to continue computing the predicate.

    – WilliamKF
    8 hours ago











  • @WilliamKF Shamefully agree with that. The other answer shows a better alternative. In case, you are interested to convert your shown algorithm to a better version by packing them into a templated function, here is an example:wandbox.org/permlink/zej1d4L0J7RoS5PH which will short circuit as in your code.

    – JeJo
    7 hours ago
















Yes, but that misses the key requirement of not counting past 2.

– WilliamKF
8 hours ago





Yes, but that misses the key requirement of not counting past 2.

– WilliamKF
8 hours ago




2




2





@WilliamKF I didn't get your point!. Could you explain!

– JeJo
8 hours ago





@WilliamKF I didn't get your point!. Could you explain!

– JeJo
8 hours ago




1




1





I mean once count reaches 2, it is unnecessary to continue computing the predicate.

– WilliamKF
8 hours ago





I mean once count reaches 2, it is unnecessary to continue computing the predicate.

– WilliamKF
8 hours ago













@WilliamKF Shamefully agree with that. The other answer shows a better alternative. In case, you are interested to convert your shown algorithm to a better version by packing them into a templated function, here is an example:wandbox.org/permlink/zej1d4L0J7RoS5PH which will short circuit as in your code.

– JeJo
7 hours ago





@WilliamKF Shamefully agree with that. The other answer shows a better alternative. In case, you are interested to convert your shown algorithm to a better version by packing them into a templated function, here is an example:wandbox.org/permlink/zej1d4L0J7RoS5PH which will short circuit as in your code.

– JeJo
7 hours ago

















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