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

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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

edited 8 hours ago

JeJo

5,91831028

asked 8 hours ago

WilliamKFWilliamKF

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
  
  

  
  
  
  

 | 
show 2 more comments

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

edited 8 hours ago

JeJo

5,91831028

asked 8 hours ago

WilliamKFWilliamKF

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
  
  

  
  
  
  

 | 
show 2 more comments

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

edited 8 hours ago

JeJo

5,91831028

asked 8 hours ago

WilliamKFWilliamKF

15.6k50149248

int count = 0;

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


return 1 == count;

share|improve this question

edited 8 hours ago

JeJo

5,91831028

asked 8 hours ago

WilliamKFWilliamKF

15.6k50149248

share|improve this question

edited 8 hours ago

JeJo

5,91831028

asked 8 hours ago

WilliamKFWilliamKF

15.6k50149248

edited 8 hours ago

JeJo

5,91831028

edited 8 hours ago

JeJo

5,91831028

asked 8 hours ago

WilliamKFWilliamKF

15.6k50149248

asked 8 hours ago

WilliamKFWilliamKF

15.6k50149248

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

edited 7 hours ago

Blastfurnace

14.4k54259

answered 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

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
  
  

  
  
  
  

 | 
show 5 more comments

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

edited 8 hours ago

answered 8 hours ago

JeJoJeJo

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
  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
  

  
  
  
  

add a comment | 

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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

edited 7 hours ago

Blastfurnace

14.4k54259

answered 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

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
  
  

  
  
  
  

 | 
show 5 more comments

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

edited 7 hours ago

Blastfurnace

14.4k54259

answered 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

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
  
  

  
  
  
  

 | 
show 5 more comments

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

edited 7 hours ago

Blastfurnace

14.4k54259

answered 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

21.4k43075

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

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

share|improve this answer

edited 7 hours ago

Blastfurnace

14.4k54259

answered 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

21.4k43075

edited 7 hours ago

Blastfurnace

14.4k54259

edited 7 hours ago

Blastfurnace

14.4k54259

answered 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

21.4k43075

answered 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

21.4k43075

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

edited 8 hours ago

answered 8 hours ago

JeJoJeJo

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
  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
  

  
  
  
  

add a comment | 

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

edited 8 hours ago

answered 8 hours ago

JeJoJeJo

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
  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
  

  
  
  
  

add a comment | 

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

edited 8 hours ago

answered 8 hours ago

JeJoJeJo

5,91831028

#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;

true

share|improve this answer

edited 8 hours ago

answered 8 hours ago

JeJoJeJo

5,91831028

answered 8 hours ago

JeJoJeJo

5,91831028

answered 8 hours ago

JeJoJeJo

5,91831028