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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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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
edited 8 hours ago
JeJo
5,91831028
asked 8 hours ago
WilliamKFWilliamKF
15.6k50149248
|
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
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
@JesperJuhlstd::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
edited 8 hours ago
JeJo
5,91831028
asked 8 hours ago
WilliamKFWilliamKF
15.6k50149248
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
edited 8 hours ago
JeJo
5,91831028
asked 8 hours ago
WilliamKFWilliamKF
15.6k50149248
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
edited 8 hours ago
JeJo
5,91831028
5,91831028
asked 8 hours ago
WilliamKFWilliamKF
15.6k50149248
asked 8 hours ago
WilliamKFWilliamKF
15.6k50149248
asked 8 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
@JesperJuhlstd::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
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
@JesperJuhlstd::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
|
show 2 more comments
2 Answers
2
active
oldest
votes
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.
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 secondstd::find_ifcould be replaced bystd::none_ofwhich returns abooldirectly without having to compare withend. It also expresses the intent of "no more matches".
– Blastfurnace
7 hours ago
|
show 5 more comments
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
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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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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 secondstd::find_ifcould be replaced bystd::none_ofwhich returns abooldirectly without having to compare withend. 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.
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 secondstd::find_ifcould be replaced bystd::none_ofwhich returns abooldirectly without having to compare withend. 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.
edited 7 hours ago
Blastfurnace
14.4k54259
answered 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
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.
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
edited 7 hours ago
Blastfurnace
14.4k54259
14.4k54259
answered 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
answered 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
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 secondstd::find_ifcould be replaced bystd::none_ofwhich returns abooldirectly without having to compare withend. It also expresses the intent of "no more matches".
– Blastfurnace
7 hours ago
|
show 5 more comments
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 secondstd::find_ifcould be replaced bystd::none_ofwhich returns abooldirectly without having to compare withend. 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
|
show 5 more comments
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
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
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
answered 8 hours ago
JeJoJeJo
5,91831028
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
answered 8 hours ago
JeJoJeJo
5,91831028
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
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 |
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
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
7 hours ago
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
@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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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