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Quickest way to find the rank of a <map>

P: n/a
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?
At the moment I'm iterating through the map and keeping count of when I
hit it.

Thanks

Steve
Feb 28 '06 #1
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13 Replies


P: n/a
Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?
At the moment I'm iterating through the map and keeping count of when I
hit it.


Check out the member functions of std::map, notably, find.

Your algorithm is O(n), the member function is probably O(log2 n).

Ben Pope
--
I'm not just a number. To many, I'm known as a string...
Feb 28 '06 #2

P: n/a
In article <44**********************@taz.nntpserver.com>,
Ben Pope <be***************@gmail.com> wrote:
Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?
At the moment I'm iterating through the map and keeping count of when I
hit it.


Check out the member functions of std::map, notably, find.

Your algorithm is O(n), the member function is probably O(log2 n).

Ben Pope


Thanks, but that returns an iterator to the element. I'm not sure how
this helps me determine its index.

Steve
Feb 28 '06 #3

P: n/a
Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?


std::map<> is an associative container with elements sorted based on
some comparison operator. In other words, it doesnot really mean
anything to find "index" of an element in std::map<> - (You cannot
really insert an element in a map at a "given position", unlike say,
std::vector<>, where you can).

If by the word "index" you actually mean the "iterator" to the element,
then the member function find() will do the trick. This function takes
a key and returns the iterator to the element with that key (and the
iterator to end() otherwise if no such element exists)

Feb 28 '06 #4

P: n/a
Steve Edwards wrote:
In article <44**********************@taz.nntpserver.com>,
Ben Pope <be***************@gmail.com> wrote:
Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?
At the moment I'm iterating through the map and keeping count of when I
hit it.

Check out the member functions of std::map, notably, find.

Your algorithm is O(n), the member function is probably O(log2 n).

Ben Pope


Thanks, but that returns an iterator to the element. I'm not sure how
this helps me determine its index.


#include <iostream>
#include <string>
#include <map>

int main()
{
std::map<int, std::string> m;

m[3] = "foo";
m[6] = "bar";
m[9] = "baz";
m[12] = "qux";
m[15] = "quux";

std::map<int, std::string>::iterator iter = m.find(3);

if(iter != m.end()) {
std::cout
<< "element with key: " << iter->first
<< " , value: " << iter->second
<< " , has index: " << std::distance(m.begin(), iter)
<< std::endl;
}
return 0;
}
Does that help?

Ben Pope
--
I'm not just a number. To many, I'm known as a string...
Feb 28 '06 #5

P: n/a

Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?
At the moment I'm iterating through the map and keeping count of when I
hit it.


No. The problem is that <map> is designed so inserting an element is
fast. Yet
if you insert a new element at rank 1, all existing items have their
rank changed.
If this rank was stored, getting the rank would be fast but insertion
would be slow.

It's a design tradeoff, and std::map wasn't designed for it. You could
try a
vector<pair<long,string>, notstd::pair_greater_first> vmapOfFreq

HTH,
Michiel Salters

Feb 28 '06 #6

P: n/a
In article <gf***********************@news.btinternet.com>,
Steve Edwards <gf*@lineone.net> wrote:
In article <44**********************@taz.nntpserver.com>,
Ben Pope <be***************@gmail.com> wrote:
Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?
At the moment I'm iterating through the map and keeping count of when I
hit it.


Check out the member functions of std::map, notably, find.

Your algorithm is O(n), the member function is probably O(log2 n).

Ben Pope


Thanks, but that returns an iterator to the element. I'm not sure how
this helps me determine its index.


Once you have the iterator, you can use std::distance.

--
Magic depends on tradition and belief. It does not welcome observation,
nor does it profit by experiment. On the other hand, science is based
on experience; it is open to correction by observation and experiment.
Feb 28 '06 #7

P: n/a
In article <44**********************@taz.nntpserver.com>,
Ben Pope <be***************@gmail.com> wrote:
std::distance(m.begin(), iter)


Thanks, that's the function I couldn't figure out.

Steve
Feb 28 '06 #8

P: n/a

Steve Edwards wrote:
In article <44**********************@taz.nntpserver.com>,
Ben Pope <be***************@gmail.com> wrote:
std::distance(m.begin(), iter)


Thanks, that's the function I couldn't figure out.

Steve


Of course this call is still O(n) - as any solution must be. Something
tells me you should instead use a std::vector for your job and keep it
sorted. This assumes that the data in the map doesn't normally change,
but if it does your "rank" is not of much use anyway.

/Peter

Feb 28 '06 #9

P: n/a
In article <11**********************@t39g2000cwt.googlegroups .com>,
Mi*************@tomtom.com wrote:
Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?
At the moment I'm iterating through the map and keeping count of when I
hit it.


No. The problem is that <map> is designed so inserting an element is
fast. Yet
if you insert a new element at rank 1, all existing items have their
rank changed.
If this rank was stored, getting the rank would be fast but insertion
would be slow.

It's a design tradeoff, and std::map wasn't designed for it. You could
try a
vector<pair<long,string>, notstd::pair_greater_first> vmapOfFreq

HTH,
Michiel Salters


Thanks, I can see why rank is not stored. But I wondered if when you did
a retrieval-by-key, the behind-the-scenes search algorithm would become
aware of the position within the sorted list, and hence be able to
provide the rank as a by-product of the search.
Feb 28 '06 #10

P: n/a
In article <11*********************@v46g2000cwv.googlegroups. com>,
"Neelesh Bodas" <ne***********@gmail.com> wrote:
Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?


std::map<> is an associative container with elements sorted based on
some comparison operator. In other words, it doesnot really mean
anything to find "index" of an element in std::map<> - (You cannot
really insert an element in a map at a "given position", unlike say,
std::vector<>, where you can).

If by the word "index" you actually mean the "iterator" to the element,
then the member function find() will do the trick. This function takes
a key and returns the iterator to the element with that key (and the
iterator to end() otherwise if no such element exists)


Thanks, I do take your point, but that's not to say that the rank isn't
a very useful bit of info. (sets that store scores often need to be
queried for rank i.e. "who came 7th?", "what rank is the one that scored
x?")
No problem, I can just iterate over them, or use
std::distance(m.begin(), iter) as Ben suggested.

Steve
Feb 28 '06 #11

P: n/a
In article <11**********************@e56g2000cwe.googlegroups .com>,
"peter koch" <pe***************@gmail.com> wrote:
Steve Edwards wrote:
In article <44**********************@taz.nntpserver.com>,
Ben Pope <be***************@gmail.com> wrote:
std::distance(m.begin(), iter)


Thanks, that's the function I couldn't figure out.

Steve


Of course this call is still O(n) - as any solution must be. Something
tells me you should instead use a std::vector for your job and keep it
sorted. This assumes that the data in the map doesn't normally change,
but if it does your "rank" is not of much use anyway.

/Peter


Yes, after reading the various posts I appreciate the problem now.
Searching by key is the most important job I have to do though, so I'll
just put up with having to determine rank manually.

Steve
Feb 28 '06 #12

P: n/a
Steve Edwards <gf*@lineone.net> wrote in
news:gf***********************@news.btinternet.com :
In article <11**********************@t39g2000cwt.googlegroups .com>,
Mi*************@tomtom.com wrote:

Thanks, I can see why rank is not stored. But I wondered if when you
did a retrieval-by-key, the behind-the-scenes search algorithm would
become aware of the position within the sorted list, and hence be able
to provide the rank as a by-product of the search.


Nope. Nothing says that the map is stored in any sort of linear way. It's
far more likely that it's stored in some sort of tree data structure.
Feb 28 '06 #13

P: n/a
Steve Edwards wrote:
Hi,

Given a map:
typedef map<long, string, greater<long> > mapOfFreq;

Is there a quicker way to find the rank (i.e. index) of the elememt that
has the long value of x?
At the moment I'm iterating through the map and keeping count of when I
hit it.


As others have pointed out, std::map<> does not support this. If performance
of this operation turns out to be critical, you can, however, design a
container very much like std::map<> based upon a balanced tree. In addition
to a key/value pair, every node would also keep a count of the size of the
subtree whose root the given node is. This way, you can obtain the position
of a node in the traversal order in log(N) time by adding some sizes along
a path. The costs for updating the size information is also logarithmic.
Thus, you can keep log(N) performance for insert/erase.
Best

Kai-Uwe Bux
Mar 1 '06 #14

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