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ways to delete the element in between with vector?

P: n/a
Hi,

is there any ways that allow us to delete an element in between?

say

int_val[1]: 1
int_val[2]: 2
int_val[3]: 3

we want to delete 2 from int_val
so that it becomes

int_val[1]: 1
int_val[2]: 3

I have heard of using iterator with erase()
but it seemed a bit complicated for me

thank you very much.

jacky

Nov 7 '06 #1
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7 Replies


P: n/a
"JH Programmer" <ho********@gmail.comwrote:
is there any ways that allow us to delete an element in between?

say

int_val[1]: 1
int_val[2]: 2
int_val[3]: 3

we want to delete 2 from int_val
so that it becomes

int_val[1]: 1
int_val[2]: 3

I have heard of using iterator with erase()
but it seemed a bit complicated for me
Do you only want to erase that one element, or do you want to erase all elements with that value?

If the former, then simply "val.erase( val.begin() + pos );" where 'pos' is the index.

If the latter, then use the "erase-remove" idiom: "val.erase( remove( val.begin(), val.end(), 2 ), val.end() );"

--
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Nov 7 '06 #2

P: n/a
Thank you so much!

###########################
Here are for others to search for this thread
=================================

If you want to delete a specific element in the vector "array"
you use this method

for (int i = 0; i < data.size(); i++){
if ( data[i] == data_buffer ){
pos = i;
}
}
data.erase(data.begin() + pos );

#############################
"Daniel T. д
"
"JH Programmer" <ho********@gmail.comwrote:
is there any ways that allow us to delete an element in between?

say

int_val[1]: 1
int_val[2]: 2
int_val[3]: 3

we want to delete 2 from int_val
so that it becomes

int_val[1]: 1
int_val[2]: 3

I have heard of using iterator with erase()
but it seemed a bit complicated for me

Do you only want to erase that one element, or do you want to erase all elements with that value?

If the former, then simply "val.erase( val.begin() + pos );" where 'pos' is the index.

If the latter, then use the "erase-remove" idiom: "val.erase( remove( val..begin(), val.end(), 2 ), val.end() );"

--
To send me email, put "sheltie" in the subject.
Nov 7 '06 #3

P: n/a
"JH Programmer" <ho********@gmail.comwrote:
Thank you so much!

###########################
Here are for others to search for this thread
=================================

If you want to delete a specific element in the vector "array"
you use this method

for (int i = 0; i < data.size(); i++){
if ( data[i] == data buffer ){
pos = i;
}
}
data.erase(data.begin() + pos );
Are you sure the above is what you want? It will erase the *last*
element that has the value data_buffer. So for example, given the array
{ 0, 1, 2, 2, 1, 0 } if you want to erase '1' then the above code will
produce { 0, 1, 2, 2, 0 } instead of { 0, 2, 2, 1, 0 }

You are using the loop to find something, so why not put it in a
function called "find". That will make your code more expressive:

template < typename T >
unsigned find( const vector<T>& vec, T buffer )
{
unsigned result = vec.size();
for ( unsigned i = 0; i < vec.size(); ++i )
if ( vec[i] == buffer )
result = i;
return result;
}

The above will return the last element that has the value 'buffer' or
vec.size() if no element has the value.

Of course, going through the whole container is needless, once you find
the element you want, you can stop looking...

template < typename T >
unsigned reverse_find( const vector<T>& vec, T buffer )
{
for ( unsigned i = vec.size(); i 0; --i )
if ( vec[i - 1] == buffer )
return i - 1;
return vec.size();
}

Notice that I'm going through the container backwards because I want to
find the *last* element with the particular value.

It just so happens that the standard already has a function like this,
defined in <algorithm>

vec.erase( find( vec.rbegin(), vec.rend(), buffer ).base() );

The above line will do the same thing your code segment does.

--
To send me email, put "sheltie" in the subject.
Nov 7 '06 #4

P: n/a
Thanks again

Since I am implementing the deletion of username in the login system,
because login name is unique, so it works fine.

but one thing I can think of is that
we can declare another vector array
vector<intpos;

so

if (data[i] == data_buf){
pos.push_back(i);
}

but to delete the elements in the array
we should delete it from the back
[ 0 1 0 2 3 4 1 0 2]
so if we want to delete [2]
pos[1] = 3
pos[2] = 8

for (int i = pos.size(); i >= 0; i--){
data.erase(data.begin() + pos[i]);
}
so delete from the back in vector "data"
will give correct output.
[0 1 0 3 4 1 0]

"Daniel T. д
"
"JH Programmer" <ho********@gmail.comwrote:
Thank you so much!

###########################
Here are for others to search for this thread
=================================

If you want to delete a specific element in the vector "array"
you use this method

for (int i = 0; i < data.size(); i++){
if ( data[i] == data buffer ){
pos = i;
}
}
data.erase(data.begin() + pos );

Are you sure the above is what you want? It will erase the *last*
element that has the value data_buffer. So for example, given the array
{ 0, 1, 2, 2, 1, 0 } if you want to erase '1' then the above code will
produce { 0, 1, 2, 2, 0 } instead of { 0, 2, 2, 1, 0 }

You are using the loop to find something, so why not put it in a
function called "find". That will make your code more expressive:

template < typename T >
unsigned find( const vector<T>& vec, T buffer )
{
unsigned result = vec.size();
for ( unsigned i = 0; i < vec.size(); ++i )
if ( vec[i] == buffer )
result = i;
return result;
}

The above will return the last element that has the value 'buffer' or
vec.size() if no element has the value.

Of course, going through the whole container is needless, once you find
the element you want, you can stop looking...

template < typename T >
unsigned reverse_find( const vector<T>& vec, T buffer )
{
for ( unsigned i = vec.size(); i 0; --i )
if ( vec[i - 1] == buffer )
return i - 1;
return vec.size();
}

Notice that I'm going through the container backwards because I want to
find the *last* element with the particular value.

It just so happens that the standard already has a function like this,
defined in <algorithm>

vec.erase( find( vec.rbegin(), vec.rend(), buffer ).base() );

The above line will do the same thing your code segment does.

--
To send me email, put "sheltie" in the subject.
Nov 7 '06 #5

P: n/a
"Daniel T." <da******@earthlink.netwrote in message
news:da****************************@news.west.eart hlink.net...
: It just so happens that the standard already has a function like this,
: defined in <algorithm>
:
: vec.erase( find( vec.rbegin(), vec.rend(), buffer ).base() );
: The above line will do the same thing your code segment does.

No! Caveat: &*revIter != &*(revIter.base()) !!

The reverse iterators had to bee shifted, because for example:
rbegin().base() == end()
but *rbegin() shall return the last element == *(end()-1)

So a corrected statement would be:
vec.erase( find( vec.rbegin(), vec.rend(), buffer ).base()-1 );
Tricky details...

hth -Ivan
--
http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form
Brainbench MVP for C++ <http://www.brainbench.com

Nov 7 '06 #6

P: n/a
"JH Programmer" <ho********@gmail.comwrote:
>
Since I am implementing the deletion of username in the login
system, because login name is unique, so it works fine.
In that case, I question the wisdom of using a vector. A std::set would
support faster deletion and insure that there is only one of each name
in existence.
but one thing I can think of is that we can declare another vector
array
vector<intpos;

so

if (data[i] == data buf){
pos.push back(i);
}

but to delete the elements in the array
we should delete it from the back
[ 0 1 0 2 3 4 1 0 2]
so if we want to delete [2]
pos[1] = 3
pos[2] = 8

for (int i = pos.size(); i >= 0; i--){
data.erase(data.begin() + pos[i]);
}
so delete from the back in vector "data"
will give correct output.
[0 1 0 3 4 1 0]
The erase-remove idiom is more effecient in general for removing all
values from a container.

pos.erase( remove( pos.begin(), pos.end(), 2 ), pos.end() );

Writing the above out would look something like this:

unsigned shift_amt = 0;
unsigned i = 0;
while ( i < pos.size() ) {
pos[i - shift_amt] = pos[i];
if ( pos[i] == value_to_remove ) {
++shift_amt;
++i;
}
pos.resize( pos.size() - shift_amt );

This way, each cell is shifted only once, no matter how may of that
value exist in the container.

--
To send me email, put "sheltie" in the subject.
Nov 7 '06 #7

P: n/a
In article <7a***************************@news.hispeed.ch>,
"Ivan Vecerina" <_I*******************@ivan.vecerina.comwrote:
"Daniel T." <da******@earthlink.netwrote in message
news:da****************************@news.west.eart hlink.net...
: It just so happens that the standard already has a function like this,
: defined in <algorithm>
:
: vec.erase( find( vec.rbegin(), vec.rend(), buffer ).base() );
: The above line will do the same thing your code segment does.

No! Caveat: &*revIter != &*(revIter.base()) !!

The reverse iterators had to bee shifted, because for example:
rbegin().base() == end()
but *rbegin() shall return the last element == *(end()-1)

So a corrected statement would be:
vec.erase( find( vec.rbegin(), vec.rend(), buffer ).base()-1 );
Tricky details...
Good catch, I forgot about that (don't use reverse iterators much.)

--
To send me email, put "sheltie" in the subject.
Nov 7 '06 #8

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