C++ 如何在 int 类型的二维向量中 push_back 数据
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/42249303/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How can I push_back data in a 2d vector of type int
提问by saru
I have a vector and want to store int data in to it at run time can I store the data in a 2D vector in this manner ?
我有一个向量，想在运行时将 int 数据存储到其中，我可以以这种方式将数据存储在 2D 向量中吗？
std::vector<std::vector <int>> normal:
for(i=0;i<10;i++){
for(j=0;j<20;j++){
normal[i].push_back(j);
}
}
回答by SingerOfTheFall
Yes, but you also need to push each of the sub-vectors:
是的，但您还需要推送每个子向量：
std::vector<std::vector<int>> normal;
for(int i=0; i<10; i++)
{
normal.push_back(std::vector<int>());
for(int j=0; j<20; j++)
{
normal[i].push_back(j);
}
}
回答by A. Monti
You are manipulating a vector of vectors.
As such, when declaring normalit is empty and does not contain any element.
您正在操纵向量的向量。因此，当声明normal它为空并且不包含任何元素时。
You can either :
你可以：
Resize the vector prior to inserting elements
在插入元素之前调整向量的大小
std::vector<std::vector<int> > normal;
normal.resize(20);
for (size_t i = 0; i < normal.size(); ++i)
{
for (size_t j = 0; j < 20; ++j)
normal[i].push_back(j);
}
This may be slightly more efficient than pushing an empty vector at each step as proposed in other answers.
这可能比在其他答案中提出的在每一步推送一个空向量更有效。
Use a flat 2D array
使用平面二维数组
If you want to store a 2D array, this is not the optimal solution, because :
如果要存储二维数组，这不是最佳解决方案，因为：
- Your array data is spread across N different dynamically allocated buffers (for N lines)
- Your array can have a different number of columns per line (because nothing enforces that
normal[i].size() == normal[j].size()
- 您的数组数据分布在 N 个不同的动态分配缓冲区（对于 N 行）
- 您的数组每行可以有不同数量的列（因为没有强制要求
normal[i].size() == normal[j].size()
Instead, you can use a vector of size N * M(where Nis the number of lines and Mthe number of columns), and access an element at line iand columns jusing the index i + j * N:
相反，您可以使用大小向量N * M（其中N是行M数和列数），并使用索引访问行i和列处的元素：ji + j * N
size_t N = 20;
size_t M = 20;
std::vector<int> normal;
normal.resize(N * M);
for (size_t i = 0; i < N; ++i)
for (size_t j = 0; j < M; ++j)
normal[i + j * N] = j;
回答by Vlad from Moscow
Here is one more approach.
这是另一种方法。
#include <iostream>
#include <iomanip>
#include <vector>
#include <numeric>
int main()
{
std::vector<std::vector <int> > normal;
normal.resize( 10, std::vector<int>( 20 ) );
for ( auto &v : normal ) std::iota( v.begin(), v.end(), 0 );
for ( const auto &v : normal )
{
for ( int x : v ) std::cout << std::setw( 2 ) << x << ' ';
std::cout << std::endl;
}
}
The program output is
程序输出是
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
You can write a corresponding function
可以写一个对应的函数
#include <iostream>
#include <iomanip>
#include <vector>
#include <numeric>
template <typename T>
T & init_2d( T &container, size_t m, size_t n )
{
container.resize( m, typename T::value_type( n ) );
for ( auto &item : container ) std::iota( item.begin(), item.end(), 0 );
return container;
}
int main()
{
std::vector<std::vector<int>> v;
for ( const auto &v : init_2d( v, 10, 20 ) )
{
for ( int x : v ) std::cout << std::setw( 2 ) << x << ' ';
std::cout << std::endl;
}
}
回答by Umang Kalra
Allocate n empty vectors, that is, empty vector for each index. Then push_back() can be applied.
为每个索引分配n个空向量，即空向量。然后可以应用 push_back()。
int main()
{
int n = 10;
std::vector<std::vector<int>> normal;
normal.resize(n); //Allocating 'n' empty vectors
for (int i = 0; i < n; i++)
{
for (int j = 0; j < 20; j++)
{
normal[i].push_back(j);
}
}
return 0;
}
回答by Cory Kramer
You cannot directly assign to [i]without allocating the outer and inner vectors first. One solution to this would be to create the inner vectors inside your for loop, then once those are populated, push_back to the outer vector.
如果不先[i]分配外部和内部向量，就不能直接分配给。对此的一种解决方案是在 for 循环内创建内部向量，然后在填充这些向量后，将 push_back 推回到外部向量。
std::vector<std::vector<int>> normal;
for(i=0;i<10;i++)
{
std::vector<int> temp;
for(j=0;j<20;j++)
{
temp.push_back(j);
}
normal.push_back(temp);
}
回答by user3853544
You have a vector of vectors.
你有一个向量的向量。
normal[i] Does not exist because you have not created it.
normal[i] 不存在，因为您尚未创建它。
std::vector<std::vector <int> > normal:
for(i=0;i<10;i++){
normal.emplace_back();
for(j=0;j<20;j++){
normal.back().push_back(j);
}
}
for(i=0;i<10;i++){
for(j=0;j<20;j++){
std::cout << normal[i][j] << " ";
}
std::cout << std::endl;
}
