C++ C++;-输出给定长度的所有可能的DNA Kmer
如果我想对了,问题实际上是将十进制整数转换为基数4,然后替换相应的基数。我想我可以使用itoa来实现这一点,但itoa不是C标准,我的编译器也不支持它。我欢迎任何聪明的想法。以下是我的示例代码:C++ C++;-输出给定长度的所有可能的DNA Kmer,c++,algorithm,bioinformatics,C++,Algorithm,Bioinformatics,如果我想对了,问题实际上是将十进制整数转换为基数4,然后替换相应的基数。我想我可以使用itoa来实现这一点,但itoa不是C标准,我的编译器也不支持它。我欢迎任何聪明的想法。以下是我的示例代码: #include <iostream> #include <string> #include <math.h> #define K 3 using namespace std; int main() { int num_kmers = pow(4,K);
#include <iostream>
#include <string>
#include <math.h>
#define K 3
using namespace std;
int main() {
int num_kmers = pow(4,K);
string* kmers = NULL;
/* Allocate memory for kmers array */
kmers = new string[num_kmers];
/* Populate kmers array */
for (int i=0; i< pow(4,K); i++) {
// POPULATE THE kmers ARRAY HERE
}
/* Display all possible kmers */
for (int i=0; i< pow(4,K); i++)
cout << kmers[i] << "\n";
delete [] kmers;
}
#包括
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#定义k3
使用名称空间std;
int main(){
int num_kmers=pow(4,K);
字符串*kmers=NULL;
/*为kmers数组分配内存*/
kmers=新字符串[num_kmers];
/*填充kmers数组*/
对于(inti=0;i cout一旦用户输入的文本被接受,您就不需要将十进制转换为任何形式
创建字符串数组也可能是一个错误,它随K
呈指数增长。只需打印输出即可
char bases[] = { 'A', 'C', 'T', 'G' };
std::vector< int > sequence( K ); // allow dynamic K
std::vector< char > output( K * ( 2 << K * 2 ) ); // flat sequence of sequences
std::vector< char >::iterator out_it = output.begin();
int i;
do {
// print current sequence
for ( i = 0; i < K; ++ i ) std::cout << bases[ sequence[ i ] ];
std::cout << '\n';
// store current sequence
for ( i = 0; i < K; ++ i ) * out_it ++ = bases[ sequence[ i ] ];
// advance to next sequence
for ( i = K; i > 0; -- i ) {
// increment the last base that we can
if ( sequence[ i - 1 ] != sizeof bases - 1 ) {
++ sequence[ i - 1 ];
break;
}
// reset bases that can't be incremented
sequence[ i - 1 ] = 0;
}
} while ( i > 0 ); // if i <= 0, failed to increment anything, we're done.
charbase[]={'A','C','T','G'};
std::vector序列(K);//允许动态K
std::vectoroutput(K*(2::iterator out_it=output.begin();
int i;
做{
//打印当前序列
对于(i=0;i
干杯。在我看来,这非常适合自定义迭代器。这样,您的主程序就可以很简单:
std::vector<std::string> v;
v.reserve(kmerIterator<4>::end() - kmerIterator<4>::begin());
std::copy(kmerIterator<4>::begin(), kmerIterator<4>::end(),
std::back_inserter(v));
std::vector v;
v、 保留(kmerIterator::end()-kmerIterator::begin());
std::copy(kmerierator::begin(),kmerierator::end(),
标准:背部插入器(v);
但是,由于我们将kmer概念实现为一个迭代器,因此我们也可以使用所有其他通用算法。由于我们将kmer迭代器实现为一个随机访问迭代器,因此找到第i个kmer非常简单:
kmerIterator<4>::begin()[i]
kmerierator::begin()[i]
以下是我的完整程序:
#include <iostream>
#include <iterator>
#include <algorithm>
#include <vector>
template<unsigned int n>
class kmerIterator :
public std::iterator<std::random_access_iterator_tag,
std::string,
unsigned int>
{
private:
typedef kmerIterator k; // 'cause I'm lazy
difference_type it;
kmerIterator(difference_type i) : it(i) {}
public:
kmerIterator() : it() {}
static k begin() {
return 0;
}
static k end() {
return difference_type(1) << n*2;
}
k& operator++() { ++it; return *this; }
k operator++(int) { return it++; }
k& operator--() { --it; return *this; }
k operator--(int) { return it--; }
k operator+(difference_type delta) { return it + delta; }
k operator-(difference_type delta) { return it - delta; }
difference_type operator-(const k& rhs) { return it - rhs.it; }
bool operator<(const k& rhs) const { return it < rhs.it; }
bool operator>(const k& rhs) const { return it > rhs.it; }
bool operator<=(const k& rhs) const { return it <= rhs.it; }
bool operator>=(const k& rhs) const { return it >= rhs.it; }
k operator+=(difference_type delta) { return it+=delta; }
k operator-=(difference_type delta) { return it-=delta; }
std::string operator[](difference_type delta) const { return *k(it+delta); }
std::string operator*() const {
std::string rz;
int i = 2*n;
do {
i -= 2;
rz += "ACGT"[(it>>i)&3];
} while(i);
return rz;
}
};
int main() {
std::copy(kmerIterator<2>::begin(),
kmerIterator<2>::end(),
std::ostream_iterator<std::string>(std::cout, "\n"));
std::vector<std::string> v;
v.reserve(kmerIterator<4>::end() - kmerIterator<4>::begin());
std::copy(kmerIterator<4>::begin(),
kmerIterator<4>::end(),
std::back_inserter(v));
std::cout << v[42] << "\n";
std::cout << kmerIterator<4>::begin()[56] << "\n";
}
#包括
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模板
kmerIterator类:
公共标准:迭代器
{
私人:
typedef kmeriator k;//因为我很懒
不同类型的it;
kmerIterator(difference_type i):它(i){
公众:
kmerIterator():它(){}
静态k begin(){
返回0;
}
静态k端(){
返回差异_type(1)rhs.it;}
bool操作符=rhs.it;}
k运算符+=(差分类型delta){返回它+=delta;}
k运算符-=(差分类型delta){返回它-=delta;}
字符串运算符[](差分类型delta)常量{return*k(it+delta);}
std::字符串运算符*()常量{
std::字符串rz;
int i=2*n;
做{
i-=2;
rz+=“ACGT”[(it>>i)和3];
}而(i),;
返回rz;
}
};
int main(){
std::copy(kmerierator::begin(),
kmerIterator::end(),
std::ostream_迭代器(std::cout,“\n”);
std::向量v;
v、 保留(kmerIterator::end()-kmerIterator::begin());
std::copy(kmerierator::begin(),
kmerIterator::end(),
标准:背部插入器(v);
std::cout哇,这不应该这么难
std::string kmers(int i, int K) {
static const char* DNA = "ACGT";
if (K==0) return std::string();
return DNA[i%4] + kmers(i / 4, K-1);
}
const int K = 2;
int main () {
for (int i = 0; i != 1<<(2*K); ++i)
std::cout << "kmers[" << i << "] = " << kmers(i,K) << std::endl;
}
std::字符串kmer(inti,intk){
静态常量char*DNA=“ACGT”;
if(K==0)返回std::string();
返回DNA[i%4]+kmer(i/4,K-1);
}
常数int K=2;
int main(){
对于(int i=0;i!=1#包括
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#包括
类kmer
{
公众:
kmer();
kmer(std::string&str,int&k):k_k,str_k(str)
{
对于(int i=0;我坚持住,我发现了一个令人讨厌的(明显的)等等。你应该真的,真的,真的使用一个向量。哦,我完全同意。但是我想写一些适合他目前存在的代码的东西。显然,如果他不小心,这会溢出。向量化我,船长。谢谢你的回答。不幸的是,这是一个更大的生物信息学应用程序的一部分n这确实需要我将结果存储在一个数组中。我很清楚大型K的危险性。@ned:你可以用它来构建一个数组,而不是打印输出。但是如果你尽可能避免使用此类数组,应用程序可能会更具可伸缩性。当涉及到大量信息时,它变得很重要将宝贵的内存用于有意义的数据,而不是“蓬松”
#include <iostream>
#include <iterator>
#include <algorithm>
#include <vector>
template<unsigned int n>
class kmerIterator :
public std::iterator<std::random_access_iterator_tag,
std::string,
unsigned int>
{
private:
typedef kmerIterator k; // 'cause I'm lazy
difference_type it;
kmerIterator(difference_type i) : it(i) {}
public:
kmerIterator() : it() {}
static k begin() {
return 0;
}
static k end() {
return difference_type(1) << n*2;
}
k& operator++() { ++it; return *this; }
k operator++(int) { return it++; }
k& operator--() { --it; return *this; }
k operator--(int) { return it--; }
k operator+(difference_type delta) { return it + delta; }
k operator-(difference_type delta) { return it - delta; }
difference_type operator-(const k& rhs) { return it - rhs.it; }
bool operator<(const k& rhs) const { return it < rhs.it; }
bool operator>(const k& rhs) const { return it > rhs.it; }
bool operator<=(const k& rhs) const { return it <= rhs.it; }
bool operator>=(const k& rhs) const { return it >= rhs.it; }
k operator+=(difference_type delta) { return it+=delta; }
k operator-=(difference_type delta) { return it-=delta; }
std::string operator[](difference_type delta) const { return *k(it+delta); }
std::string operator*() const {
std::string rz;
int i = 2*n;
do {
i -= 2;
rz += "ACGT"[(it>>i)&3];
} while(i);
return rz;
}
};
int main() {
std::copy(kmerIterator<2>::begin(),
kmerIterator<2>::end(),
std::ostream_iterator<std::string>(std::cout, "\n"));
std::vector<std::string> v;
v.reserve(kmerIterator<4>::end() - kmerIterator<4>::begin());
std::copy(kmerIterator<4>::begin(),
kmerIterator<4>::end(),
std::back_inserter(v));
std::cout << v[42] << "\n";
std::cout << kmerIterator<4>::begin()[56] << "\n";
}
std::string kmers(int i, int K) {
static const char* DNA = "ACGT";
if (K==0) return std::string();
return DNA[i%4] + kmers(i / 4, K-1);
}
const int K = 2;
int main () {
for (int i = 0; i != 1<<(2*K); ++i)
std::cout << "kmers[" << i << "] = " << kmers(i,K) << std::endl;
}
#include <iostream>
#include <string>
#include <string_view>
#include <vector>
class kmer
{
public:
kmer();
kmer(std::string &str, int &k) : k_(k), str_(str)
{
for(int i = 0; i <= str_.size() - k_; ++i)
value_.push_back(str_.substr(i, k_));
}
auto value() { return value_; }
auto size() { return value_.size(); }
private:
std::vector<std::string_view> value_;
std::string_view str_;
int k_;
};
int main()
{
std::string seq{"AGCTAGCT"};
kmer k4_mer(seq, 4);
std::cout << "Total " << k4_mer.size() << " mers\n";
for(auto &i : k4_mer.value())
std::cout << i << "\n";
return 0;
}