问题描述
这是我发布的问题的进展c++ 程序，用于将未知大小的 csv 文件(仅填充有浮点数)具有恒定(但未知)的列数读取到数组中.我现在转向真正的应用程序，我在其中读取文件，例如:
this is a progression from the question I posted c++ program for reading an unknown size csv file (filled only with floats) with constant (but unknown) number of columns into an array. Im now moving onto the real application where I read in a file like:
MESH2D
MESHNAME "default coverage"
NUM_MATERIALS_PER_ELEM 1
E4Q 1 19 20 14 16 2
E4Q 2 17 16 15 23 2
E4Q 3 22 15 14 21 2
E4Q 4 4 3 21 20 1
E4Q 5 6 20 19 7 1
E4Q 6 18 17 10 9 1
E4Q 7 17 23 12 11 1
E4Q 8 7 19 18 8 1
E4Q 9 22 1 13 23 1
E3T 10 14 20 21 2
E3T 11 21 2 22 1
E3T 12 21 3 2 1
E3T 13 22 2 1 1
E3T 14 5 20 6 1
E3T 15 20 5 4 1
E3T 16 16 14 15 2
E3T 17 23 13 12 1
E3T 18 22 23 15 2
E3T 19 17 11 10 1
E3T 20 17 18 16 2
E3T 21 8 18 9 1
E3T 22 18 19 16 2
ND 1 -3.25811078e+002 7.70285567e+001 0.00000000e+000
ND 2 -3.24209146e+002 7.60394871e+001 0.00000000e+000
ND 3 -3.23012110e+002 7.44783503e+001 0.00000000e+000
ND 4 -3.22754089e+002 7.25326647e+001 0.00000000e+000
ND 5 -3.23617358e+002 7.08079432e+001 0.00000000e+000
ND 6 -3.25161538e+002 6.98134116e+001 0.00000000e+000
ND 7 -3.27128620e+002 6.98759747e+001 0.00000000e+000
ND 8 -3.29095703e+002 6.99385378e+001 0.00000000e+000
ND 9 -3.30301095e+002 7.14667646e+001 0.00000000e+000
ND 10 -3.30786908e+002 7.33241555e+001 0.00000000e+000
ND 11 -3.30835733e+002 7.52916270e+001 0.00000000e+000
ND 12 -3.29587322e+002 7.65401204e+001 0.00000000e+000
ND 13 -3.27743000e+002 7.72270000e+001 0.00000000e+000
ND 14 -3.26108525e+002 7.32067724e+001 0.00000000e+000
ND 15 -3.27041416e+002 7.42070316e+001 0.00000000e+000
ND 16 -3.27350377e+002 7.31716751e+001 0.00000000e+000
ND 17 -3.29153676e+002 7.40024406e+001 0.00000000e+000
ND 18 -3.28659180e+002 7.19967464e+001 0.00000000e+000
ND 19 -3.26845856e+002 7.14062637e+001 0.00000000e+000
ND 20 -3.25000347e+002 7.20534611e+001 0.00000000e+000
ND 21 -3.24701329e+002 7.39638966e+001 0.00000000e+000
ND 22 -3.26167714e+002 7.53360591e+001 0.00000000e+000
ND 23 -3.28060316e+002 7.54194849e+001 0.00000000e+000
BEGPARAMDEF
GM "Mesh"
SI 0
DY 0
TU ""
TD 0 0
NUME 3
BCPGC 0
DISP_OPTS entity 0 0 0 0 1 0 0 0
DISP_OPTS inactive 0 0 0 0 1 0 0 0
DISP_OPTS multiple 0 0 0 0 1 0 0 0
BEFONT 0 1
DISP_OPTS entity 1 0 0 0 1 0 0 0
DISP_OPTS inactive 1 0 0 0 1 0 1 0
DISP_OPTS multiple 1 0 0 0 1 0 1 0
BEFONT 1 1
DISP_OPTS entity 2 0 0 0 1 0 0 0
DISP_OPTS inactive 2 0 0 0 1 0 1 0
DISP_OPTS multiple 2 0 0 0 1 0 1 0
BEFONT 2 1
MAT 1 "material 01"
MAT 2 "material 02"
MAT_MULTI 0
ENDPARAMDEF
BEG2DMBC
END2DMBC
BEGCURVE Version: 1
ENDCURVE
调用example.2dm"并尝试为以 E4Q E3T 和 ND 开头的行中存储的数据编写 3 个二维向量(或可能(最好)数组).我想使用这些单元格定义来打印带有单元格中心的文件.
called 'example.2dm' and try and write 3 2d vectors (or possibly (preferably) arrays) for the data stored in lines starting with E4Q E3T and ND. I want to use these cell difinitions to print a file with cell centres.
我认为在此记录此类问题非常有用，因为我的最后一个问题不断获得关注.
I think this sort of problem is very usefull to be documented on here as my last question keeps getting views.
到目前为止我的代码是:
My code so far is:
// Read in CSV
//
// Alex Byasse
#include <algorithm>
#include <fstream>
#include <iostream>
#include <iterator>
#include <sstream>
#include <string>
#include <vector>
#include <cstdlib>
enum case_name {e3t, e4q, nd};
case_name checkit(std::string const& inString)
{
if (inString == "E3T") return e3t;
if (inString == "E4Q") return e4q;
if (inString == "ND") return nd;
}
int main()
{
std::vector<std::vector<std::string>> values;
std::ifstream fin("example.2dm");
for (std::string line; std::getline(fin, line); )
{
std::istringstream in(line);
values.push_back(
std::vector<std::string>(std::istream_iterator<std::string>(in),
std::istream_iterator<std::string>()));
}
int nc = 0;
int nn = 0;
std::vector<std::vector<double>> cells;
std::vector<std::vector<double>> nodes;
for (int i = 0; i < values.size() - 1; i++) {
switch (checkit(values[i][0])){
case e3t:
cells[nc].push_back(std::stod(values[i][1]));
cells[nc].push_back(std::stod(values[i][2]));
cells[nc].push_back(std::stod(values[i][3]));
cells[nc].push_back(std::stod(values[i][4]));
cells[nc].push_back(std::stod(values[i][2]));
cells[nc].push_back(std::stod(values[i][5]));
nc++;
break;
case e4q:
cells[nc].push_back(std::stod(values[i][1]));
cells[nc].push_back(std::stod(values[i][2]));
cells[nc].push_back(std::stod(values[i][3]));
cells[nc].push_back(std::stod(values[i][4]));
cells[nc].push_back(std::stod(values[i][5]));
cells[nc].push_back(std::stod(values[i][6]));
nc++;
break;
case nd:
nodes[nn].push_back(std::stod(values[i][1]));
nodes[nn].push_back(std::stod(values[i][2]));
nodes[nn].push_back(std::stod(values[i][3]));
nn++;
break;
}
}
}
我能帮我修复这段代码吗?这是正确的吗?
Can I get some help fixing this code and also.Is this correct?
#include <cstdlib>
我收到以下编译错误:
$ g++ read_csv2.cpp -std=c++0x
read_csv2.cpp: In function ‘int main()’:
read_csv2.cpp:44:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:45:33: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:46:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:47:33: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:48:33: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:49:33: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:53:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:54:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:55:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:56:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:57:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:58:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:62:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:63:26: error: ‘stod’ is not a member of ‘std’
read_csv2.cpp:64:26: error: ‘stod’ is not a member of ‘std’
我在 cygwin 上运行 g++.我检查了手册页，-std=c++11 是一个可用的选项，我已经尝试过了，但还是出现了同样的错误?
I'm running g++ on cygwin. I've checked the man pages and -std=c++11 is an available option and I've tryed it yet I get the same error?
推荐答案
cells[nc] 和 nodes[nd] 的地方被破坏了，因为那些向量永远不会增加大小.
The places where you have cells[nc] and nodes[nd] are broken because those vectors are never increased in size.
这里有一些不需要 std::stod 就可以做你需要的东西.我将其更改为使用 ND 行中的所有值，假设这就是您的意思，如果不是，请在 GetValues 调用中将 4 更改为 3.
Here's something that may do what you need without requiring std::stod. I changed it to use all of the values from the ND lines assuming that was what you meant, if not, change the 4 to a 3 in that GetValues call.
#include <algorithm>
#include <fstream>
#include <iostream>
#include <iterator>
#include <sstream>
#include <string>
#include <vector>
#include <cstdlib>
double ToDouble(const std::string& str)
{
std::stringstream s(str);
double d;
s >> d;
return d;
}
std::vector<double> GetValues(const std::vector<std::string>& src, int start, int end)
{
std::vector<double> ret;
for(int i = start; i <= end; ++i)
{
ret.push_back(ToDouble(src[i]));
}
return ret;
}
void PrintValues(const std::string& title, std::vector<std::vector<double>>& v)
{
std::cout << title << std::endl;
for(size_t line = 0; line < v.size(); ++line)
{
for(size_t val = 0; val < v[line].size(); ++val)
{
std::cout << v[line][val] << " ";
}
std::cout << std::endl;
}
std::cout << std::endl;
}
int main()
{
std::vector<std::vector<std::string>> values;
std::ifstream fin("example.2dm");
for (std::string line; std::getline(fin, line); )
{
std::istringstream in(line);
values.push_back(
std::vector<std::string>(std::istream_iterator<std::string>(in),
std::istream_iterator<std::string>()));
}
std::vector<std::vector<double>> cells;
std::vector<std::vector<double>> nodes;
for (size_t i = 0; i < values.size(); ++i)
{
if(values[i][0] == "E3T")
{
cells.push_back(GetValues(values[i], 1, 5));
}
else if(values[i][0] == "E4Q")
{
cells.push_back(GetValues(values[i], 1, 6));
}
else if(values[i][0] == "ND")
{
nodes.push_back(GetValues(values[i], 1, 4));
}
}
PrintValues("Cells", cells);
PrintValues("Nodes", nodes);
return 0;
}
这篇关于c ++将文本文件读入向量<vector>然后根据内部向量中的第一个单词写入向量或数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！