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
这是我发布的问题的进展c++ 程序，用于读取具有恒定(但未知)列数的未知大小的 csv 文件(仅填充浮点数)到数组中.我现在进入真正的应用程序，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"并尝试为存储在以 E4Q E3T 和 ND 开头的行中的数据编写 3 个 2d 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.
我认为在此处记录此类问题非常有用，因为我的最后一个问题不断获得意见.
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;
}
}
}
我能得到一些帮助来修复这个代码吗?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++.我检查了手册页并且 -std=c++11 是一个可用的选项，我已经尝试过但我得到了同样的错误?
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
你有 cells[nc] 和 nodes[nd] 的地方被破坏了，因为那些向量尺寸永远不会增加.
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 就可以满足你的需求.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，则在 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;
}
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