按列对 CSV 文件进行排序并与 C++ 中的另一列进行比较
Sort CSV file by column and compare with another column in C++
我有一些 CSV 文件可以导入到 C++ 中，分成列并打印它们，但我无法执行我需要的分析。我希望能够对每一列进行排序（升序或降序），然后在单独的 1 或 0 列中找到一个分组。这是我到目前为止的代码，但似乎每次创建新行时我都在替换变量。
#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <vector>
using namespace std;
struct sampleData { //Create set of variables that are used in the vector
//to print sampleData
float first, second, third, fourth;
};
void printSample(sampleData sample) // Function that prints the vectors.
{
cout << sample.first << " " << sample.second << " " << sample.third << "
" << sample.fourth endl;
}
int main()
{
ifstream myFile("file.csv"); //Open file with ifstream constructor.
if (myFile.is_open())
{
vector<sampleData> sample; //Create vector that stores the variables declared in the struct.
float first, second, third, fourth;
char delim;
while (myFile >> first >> delim >> second >> delim >> third >> delim >> fourth) { //Places each value of csv file into individual variables in the vector.
sample.push_back({ first, second, third, fourth });
}
cout << "First value " << " Second value: " << " Third Value: " << " Fourth Value: " << endl; //Create column headers.
for (int x(0); x < sample.size(); ++x)
{
printSample(sample.at(x));
}
}
else
{
cout << "The file did not open."; //Let's me know if file has not been opened.
}
system("pause");
return 0;
}
下面是我需要的示例。我想对每一列 (1-3) 进行排序，并将它们与第四列的 1 和 0 进行比较，以找到至少有七个 1 且平均值至少为 .70 的分组。最好创建一个二维数组或二维向量吗？如果是这样，将如何对其进行排序和比较？
感谢大家的帮助。
> -40.31945 -20.71259 4.024558 1
> -8.428544 -1.173988 13.55221 1
> -9.99227 -1.964128 22.35553 1
> -6.227934 -0.6318588 11.28533 0
> -7.350101 -4.340335 9.932037 1
> -11.32407 -3.242851 15.07184 1
> -15.81499 -5.500328 15.33309 0
> -6.112404 -1.504377 24.17496 1
> -7.5483 -3.147136 17.5016 1
> -9.895069 -6.141642 17.70264 1
> -6.691729 -5.821645 41.11068 1
> -9.520897 -4.83869 12.83501 0
> -6.09901 -1.291806 22.62663 1
> -2.136172 -0.7562032 34.48225 1
> -5.813394 -2.087043 26.70455 0
> -2.359689 -0.04058313 68.30959 0
> -4.093154 -2.890539 32.40205 0
> -7.326787 -8.31641 23.47626 0
> -5.842336 -4.699064 32.14418 0
> -1.26901 -1.150853 54.72232 1
> -4.532993 -1.921023 27.54052 0
> -13.04364 -12.8271 17.78159 1
> -22.29973 -18.63197 10.62449 1
> -13.097 -11.09199 9.261793 0
> -6.73371 -4.044 24.63213 1
> -8.487038 -5.855842 20.65492 1
> -1.271804 -0.1592398 73.54436 0
> -5.903441 -2.511718 2.906148 0
> -6.569601 -3.63947 14.92872 0
> -2.671139 -1.596091 61.78936 1
> -0.67129 -0.1758051 35.63146 0
> -10.33999 -10.25158 19.83222 0
> -5.900752 -4.774312 22.25315 0
> -3.473342 -2.116564 60.31918 0
> -5.51118 -8.684725 45.30108 1
> -4.393883 -3.597137 21.0572 0
> -3.671957 -3.355143 51.05236 1
> -7.700621 -7.257176 29.59876 1
> -6.959113 -5.834087 21.52065 1
> -6.978306 -6.291922 26.17615 0
> -3.525233 -0.2435265 39.66356 0
> -8.017325 -7.190228 16.78984 1
> -9.686805 -6.356866 24.96812 1
> -5.841892 -4.090017 12.90826 1
> -4.101501 -0.8392091 29.49425 1
> -0.50966 -0.6248183 72.55316 0
> -2.747329 -3.107922 70.82893 1
> -3.682684 -5.461088 7.237332 0
> -1.726765 -1.030436 51.13756 0
> -5.065511 -5.105534 48.8038 1
> -3.490172 -0.8473139 54.89489 1
> -14.56848 -13.29985 8.508147 1
> -5.511615 -2.257046 26.53605 1
> -0.80373 -1.259443 54.58532 1
> -11.76727 -10.51294 19.43544 0
> -4.924498 -5.660692 64.22583 1
> -1.662102 -1.329681 68.50871 0
> -2.225776 -1.191363 46.14959 1
> -11.97834 -1.471152 18.86225 0
> -9.986734 -8.210676 15.11784 1
> -0.78368 -0.2543859 64.04224 1
> -11.41681 -13.24663 9.016961 1
> -10.73357 -13.46118 31.8038 1
> -2.443766 -0.841536 35.3982 1
> -3.112007 -1.327887 32.61596 1
> -1.647414 -0.9874625 65.37144 0
> -3.771582 -2.685039 42.65498 0
> -5.503803 -6.65314 15.60404 1
> -6.844056 -10.59976 22.71807 1
> -3.977231 -6.444871 47.65485 1
> -0.43918 -1.813655 35.90933 1
> -4.520459 -3.337119 17.47536 1
> -3.102405 -2.276846 15.49771 1
> -3.173711 -4.548148 54.85541 1
> -4.157713 -2.368944 36.82358 1
> -6.671762 -6.863191 33.18528 1
> -5.806525 -8.300102 38.04575 1
> -9.137906 -10.43044 20.62558 1
> -4.830114 -5.035967 80.04454 1
> -6.717423 -7.807728 18.62613 1
> -1.654782 -2.814744 69.35754 1
> -5.718936 -5.041555 19.44518 1
> -1.139612 -1.246455 31.46728 1
> -5.193422 -4.141603 49.06763 0
> -0.72360 -1.519114 68.06107 1
> -3.45456 -2.324488 24.8586 1
> -3.946017 -1.809939 26.39728 1
> -1.373865 -1.385224 59.31034 0
> -12.91463 -16.81217 21.9325 1
> -7.101114 -4.463167 24.6039 1
> -11.19178 -7.923832 11.70692 1
> -6.337176 -3.290151 46.2829 1
> -6.034304 -6.688771 12.98928 1
> -10.72616 -16.16286 27.24244 1
> -10.01076 -11.90333 16.67032 1
> -2.85405 -1.064295 18.82794 1
> -3.582814 -3.041154 34.58895 0
> -0.88143 -2.513154 72.57123 0
> -2.936312 -2.92483 32.65664 0
> -2.859565 -7.337652 31.87842 1
> -4.467122 -6.427214 56.81916 0
> -6.340817 -6.706052 9.87694 1
> -1.40155 -2.738037 35.32452 1
> -10.92032 -11.05833 30.82691 1
> -7.330603 -6.257256 22.16484 1
> -2.714168 -2.258151 36.30459 0
> -2.793682 -2.935043 56.51117 1
> -6.706202 -11.04426 11.10245 0
> -6.113976 -7.36745 11.36128 1
> -9.845764 -10.35044 37.52305 0
> -7.786937 -10.70406 21.68431 1
> -0.54450 -3.818708 64.34981 1
> -1.402748 -4.612042 52.94871 0
> -1.771809 -3.918717 41.45876 1
> -4.142132 -7.088901 45.44987 1
> -1.640578 -4.787658 40.82234 1
> -1.050637 -2.535334 42.87785 1
> -0.32151 -3.315413 40.40543 1
您需要提供 comparitor 函数，用于在排序期间比较单个 行 。希望你有一个现代版本的 C++ 并且可以使用 lambdas:
// Sort by first column
std::sort( samples.begin(), samples.end(),
[]( const sampleData& a, const sampleData& b )
{
return a.first < b.first;
}
);
按特定列排序后，您可以遍历序列并计算连续的第四列 1s。
编辑
lambda 是一种即时创建函数对象的方法。以上等同于：
struct unnamed_function
{
bool operator () ( const sampleData& a, const sampleData& b ) const
{
return a.first < b.first;
}
};
...
std::sort( samples.begin(), samples.end(), unnamed_function() );
[]是引入lambda的“捕获列表”
Read more about lambdas.
（抱歉，我还没有更好的常见问题解答...）
* 大致如此。幕后其实比这个复杂一点。
我有一些 CSV 文件可以导入到 C++ 中，分成列并打印它们，但我无法执行我需要的分析。我希望能够对每一列进行排序（升序或降序），然后在单独的 1 或 0 列中找到一个分组。这是我到目前为止的代码，但似乎每次创建新行时我都在替换变量。
#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <vector>
using namespace std;
struct sampleData { //Create set of variables that are used in the vector
//to print sampleData
float first, second, third, fourth;
};
void printSample(sampleData sample) // Function that prints the vectors.
{
cout << sample.first << " " << sample.second << " " << sample.third << "
" << sample.fourth endl;
}
int main()
{
ifstream myFile("file.csv"); //Open file with ifstream constructor.
if (myFile.is_open())
{
vector<sampleData> sample; //Create vector that stores the variables declared in the struct.
float first, second, third, fourth;
char delim;
while (myFile >> first >> delim >> second >> delim >> third >> delim >> fourth) { //Places each value of csv file into individual variables in the vector.
sample.push_back({ first, second, third, fourth });
}
cout << "First value " << " Second value: " << " Third Value: " << " Fourth Value: " << endl; //Create column headers.
for (int x(0); x < sample.size(); ++x)
{
printSample(sample.at(x));
}
}
else
{
cout << "The file did not open."; //Let's me know if file has not been opened.
}
system("pause");
return 0;
}
下面是我需要的示例。我想对每一列 (1-3) 进行排序，并将它们与第四列的 1 和 0 进行比较，以找到至少有七个 1 且平均值至少为 .70 的分组。最好创建一个二维数组或二维向量吗？如果是这样，将如何对其进行排序和比较？
感谢大家的帮助。
> -40.31945 -20.71259 4.024558 1
> -8.428544 -1.173988 13.55221 1
> -9.99227 -1.964128 22.35553 1
> -6.227934 -0.6318588 11.28533 0
> -7.350101 -4.340335 9.932037 1
> -11.32407 -3.242851 15.07184 1
> -15.81499 -5.500328 15.33309 0
> -6.112404 -1.504377 24.17496 1
> -7.5483 -3.147136 17.5016 1
> -9.895069 -6.141642 17.70264 1
> -6.691729 -5.821645 41.11068 1
> -9.520897 -4.83869 12.83501 0
> -6.09901 -1.291806 22.62663 1
> -2.136172 -0.7562032 34.48225 1
> -5.813394 -2.087043 26.70455 0
> -2.359689 -0.04058313 68.30959 0
> -4.093154 -2.890539 32.40205 0
> -7.326787 -8.31641 23.47626 0
> -5.842336 -4.699064 32.14418 0
> -1.26901 -1.150853 54.72232 1
> -4.532993 -1.921023 27.54052 0
> -13.04364 -12.8271 17.78159 1
> -22.29973 -18.63197 10.62449 1
> -13.097 -11.09199 9.261793 0
> -6.73371 -4.044 24.63213 1
> -8.487038 -5.855842 20.65492 1
> -1.271804 -0.1592398 73.54436 0
> -5.903441 -2.511718 2.906148 0
> -6.569601 -3.63947 14.92872 0
> -2.671139 -1.596091 61.78936 1
> -0.67129 -0.1758051 35.63146 0
> -10.33999 -10.25158 19.83222 0
> -5.900752 -4.774312 22.25315 0
> -3.473342 -2.116564 60.31918 0
> -5.51118 -8.684725 45.30108 1
> -4.393883 -3.597137 21.0572 0
> -3.671957 -3.355143 51.05236 1
> -7.700621 -7.257176 29.59876 1
> -6.959113 -5.834087 21.52065 1
> -6.978306 -6.291922 26.17615 0
> -3.525233 -0.2435265 39.66356 0
> -8.017325 -7.190228 16.78984 1
> -9.686805 -6.356866 24.96812 1
> -5.841892 -4.090017 12.90826 1
> -4.101501 -0.8392091 29.49425 1
> -0.50966 -0.6248183 72.55316 0
> -2.747329 -3.107922 70.82893 1
> -3.682684 -5.461088 7.237332 0
> -1.726765 -1.030436 51.13756 0
> -5.065511 -5.105534 48.8038 1
> -3.490172 -0.8473139 54.89489 1
> -14.56848 -13.29985 8.508147 1
> -5.511615 -2.257046 26.53605 1
> -0.80373 -1.259443 54.58532 1
> -11.76727 -10.51294 19.43544 0
> -4.924498 -5.660692 64.22583 1
> -1.662102 -1.329681 68.50871 0
> -2.225776 -1.191363 46.14959 1
> -11.97834 -1.471152 18.86225 0
> -9.986734 -8.210676 15.11784 1
> -0.78368 -0.2543859 64.04224 1
> -11.41681 -13.24663 9.016961 1
> -10.73357 -13.46118 31.8038 1
> -2.443766 -0.841536 35.3982 1
> -3.112007 -1.327887 32.61596 1
> -1.647414 -0.9874625 65.37144 0
> -3.771582 -2.685039 42.65498 0
> -5.503803 -6.65314 15.60404 1
> -6.844056 -10.59976 22.71807 1
> -3.977231 -6.444871 47.65485 1
> -0.43918 -1.813655 35.90933 1
> -4.520459 -3.337119 17.47536 1
> -3.102405 -2.276846 15.49771 1
> -3.173711 -4.548148 54.85541 1
> -4.157713 -2.368944 36.82358 1
> -6.671762 -6.863191 33.18528 1
> -5.806525 -8.300102 38.04575 1
> -9.137906 -10.43044 20.62558 1
> -4.830114 -5.035967 80.04454 1
> -6.717423 -7.807728 18.62613 1
> -1.654782 -2.814744 69.35754 1
> -5.718936 -5.041555 19.44518 1
> -1.139612 -1.246455 31.46728 1
> -5.193422 -4.141603 49.06763 0
> -0.72360 -1.519114 68.06107 1
> -3.45456 -2.324488 24.8586 1
> -3.946017 -1.809939 26.39728 1
> -1.373865 -1.385224 59.31034 0
> -12.91463 -16.81217 21.9325 1
> -7.101114 -4.463167 24.6039 1
> -11.19178 -7.923832 11.70692 1
> -6.337176 -3.290151 46.2829 1
> -6.034304 -6.688771 12.98928 1
> -10.72616 -16.16286 27.24244 1
> -10.01076 -11.90333 16.67032 1
> -2.85405 -1.064295 18.82794 1
> -3.582814 -3.041154 34.58895 0
> -0.88143 -2.513154 72.57123 0
> -2.936312 -2.92483 32.65664 0
> -2.859565 -7.337652 31.87842 1
> -4.467122 -6.427214 56.81916 0
> -6.340817 -6.706052 9.87694 1
> -1.40155 -2.738037 35.32452 1
> -10.92032 -11.05833 30.82691 1
> -7.330603 -6.257256 22.16484 1
> -2.714168 -2.258151 36.30459 0
> -2.793682 -2.935043 56.51117 1
> -6.706202 -11.04426 11.10245 0
> -6.113976 -7.36745 11.36128 1
> -9.845764 -10.35044 37.52305 0
> -7.786937 -10.70406 21.68431 1
> -0.54450 -3.818708 64.34981 1
> -1.402748 -4.612042 52.94871 0
> -1.771809 -3.918717 41.45876 1
> -4.142132 -7.088901 45.44987 1
> -1.640578 -4.787658 40.82234 1
> -1.050637 -2.535334 42.87785 1
> -0.32151 -3.315413 40.40543 1
您需要提供 comparitor 函数，用于在排序期间比较单个 行 。希望你有一个现代版本的 C++ 并且可以使用 lambdas:
// Sort by first column
std::sort( samples.begin(), samples.end(),
[]( const sampleData& a, const sampleData& b )
{
return a.first < b.first;
}
);
按特定列排序后，您可以遍历序列并计算连续的第四列 1s。
编辑
lambda 是一种即时创建函数对象的方法。以上等同于：
struct unnamed_function
{
bool operator () ( const sampleData& a, const sampleData& b ) const
{
return a.first < b.first;
}
};
...
std::sort( samples.begin(), samples.end(), unnamed_function() );
[]是引入lambda的“捕获列表”
Read more about lambdas. （抱歉，我还没有更好的常见问题解答...）
* 大致如此。幕后其实比这个复杂一点。