Question

有没有更好的方法将循环中生成的数据存储为data.frame（但仍使用循环）？

脚本：

import javafx.application.Application;
import javafx.geometry.Insets;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.image.ImageView;
import javafx.scene.layout.Background;
import javafx.scene.layout.BackgroundFill;
import javafx.scene.layout.CornerRadii;
import javafx.scene.layout.GridPane;
import javafx.scene.layout.HBox;
import javafx.scene.layout.Priority;
import javafx.scene.layout.RowConstraints;
import javafx.scene.layout.VBox;
import javafx.scene.paint.Color;
import javafx.stage.Stage;

public class Main extends Application {

    @Override
    public void start(Stage primaryStage) {
        GridPane root = new GridPane();

        RowConstraints rc = new RowConstraints();
        rc.setVgrow(Priority.ALWAYS);

        root.getRowConstraints().add(rc);

        root.setBackground(new Background(new BackgroundFill(Color.RED, CornerRadii.EMPTY, Insets.EMPTY)));
        HBox hBox = new HBox();
        hBox.setBackground(new Background(new BackgroundFill(Color.GREEN, CornerRadii.EMPTY, Insets.EMPTY)));
        hBox.setMinHeight(0);
        ImageView imageView = new ImageView("http://www.calgary.ca/CA/city-manager/scripts/about-us/webparts/images/ourHistory_retina.jpg");
        imageView.fitHeightProperty().bind(hBox.heightProperty());
        imageView.setPreserveRatio(true);
        hBox.getChildren().add(imageView);
        root.add(hBox, 0, 0);         

        primaryStage.setScene(new Scene(root, 300, 275));
        primaryStage.show();
    }


    public static void main(String[] args) {
        launch(args);
    }
}

Answer 1

如果你真的想使用一个循环，猜猜这个有效

Results <- data.frame()
Mystep <- 2 

for(i in seq(0, 10, by = Mystep  )){
  xy = c(i*10, i/10)

  Results = rbind(Results, xy)    
}

names(Results) = c("X", "Y")
> Results
    X  Y
1   0  0.0
2  20  0.2
3  40  0.4
4  60  0.6
5  80  0.8
6 100  1.0

Answer 2

使用循环不能有效地使用R.矢量化解决方案是正确的答案。但由于这已不在考虑范围之内，我认为双循环解决方案并不符合要求：

data.frame(
  sapply(seq(0, 10, by = Mystep), `*`, 10)),
  sapply(seq(0, 10, by = Mystep), `/`, 10))
)

或（清洁）

itr <- seq(0, 10, by = Mystep)
data.frame(sapply(itr, `*`, 10)), sapply(itr, `/`, 10)))

最后是一个for循环：

output <- data.frame(numeric(length(itr)), numeric(length(itr)))
for(i in itr) {
  output[i, 1] <- itr*10
  output[i, 2] <- itr/10
}

R-将数据作为data.frame存储在循环中

2 个答案: