Question

我有一些看起来像这样的数据：

  Year Revenue Cost Rent
1 2016    3000    4  100
2 2017    4000    5  100
3 2018    5000    6  100

df <- data.frame(stringsAsFactors=FALSE,
        Year = c(2016L, 2017L, 2018L),
     Revenue = c(3000L, 4000L, 5000L),
        Cost = c(4L, 5L, 6L),
        Rent = c(100L, 100L, 100L)
)

我想将所有说的话除以Rent的百分比：

library(dplyr)
df <- df %>% mutate_at(vars(Revenue:Rent), funs(. /Rent))

效果很好。

  Year Revenue Cost Rent
1 2016      30 0.04    1
2 2017      40 0.05    1
3 2018      50 0.06    1

唯一的是：我已经失去了原来的专栏。

如何进行mutate_all，以便有新的列，称为Revenue_percentage_of_rent，Cost_percentage_of_rent？

Answer 1

在public class StageDrag extends Application { @Override public void start(Stage primaryStage) { Pane root = new Pane(); Scene scene = new Scene(root, 300, 250); primaryStage.setTitle("Hello World!"); primaryStage.setScene(scene); primaryStage.addEventFilter(MouseEvent.MOUSE_RELEASED, e ->{System.out.println("released");}); //scene.addEventHandler(MouseEvent.MOUSE_RELEASED, e ->{System.out.println("released2");}); //root.addEventHandler(MouseEvent.MOUSE_RELEASED, e ->{System.out.println("released3");}); primaryStage.show(); } /** * @param args the command line arguments */ public static void main(String[] args) { launch(args); }

中命名函数中的列

#include <stdio.h>

int main() {
    int n;
    printf("Enter width: ");
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
       for (int j = 0; j < n; j++) {
           if (i == j || (i + j + 1) == n) {
               printf("X");
           } else {
               printf(" ");
           }
       }
       printf("\n");
    }
    return 0;
}

您可以使用void recursiveProblem(int num, int max) { if (num <= (max / 2 + 1)) ... }来做到这一点，但是它还会将import csv import pandas as pd file_name = ['sample_file1.csv', 'sample_file2.csv', 'sample_file3.csv'] content = [] for n in file_name: with open(n) as f: reader = csv.reader(f, delimiter=',') content += list(reader)[1:] # If the first row contains headers df = pd.DataFrame(content) df.columns = ['Sensor ID' , 'Time Instant' , 'Measurement']列除以mutate_at，我想您不需要。

使用library(dplyr) df %>% mutate_at(vars(Revenue:Rent), funs(percentage_of_rent = . /Rent))的解决方法是

mutate_all

但是您在此处失去了Year列。

Answer 2

不赞成使用funs，而推荐使用list中的dplyr_0.8.0，因此，选择是

library(dplyr)
df %>%
    mutate_at(vars(Revenue:Rent), list(percentage_of_rent = ~  ./Rent))
#  Year Revenue Cost Rent Revenue_percentage_of_rent Cost_percentage_of_rent Rent_percentage_of_rent
#1 2016    3000    4  100                         30                    0.04                       1
#2 2017    4000    5  100                         40                    0.05                       1
#3 2018    5000    6  100                         50                    0.06                       1

使用mutate_all

2 个答案: