Question

我不确定对表进行硬编码的方式是否是最好的方式。但是无论如何，我正在尝试将“总积分”列除以获取每分钟积分所需的时间，然后为该计算创建一个新列，但我似乎无法使其工作。

table <- data.frame(list(Question=c("Q1", "Q2", "Q3", "Q4" , "Q5" , "Q6" , "Q7" , "Q8", "Q9"), 
                        Total_Points=c("21","5","10","14","5","5","10","5","5"), Time_needed=c("24","7","15","12","4","3","10","5","6")))
table <- transform(table, Points_per_min = table$Total_Points / table$Time_needed)

给我一个错误“警告消息：在Ops.factor（table $ Total_Points，table $ Time_needed）中： ‘/’对因素没有意义”

非常感谢您的帮助！

Answer 1

此处的警告是有益的，您可以通过检查str()中的table来看到。它告诉您所有变量都是类型因子。 Check here可以快速介绍不同的数据类型。

将您的数据转换为数字，然后进行数学计算：

table <- data.frame(list(Question=c("Q1", "Q2", "Q3", "Q4" , "Q5" , "Q6" , "Q7" , "Q8", "Q9"), 
                         Total_Points=c("21","5","10","14","5","5","10","5","5"), Time_needed=c("24","7","15","12","4","3","10","5","6")))
str(table)
#> 'data.frame':    9 obs. of  3 variables:
#>  $ Question    : Factor w/ 9 levels "Q1","Q2","Q3",..: 1 2 3 4 5 6 7 8 9
#>  $ Total_Points: Factor w/ 4 levels "10","14","21",..: 3 4 1 2 4 4 1 4 4
#>  $ Time_needed : Factor w/ 9 levels "10","12","15",..: 4 9 3 2 6 5 1 7 8
table$Total_Points <- as.numeric(as.character(table$Total_Points))
table$Time_needed <- as.numeric(as.character(table$Time_needed))
table$Points_per_min <- table$Total_Points / table$Time_needed
table
#>   Question Total_Points Time_needed Points_per_min
#> 1       Q1           21          24      0.8750000
#> 2       Q2            5           7      0.7142857
#> 3       Q3           10          15      0.6666667
#> 4       Q4           14          12      1.1666667
#> 5       Q5            5           4      1.2500000
#> 6       Q6            5           3      1.6666667
#> 7       Q7           10          10      1.0000000
#> 8       Q8            5           5      1.0000000
#> 9       Q9            5           6      0.8333333

^{由reprex package（v0.2.1）于2019-02-10创建}

Answer 2

@Chase已经解释了为什么以前无法创建新列，但是如果您想要一个涉及更少代码行的解决方案，则可以使用public class LambdaHandler implements RequestHandler<S3Event, String> { // Code to fetch S3 object name here which is needed in application processing logic further }包（我正在使用dplyr版本0.7.6）。

如@Ozgur Yigit所述，您无需将数据作为列表传递。实际上，您可以仅将每一列作为dplyr的参数来传递。

编辑

data.frame(...)

有关library(dplyr) table <- data.frame(Question = c("Q1", "Q2", "Q3", "Q4" , "Q5" , "Q6" , "Q7" , "Q8", "Q9"), Total_Points = c("21","5","10","14","5","5","10","5","5"), Time_needed = c("24","7","15","12","4","3","10","5","6")) table %>% mutate_at(vars(Total_Points, Time_needed), funs(as.numeric(as.character(.)))) %>% # converts the `Total_Points` and `Time_needed` cols to character first (b/c factors coded as integers under the hood) then to numeric mutate(Points_per_min = Total_Points / Time_needed)的更多信息：https://r4ds.had.co.nz/transform.html#add-new-variables-with-mutate

如果我的回答无济于事或不在SO响应范围之内，则表示歉意。我仍在学习如何在SO上写答案。

Answer 3

您可以使用tidyverse，但只需要一个mutate语句-与其他应答者几乎没有什么不同：

table2 <- table %>%
     mutate(Total_Points = as.numeric(as.character(Total_Points)),
     Time_needed  = as.numeric(as.character(Time_needed)),
     Points_per_min = Total_Points / Time_needed)

  Question Total_Points Time_needed Points_per_min
1       Q1           21          24      0.8750000
2       Q2            5           7      0.7142857
3       Q3           10          15      0.6666667
4       Q4           14          12      1.1666667
5       Q5            5           4      1.2500000
6       Q6            5           3      1.6666667
7       Q7           10          10      1.0000000
8       Q8            5           5      1.0000000
9       Q9            5           6      0.8333333

如何获取两列数据，执行计算并使用该数据创建新列

3 个答案: