用户在r中输入函数参数



我试图接受用户输入的函数作为参数,但我很难使其工作。我的实际代码是:

t.student.2code <- function()
{
  f <- readline(prompt="Select column group A:")
  y <- readline(prompt="Select column group B:")

  t.test(f,y)

  }

正常t.test

t.test(beaver1$time,beaver1$temp)
    Welch Two Sample t-test
data:  beaver1$time and beaver1$temp
t = 19.399, df = 113, p-value < 2.2e-16
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 1144.924 1405.387
sample estimates:
mean of x  mean of y 
1312.01754   36.86219 

在我的函数中使用beaver1$time和beaver1$1temp作为参数,不起作用

Error in t.test.default(x, y) : not enough 'x' observations
Inoltre: Warning messages:
1: In mean.default(x) : argument is not numeric or logical: returning NA
2: In var(x) : si è prodotto un NA per coercizione

您的代码看起来不错。请注意,readline()返回字符串,因此函数t.test必须接受两个字符串参数。

简单示例:

testinp <- function() {
  c1 <- readline()
  c2 <- readline()
  sum(as.numeric(c1), as.numeric(c2))
}

如果您从stats包中管理t.test,那么传递字符串将不会像它期望的数字向量那样工作。也许您想指定变量名,如下面的示例?

c1 <- c(1,2,3)
c2 <- c(2,4,5)
testinp <- function() {
  tmp1 <- readline()
  tmp2 <- readline()
  t.test(get(tmp1), get(tmp2))
}
testinp() # enter "c1" and "c2"

要处理您的问题示例:

testinp <- function() {
  c1 <- readline()
  c2 <- readline()
  t.test(eval(parse(text=c1)), eval(parse(text=c2)))
}
testinp() # enter "beaver1$time" and "beaver1$temp"
# default data frame (df)
f <- c(1,2,3)
g <- c(3,5,6)
df <- data.frame(f,g)
tfunc1 <- function() {
    print("Available column names:")
    print(names(df))
    col1 <- readline(prompt="Enter column 1 name:n")[[1]]
    col2 <- readline(prompt="Enter column 2 name:n")[[1]]
    t.test(df[[col1]], df[[col2]])
}
tfunc2 <- function(localdf) {
    print("Available column names:")
    print(names(localdf))
    col1 <- readline(prompt="Enter column 1 name:n")[[1]]
    col2 <- readline(prompt="Enter column 2 name:n")[[1]]
    t.test(localdf[[col1]], localdf[[col2]])
}
tfunc1()
[1] "Available column names:"
[1] "f" "g"
Enter column 1 name:
f
Enter column 2 name:
g
    Welch Two Sample t-test
data:  df[[col1]] and df[[col2]]
t = -2.5298, df = 3.4483, p-value = 0.07459
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -5.7875506  0.4542173
sample estimates:
mean of x mean of y 
 2.000000  4.666667 
tfunc2(df)
[1] "Available column names:"
[1] "f" "g"
Enter column 1 name:
g
Enter column 2 name:
f
    Welch Two Sample t-test
data:  localdf[[col1]] and localdf[[col2]]
t = 2.5298, df = 3.4483, p-value = 0.07459
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.4542173  5.7875506
sample estimates:
mean of x mean of y 
 4.666667  2.000000 

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