r语言 - expect_equal() 按一个顺序传递参数，如果交换参数，则失败

Question

当尝试将testthat::expect_equal()与两个数字和一个容差参数一起使用时，当参数按特定顺序时，它会通过，但如果两个数字交换参数位置，则会失败。 我注意到这个函数使用基本包中的all.equal()引用，并且在交换参数时该函数也具有通过/失败差异。

无论两个函数的前两个参数的顺序如何，我都期望得到相同的答案。 如果这不是正确的期望，请告诉我。

library(testthat)
# expect_equal does not throw error with one pair of numbers to compare
expect_equal(5, 1, tolerance=1)
# But does when the two numbers are reversed in their arguments
tryCatch(expect_equal(1, 5, tolerance=1), expectation_failure=conditionMessage)
#> [1] "1 not equal to 5.n1/1 mismatchesn[1] 1 - 5 == -4n"
# Since this seems to reference `all.equal()` I tried there too, and see an issue:
all.equal(1, 5, tolerance=1)
#> [1] "Mean absolute difference: 4"
all.equal(5, 1, tolerance=1)
#> [1] TRUE
# My session info:
sessionInfo()
#> R version 3.3.3 (2017-03-06)
#> Platform: x86_64-w64-mingw32/x64 (64-bit)
#> Running under: Windows 7 x64 (build 7601) Service Pack 1
#> 
#> locale:
#> [1] LC_COLLATE=English_United States.1252 
#> [2] LC_CTYPE=English_United States.1252   
#> [3] LC_MONETARY=English_United States.1252
#> [4] LC_NUMERIC=C                          
#> [5] LC_TIME=English_United States.1252    
#> 
#> attached base packages:
#> [1] stats     graphics  grDevices utils     datasets  methods   base     
#> 
#> other attached packages:
#> [1] testthat_1.0.2
#> 
#> loaded via a namespace (and not attached):
#>  [1] backports_1.0.5 R6_2.2.1        magrittr_1.5    rprojroot_1.2  
#>  [5] tools_3.3.3     htmltools_0.3.6 yaml_2.1.14     crayon_1.3.2   
#>  [9] Rcpp_0.12.10    stringi_1.1.5   rmarkdown_1.5   knitr_1.15.1   
#> [13] stringr_1.2.0   digest_0.6.12   evaluate_0.10

Answer 1

tl;博士

您的容忍度大于

mean(abs(target-current)) / abs(target)
mean(5 - 1) / 5
4 / 5
0.8  

因此，该函数在比较0.8 < 1后返回，因为您已允许它通过此公差检查

---

完整答案

all.equal()函数具有参数target和current

如果您深入研究代码，当target为 5 且current为 1 时，它会计算平均绝对差值

mean(abs(target - current))
# 4

然后是目标和公差之间的比较，在这种情况下是

5 > 1
## TRUE

由于这是 TRUE，因此它会计算相对差异

4 / 5 
## 0.8

在这里，0.8 不大于公差，因此它返回带有TRUE的函数，即

all.equal(5, 1, tolerance = 1)
# [1] TRUE

因此，您的tolerance值用于比较相对差异。这就是为什么公差默认为较小的值1.5e-8

在这里，我从all.equal.numeric中获取了相关代码，仅删除了感兴趣的部分，以便您可以更清楚地看到它

allEqual <- function (target, current, tolerance = sqrt(.Machine$double.eps)) 
{
msg <- NULL
target <- as.vector(target)
current <- as.vector(current)
out <- is.na(target)
out <- out | target == current
if (all(out)) 
return(if (is.null(msg)) TRUE else msg)
target <- target[!out]
current <- current[!out]
xy <- mean(abs(target - current))
## THIS BIT HERE vv
what <- {
xn <- mean(abs(target))
# print(paste0("xn: ", xn))
if (is.finite(xn) && xn > tolerance) {
#print("xn (target) is GREATER than the tolerance")
xy <- xy/xn
#print(paste0("relative difference: ", xy))
"relative"
}
else{
# print("xn (target) is SMALLER than the tolerance")
"absolute"
} 
}
## THIS BIT HERE ^^
if (is.na(xy) || xy > tolerance) 
#print("xy is GREATER than the tolerance")
msg <- c(msg, paste("Mean", what, "difference:", format(xy)))
if (is.null(msg)) 
TRUE
else msg
}

---

allEqual(5, 1, tolerance = 0.79)
# [1] "Mean relative difference: 0.8"
allEqual(5, 1, tolerance = 0.81)
# [1] TRUE
allEqual(5, 1, tolerance = 1)
# [1] TRUE
allEqual(1, 5, tolerance = 1)
# [1] "Mean absolute difference: 4"