检查列表中的所有元素是否在R中相等-checkwhetherallelementsofalistareinequalinR

13  

How about

allSame <- function(x) length(unique(x)) == 1

allSame(test_true)
# [1] TRUE
allSame(test_false)
# [1] FALSE

As @JoshuaUlrich pointed out below, unique may be slow on lists. Also, identical and unique may use different criteria. Reduce is a function I recently learned about for extending pairwise operations:

正如@JoshuaUlrich在下面指出的那样,名单上的独特可能会很慢。而且,相同且唯一的可以使用不同的标准。 Reduce是我最近学习的用于扩展成对操作的函数:

identicalValue <- function(x,y) if (identical(x,y)) x else FALSE
Reduce(identicalValue,test_true)
# [1] 1 2 3
Reduce(identicalValue,test_false)
# [1] FALSE

This inefficiently continues making comparisons after finding one non-match. My crude solution to that would be to write else break instead of else FALSE, throwing an error.

在找到一个不匹配后,这无效地继续进行比较。我的原始解决方案是写else break而不是FALSE,抛出一个错误。

3  

I woud do:

我做了:

all.identical <- function(l) all(mapply(identical, head(l, 1), tail(l, -1)))

all.identical(test_true)
# [1] TRUE
all.identical(test_false)
# [1] FALSE

1  

To summarize the solutions. Data for the tests:

总结解决方案。测试数据:

x1 <- as.list(as.data.frame(replicate(1000, 1:100)))
x2 <- as.list(as.data.frame(replicate(1000, sample(1:100, 100))))

Solutions:

comp_list1 <- function(x) length(unique.default(x)) == 1L
comp_list2 <- function(x) all(vapply(x[-1], identical, logical(1L), x = x[[1]]))
comp_list3 <- function(x) all(vapply(x[-1], function(x2) all(x[[1]] == x2), logical(1L)))
comp_list4 <- function(x) sum(duplicated.default(x)) == length(x) - 1L

Test on the data:

测试数据:

for (i in 1:4) cat(match.fun(paste0("comp_list", i))(x1), " ")
#> TRUE  TRUE  TRUE  TRUE   
for (i in 1:4) cat(match.fun(paste0("comp_list", i))(x2), " ")
#> FALSE  FALSE  FALSE  FALSE  

Benchmarks:

library(microbenchmark)
microbenchmark(comp_list1(x1), comp_list2(x1), comp_list3(x1), comp_list4(x1))
#> Unit: microseconds
#>            expr      min        lq      mean   median        uq      max neval cld
#>  comp_list1(x1)  138.327  148.5955  171.9481  162.013  188.9315  269.342   100 a  
#>  comp_list2(x1) 1023.932 1125.2210 1387.6268 1255.985 1403.1885 3458.597   100  b 
#>  comp_list3(x1) 1130.275 1275.9940 1511.7916 1378.789 1550.8240 3254.292   100   c
#>  comp_list4(x1)  138.075  144.8635  169.7833  159.954  185.1515  298.282   100 a  
microbenchmark(comp_list1(x2), comp_list2(x2), comp_list3(x2), comp_list4(x2))
#> Unit: microseconds
#>            expr     min        lq      mean   median        uq      max neval cld
#>  comp_list1(x2) 139.492  140.3540  147.7695  145.380  149.6495  218.800   100  a 
#>  comp_list2(x2) 995.373 1030.4325 1179.2274 1054.711 1136.5050 3763.506   100   b
#>  comp_list3(x2) 977.805 1029.7310 1134.3650 1049.684 1086.0730 2846.592   100   b
#>  comp_list4(x2) 135.516  136.4685  150.7185  139.030  146.7170  345.985   100  a

As we see the most efficient solutions based on the duplicated and unique functions.

我们看到基于重复和独特功能的最有效解决方案。

-1  

this also works

这也行得通

m <- combn(length(test_true),2)

for(i in 1:ncol(m)){
    print(all(test_true[[m[,i][1]]] == test_true[[m[,i][2]]]))
    }

-1  

PUtting in my self-promoting suggestion for cgwtools::approxeq which essentially does what all.equal does but returns a vector of logical values indicating equality or not.

在我对cgwtools :: approxeq的自我推销建议中提出的,它基本上执行all.equal所做的但是返回一个逻辑值向量,表示是否相等。

So: depends whether you want exact equality or floating-point-representational equality.

所以:取决于您是否需要精确相等或浮点表示相等。