Question

我使用ks.test（）比较两个向量（data_A_score，data_B_score）和另一个向量K1，我得到这个结果：

score_ref_k1 <- c(0.09651, 0.09543, 0.09122, 0.09458, 0.09382, 0.10158, 0.10339, 
  0.13594, 0.09458, 0.09296)
data_A_score_src <- c(0.09293, 0.09838, 0.09866, 0.10866, 0.09726, 0.10731, 
  0.09866, 0.09398, 0.10007, 0.10408)
data_B_score_src <- c(0.04741, 0.0621, 0.09606, 0.08851, 0.05063, 0.39775, 0.05509, 
  0.10784, 0.0468, 0.04782)
ks.test(data_A_score_src, score_ref_k1, exact = FALSE, alternative = "g")
#> Warning in ks.test(data_A_score_src, score_ref_k1, exact = FALSE,
#> alternative = "g"): p-value will be approximate in the presence of ties
#> 
#>  Two-sample Kolmogorov-Smirnov test
#> 
#> data:  data_A_score_src and score_ref_k1
#> D^+ = 0.1, p-value = 0.9048
#> alternative hypothesis: the CDF of x lies above that of y

ks.test(data_B_score_src, score_ref_k1, exact = FALSE, alternative = "g")
#> Warning in ks.test(data_B_score_src, score_ref_k1, exact = FALSE,
#> alternative = "g"): p-value will be approximate in the presence of ties
#> 
#>  Two-sample Kolmogorov-Smirnov test
#> 
#> data:  data_B_score_src and score_ref_k1
#> D^+ = 0.7, p-value = 0.007447
#> alternative hypothesis: the CDF of x lies above that of y

然后我尝试用purrr::map做同样的方法。首先这是数据：

library(tidyverse)

all_comb <- structure(list(src = structure(1:2, .Label = c("data_B", "data_A"), 
  class = "factor"), ref = structure(c(1L, 1L), .Label = "K1", class = "factor"), 
  data = list(structure(list(score_ref = c(0.09651, 0.09543, 0.09122, 0.09458, 
    0.09382, 0.10158, 0.10339, 0.13594, 0.09458, 0.09296), score_src = c(0.04741, 
    0.0621, 0.09606, 0.08851, 0.05063, 0.39775, 0.05509, 0.10784, 0.0468, 
    0.04782)), row.names = c(NA, -10L), class = c("tbl_df", "tbl", "data.frame"), 
    .Names = c("score_ref", "score_src")), structure(list(score_ref = c(0.09651, 
    0.09543, 0.09122, 0.09458, 0.09382, 0.10158, 0.10339, 0.13594, 0.09458, 
    0.09296), score_src = c(0.09293, 0.09838, 0.09866, 0.10866, 0.09726, 
    0.10731, 0.09866, 0.09398, 0.10007, 0.10408)), row.names = c(NA, -10L), 
    class = c("tbl_df", "tbl", "data.frame"), .Names = c("score_ref", "score_src")))), 
  class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -2L), .Names = c("src", 
    "ref", "data"))

# just showing the content
all_comb %>% unnest(data)
#> # A tibble: 20 × 4
#>       src    ref score_ref score_src
#>    <fctr> <fctr>     <dbl>     <dbl>
#> 1  data_B     K1   0.09651   0.04741
#> 2  data_B     K1   0.09543   0.06210
#> 3  data_B     K1   0.09122   0.09606
#> 4  data_B     K1   0.09458   0.08851
#> 5  data_B     K1   0.09382   0.05063
#> 6  data_B     K1   0.10158   0.39775
#> 7  data_B     K1   0.10339   0.05509
#> 8  data_B     K1   0.13594   0.10784
#> 9  data_B     K1   0.09458   0.04680
#> 10 data_B     K1   0.09296   0.04782
#> 11 data_A     K1   0.09651   0.09293
#> 12 data_A     K1   0.09543   0.09838
#> 13 data_A     K1   0.09122   0.09866
#> 14 data_A     K1   0.09458   0.10866
#> 15 data_A     K1   0.09382   0.09726
#> 16 data_A     K1   0.10158   0.10731
#> 17 data_A     K1   0.10339   0.09866
#> 18 data_A     K1   0.13594   0.09398
#> 19 data_A     K1   0.09458   0.10007
#> 20 data_A     K1   0.09296   0.10408

然后我用这个计算相同的测试：

all_comb %>%
      mutate(vt = purrr::map(data,  ~ks.test(.$score_ref, .$score_src, exact=FALSE, alternative="g")),
      tidied = purrr::map(vt, broom::tidy)) %>%
      tidyr::unnest(tidied) %>%
      select(src, ref, p.value)

给出了：

# A tibble: 2 × 3
     src    ref   p.value
  <fctr> <fctr>     <dbl>
1 data_B     K1 0.9048374
2 data_A     K1 0.0820850

注意与最高结果的差异：

#> data:  data_A_score_src and score_ref_k1
#> D^+ = 0.1, p-value = 0.9048
#> data:  data_B_score_src and score_ref_k1
#> D^+ = 0.7, p-value = 0.007447

为什么？我怎样才能修复我的purrr，dplyr方法？

Answer 1

如果你在ks.test中切换得分和测试，你会得到相同的值：

all_comb %>%
      mutate(vt = purrr::map(data,  ~ks.test(.$score_src,.$score_ref,  exact=FALSE, alternative="g")),
      tidied = purrr::map(vt, broom::tidy)) %>%
      tidyr::unnest(tidied) %>%
      select(src, ref, p.value) 
# A tibble: 2 <U+00D7> 3
      src    ref     p.value
   <fctr> <fctr>       <dbl>
1 bcdidsp     K1 0.007446583
2 bcdipsp     K1 0.904837418

purrr :: map和dplyr的组合使用简单的统计测试得出不一致的结果

1 个答案: