我有两个数据框：
src_tbl <- structure(list(Sample_name = c("S1", "S2", "S1", "S2", "S1",
"S2"), crt = c(0.079, 0.082, 0.079, 0.082, 0.079, 0.082), sr = c(0.592,
0.549, 0.592, 0.549, 0.592, 0.549), condition = c("x1", "x1",
"x2", "x2", "x3", "x3"), score = c("0.077", "0.075", "0.483",
"0.268", "0.555", "0.120")), row.names = c(NA, -6L), .Names = c("Sample_name",
"crt", "sr", "condition", "score"), class = c("tbl_df",
"tbl", "data.frame"))
src_tbl
#> Sample_name crt sr condition score
#> 1 S1 0.079 0.592 x1 0.077
#> 2 S2 0.082 0.549 x1 0.075
#> 3 S1 0.079 0.592 x2 0.483
#> 4 S2 0.082 0.549 x2 0.268
#> 5 S1 0.079 0.592 x3 0.555
#> 6 S2 0.082 0.549 x3 0.120
ref_tbl <- structure(list(Sample_name = c("P1", "P2", "P3", "P1", "P2",
"P3", "P1", "P2", "P3"), crt = c(1, 1, 1, 1, 1, 1, 1, 1, 1),
sr = c(2, 2, 2, 2, 2, 2, 2, 2, 2), condition = c("r1", "r1",
"r1", "r2", "r2", "r2", "r3", "r3", "r3"), score = c("0.200",
"0.201", "0.199", "0.200", "0.202", "0.200", "0.200", "0.204",
"0.197")), row.names = c(NA, -9L), .Names = c("Sample_name",
"crt", "sr", "condition", "score"), class = c("tbl_df",
"tbl", "data.frame"))
ref_tbl
#> Sample_name crt sr condition score
#> 1 P1 1 2 r1 0.200
#> 2 P2 1 2 r1 0.201
#> 3 P3 1 2 r1 0.199
#> 4 P1 1 2 r2 0.200
#> 5 P2 1 2 r2 0.202
#> 6 P3 1 2 r2 0.200
#> 7 P1 1 2 r3 0.200
#> 8 P2 1 2 r3 0.204
#> 9 P3 1 2 r3 0.197
我想要做的是在两个数据框中按ks.test()分组的score列执行操作（Sample_name）。例如，S1和P1的KS测试的p值是：
# in src_tbl
s1 <- c(0.077,0.483,0.555)
#in ref_tbl
p1 <- c(0.200,0.200,0.200)
testout <- ks.test(s1,p1)
#> Warning in ks.test(s1, p1): cannot compute exact p-value with ties
broom::tidy(testout)
#> statistic p.value method alternative
#> 1 0.6666667 0.5175508 Two-sample Kolmogorov-Smirnov test two-sided
我想做的是对所有操作执行所有操作，以便最终得到这样的表
src ref p.value
S1 P1 0.5175508
S1 P2 0.6
S1 P3 0.6
S2 P1 0.5175508
S2 P2 0.6
S2 P3 0.6
我该怎么做？由于ref_table中的样本数量可能很大（P1，P2 ...... P10k），因此可以更快。
答案 0 :(得分：5)
以下是tidyverse中的解决方案。我首先在每个源数据集中嵌套得分：
ref_tbl <- ref_tbl %>%
mutate(ref = as.factor(Sample_name),
score_ref = as.numeric(score)) %>%
select(ref, score_ref) %>%
tidyr::nest(score_ref)
ref_tbl
# A tibble: 3 x 2
ref data
<fctr> <list>
1 P1 <tibble [3 x 1]>
2 P2 <tibble [3 x 1]>
3 P3 <tibble [3 x 1]>
src_tbl <- src_tbl %>%
mutate(src = as.factor(Sample_name),
score_src = as.numeric(score)) %>%
select(src, score_src) %>%
tidyr::nest(score_src)
src_tbl
# A tibble: 2 x 2
src data
<fctr> <list>
1 S1 <tibble [3 x 1]>
2 S2 <tibble [3 x 1]>
然后我创建一个包含所有样本名称组合的网格：
all_comb <- as_data_frame(expand.grid(src = src_tbl$src, ref = ref_tbl$ref))
all_comb
# A tibble: 6 x 2
src ref
<fctr> <fctr>
1 S1 P1
2 S2 P1
3 S1 P2
4 S2 P2
5 S1 P3
6 S2 P3
现在，我们可以使用嵌套数据加入，然后绑定列，因此每个组合都必须有一个包含分数的列表列。
all_comb <- all_comb %>%
left_join(ref_tbl, by = "ref") %>%
left_join(src_tbl, by = "src") %>%
mutate(data = purrr::map2(data.x, data.y, bind_cols)) %>%
select(-data.x, -data.y)
all_comb
# A tibble: 6 x 3
src ref data
<fctr> <fctr> <list>
1 S1 P1 <tibble [3 x 2]>
2 S2 P1 <tibble [3 x 2]>
3 S1 P2 <tibble [3 x 2]>
4 S2 P2 <tibble [3 x 2]>
5 S1 P3 <tibble [3 x 2]>
6 S2 P3 <tibble [3 x 2]>
最后，如果每个数据集都映射ks.test，请使用扫帚获取请求的p.value。
final <- all_comb %>%
mutate(ks = purrr::map(data, ~ks.test(.$score_ref, .$score_src)),
tidied = purrr::map(ks, broom::tidy)) %>%
tidyr::unnest(tidied) %>%
select(src, ref, p.value)
Warning message: cannot compute exact p-value with ties
Warning message: cannot compute exact p-value with ties
final
# A tibble: 6 x 3
src ref p.value
<fctr> <fctr> <dbl>
1 S1 P1 0.5175508
2 S2 P1 0.5175508
3 S1 P2 0.6000000
4 S2 P2 0.6000000
5 S1 P3 0.6000000
6 S2 P3 0.6000000
答案 1 :(得分：1)
好吧花了一段时间，但我拼凑了一个hacky解决方案。我确信有ddply这样的优雅方式，但这超出了我。 （注意，当我缩短其中一个数据框时，我的p值与你的p值略有不同）
library(dplyr)
library(tidyr)
ref_tbl<-ref_tbl[1:6,]#make equal rows for this example
dd<-as.data.frame(cbind(paste(src_tbl$Sample_name,'-', src_tbl$score),
paste(ref_tbl$Sample_name,'-',ref_tbl$score)))#concatenate sample names with their scores
ex<-expand.grid(x = levels(dd$V1), y = levels(dd$V2))#obtain all combinations
all<-ex %>%
separate(x, c("S","svalue"),"-")%>%
separate(y, c("P","pvalue"),"-")#unseparate now that we have the combinations
all$svalue<-as.numeric(all$svalue)#change to numeric for ks.test
all$pvalue<-as.numeric(all$pvalue)
x<-split(all,list(all$S,all$P))#split into a list of dataframes showing individual combinations
ks<-lapply(x,function(x)ks.test(x[,2],x[,4]))#apply ks.test to each individual combination
pval<-lapply(ks, '[[', 'p.value')#extract pvalues
do.call(rbind,pval)#final result at last!
# [,1]
#S1 .P1 0.5175508
#S2 .P1 0.5175508
#S1 .P2 0.1389203
#S2 .P2 0.1389203
#S1 .P3 0.1389203
#S2 .P3 0.1389203