我有长度15000的第2个数值向量,met
并且nor
其中一些相似的名称存在。例如:
head(met)
ALB IGKJ1 IGKC IGKJ4 IGKJ2 IGHG2
25.75415 20.55957 18.28749 17.87589 17.22944 16.60235
head(nor)
SAA1 CRP RNVU SNORD68 CYP1A2 IGKJ1
25.74548 24.05058 16.72566 15.05746 13.75348 10.74111
我想对met
是否存在剂量nor
以及每个met
值是否1.5*nor
大于其对应nor
值进行子集化。
在上面的例子中,通过比较,我想要的IGKJ1
将是唯一的输出。
我应该如何编码?
library(dplyr)
# get named vectors
met = c(25.75415, 20.55957, 18.28749, 17.87589, 17.22944, 16.60235)
names(met) = c("ALB", "IGKJ1", "IGKC", "IGKJ4", "IGKJ2", "IGHG2")
nor = c(25.74548, 24.05058, 16.72566, 15.05746, 13.75348, 10.74111)
names(nor) = c("SAA1", "CRP", "RNVU", "SNORD68", "CYP1A2", "IGKJ1")
# transform them as data frames
dt_met = data.frame(v_met = met)
dt_met$names = row.names(dt_met)
dt_nor = data.frame(v_nor = nor)
dt_nor$names = row.names(dt_nor)
将名称和两个值保留为新数据框行的第一个选项:
# keep names as a dataset
dt_met %>%
inner_join(dt_nor, by="names") %>% # keep names that exist in both datsets
filter(v_met > 1.5*v_nor) %>% # keep rows where the condition is satisfied
select(names, everything()) # order columns
# names v_met v_nor
# 1 IGKJ1 20.55957 10.74111
第二个选项是仅保留通过您的标准的名称,然后使用它们来对原始向量进行子集:
# save names as a vector
dt_met %>%
inner_join(dt_nor, by="names") %>%
filter(v_met > 1.5*v_nor) %>%
pull(names) -> new_names
# subset met using those names
met[names(met) %in% new_names]
# IGKJ1
# 20.55957
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