问题描述
BV.MC.Entry.data.AB = data.frame(unique_female_id=c("abc123","ade456","dde345","abc1234","abc12"))
BV.MC.Entry = data.frame(unique_id=c("abc123","dde345"))
to_search_in.female <- data.table(BV.MC.Entry.data.AB[!duplicated(BV.MC.Entry.data.AB$unique_female_id),c(1)])
colnames(to_search_in.female)=c("unique_female")
to_search_with.female <- tibble(BV.MC.Entry[!duplicated(BV.MC.Entry$unique_id),c(1)])
colnames(to_search_with.female) = "unique_female_id"
dim(to_search_in.female);dim(to_search_in.male);dim(to_search_with.male);dim(to_search_with.female)
linked.female.peds = to_search_with.female %>%
mutate(data = list(to_search_in.female)) %>%
unnest(data) %>%
filter(str_detect(unique_female,fixed(unique_female_id))) %>%
#select(unique_female,female.pedigree,unique_female_id) %>%
group_by(unique_female_id) %>%
summarise(strings = str_c(unique_female,collapse = ","))
linked.female.peds = linked.female.peds %>% tidyr::separate("strings",sep="[,][ ]",c("match1","match2","match3","match4","match5","match6","match7","match8","match9","match10","match11","match12","match13","match14","match15","match16"),extra="merge",remove=F)
linked.female.peds
# A tibble: 3 x 18
unique_female_id strings match1 match2 match3 match4
<fct> <chr> <chr> <chr> <chr> <chr>
1 abc123 abc123~ abc123 abc12~ NA NA
2 ade456 ade456 ade456 NA NA NA
3 dde345 dde345 dde345 NA NA NA
我希望结果返回此数据帧:
所以它将代替abc12前进,并且match2列将等于abc1234,NA,NA。
解决方法
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