我有大量数据,其中包含许多个体的病理学测试数据。我提供了按比例缩小的数据集,描述了案件的类型。
library(plyr)
library(tidyr)
library(dplyr)
library(lubridate)
options(stringsAsFactors = FALSE)
dat <- structure(list(PersID = c("am1", "am2", "am2", "am3", "am3", "am4", "am4", "am4", "am4", "am4", "am4"), Sex = c("M", "F","F", "M", "M", "F", "F", "F", "F", "F", "F"), DateTested = c("21/10/2015", "9/07/2010", "24/09/2010", "23/10/2013", "25/10/2013", "28/04/2010", "23/06/2010", "21/07/2010", "20/10/2010", "4/03/2011", "2/12/2011"), Res = c("NR", "R", "R", "NR", "R", "R", "R", "R", "R", "R", "R"), Status = c("Yes", "No", "No", "Yes", "Yes", "No", "No", "No", "No", "No", "No"), DateOrder = c(1L, 1L, 2L, 1L, 2L, 1L, 2L, 3L, 4L, 5L, 6L)), .Names = c("PersID", "Sex", "DateTested", "Res", "Status", "DateOrder"), class = "data.frame", row.names = c(NA, -11L))
数据描述了三种类型的人(1)仅具有单个结果的人(2)具有2个结果的人,以及(3)具有许多结果的人。
我的目标是提出一个脚本,该脚本将根据一组条件仅包含针对个人的行。从技术上讲,这是一种仅对个体的行进行计数的方法,前提是其后续结果在指定的重新感染期限(30天)内。
我已将数据转换为列表,并向其传递了许多函数以开始处理数据。
dat$DateTested <- dmy(dat$DateTested)
datList <- dlply(.data=dat, .variables=c('PersID'))
到目前为止,我所做的是:
选择所有每人只有一个结果的行
fnSingleTests <- function(y){
y <- y[length(y$DateOrder)==1,]
}
singleTests <- ldply(datList, fnSingleTests, .id = NULL)
将数据框转换为列表并传递一个函数,该函数确定(a)在30天的再感染期内每人是否有两行,然后选择第一行,以及(b)如果每人有两行以上,最后一条记录和第一条记录都在30天内,请保留第一条记录。
fnMultiTests <- function(y){
y <- y[length(y$DateOrder) > 1,]
}
multiTests <- llply(datList, fnMultiTests)
fnMultiTestsSplit <- function(y){
test <- difftime(y$DateTested[length(y$DateTested)], y$DateTested[1], units='days')
if (nrow(y) <=2){
if (test < 31){
y <- y[y$DateOrder == 1, ]
y <- y[!is.na(y$PerdID), ]
} else {
y <- y[y$DateOrder %in% 1:2, ]
y <- y[!is.na(y$PersID), ]
}
} else {
if (test < 31){
y <- y[y$DateOrder == 1, ]
y <- y[!is.na(y$PersID), ]
} else {
break()
}
}
}
finalTests <- ldply(multiTests, failwith(NULL, fnMultiTestsSplit, quiet = TRUE), .id = NULL)
然后,我可以将数据框与rbind结合使用:
allFinalTests <- rbind(singleTests, finalTests)
我受困的地方是每人多于两排的情况,并且在连续的排内可能存在一段时间,其时间大于30天的再感染时间。
谁能建议我如何扩展此代码,以PersID
仅包括两个以上的案例,然后仅包含在30天再感染期之后发生后续案例的结果。
具体而言,从最旧的案例开始,如果下一个案例在30天内,则排除第二个案例,或者如果第二个案例距上一个案例超过30天,则包括这两个案例。它应该在所有情况下都这样做PersID
在此示例中,我正在寻找的最终输出是:
PersID Sex DateTested Res Status DateOrder
am1 M 21/10/2015 NR Yes 1
am2 F 9/07/2010 R No 1
am2 F 24/09/2010 R No 2
am3 M 23/10/2013 NR Yes 1
am4 F 28/04/2010 R No 1
am4 F 23/06/2010 R No 2
am4 F 20/10/2010 R No 4
am4 F 4/03/2011 R No 5
am4 F 2/12/2011 R No 6
在基数R中,我将采用以下方法:
# convert the 'DateTested' column to a date-format
dat$DateTested <- as.Date(dat$DateTested, format = "%d/%m/%Y")
# calculate the difference in days with the previous observation in the group
dat$tdiff <- unlist(tapply(dat$DateTested, INDEX = dat$PersID,
FUN = function(x) c(0, `units<-`(diff(x), "days"))))
# filter the observations that have either a timedifference of zero or more
dat[(dat[,"tdiff"]==0 | dat[,"tdiff"] > 30),]
这使:
PersID Sex DateTested Res Status DateOrder tdiff
1 am1 M 2015-10-21 NR Yes 1 0
2 am2 F 2010-07-09 R No 1 0
3 am2 F 2010-09-24 R No 2 77
4 am3 M 2013-10-23 NR Yes 1 0
6 am4 F 2010-04-28 R No 1 0
7 am4 F 2010-06-23 R No 2 56
9 am4 F 2010-10-20 R No 4 91
10 am4 F 2011-03-04 R No 5 135
11 am4 F 2011-12-02 R No 6 273
使用data.table包:
library(data.table)
# convert the 'data.frame' to a 'data.table'
# and convert the 'DateTested' column to a date-format
setDT(dat)[, DateTested := as.Date(DateTested, format = "%d/%m/%Y")]
# calculate the difference in days with the previous observation in the group
dat[, tdiff := c(0, `units<-`(diff(DateTested), "days")), PersID]
# filter the observations that have either a timedifference of zero or more than 30 days
dat[(tdiff==0 | tdiff > 30)]
这将给您相同的结果。您还可以按如下所示将它们链接在一起:
setDT(dat)[, DateTested := as.Date(DateTested, format = "%d/%m/%Y")
][, tdiff := c(0, `units<-`(diff(DateTested), "days")), by = PersID
][(tdiff==0 | tdiff > 30)]
并使用dplyr:
library(dplyr)
dat %>%
mutate(DateTested = as.Date(DateTested, format = "%d/%m/%Y")) %>%
group_by(PersID) %>%
mutate(tdiff = c(0, `units<-`(diff(DateTested), "days"))) %>%
filter(tdiff == 0 | tdiff > 30)
这也会给您相同的结果。
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