说我的访问中有(假的)患者数据:
## Create a fake dataframe
foo <- data.frame(PatientNumber=c(11,11,11,22,22,33,33,33,44,55,55),
VisitDate=c("11/03/07","11/03/07","11/20/07","12/20/08",
"12/30/09","09/20/12","09/20/12","10/25/07","05/09/08","06/09/13","06/09/13"),
ICD9=c(10,15,10,30,30,25,60,25,14,40,13))
给出了:
PatientNumber VisitDate ICD9
1 11 11/03/07 10
2 11 11/03/07 15
3 11 11/20/07 10
4 22 12/20/08 30
5 22 12/30/09 30
6 33 09/20/12 25
7 33 09/20/12 60
8 33 10/25/07 25
9 44 05/09/08 14
10 55 06/09/13 40
11 55 06/09/13 13
我希望在给定的访问日期为每位患者提供一个独特的行。如果患者有多个日期代码,我希望在该访问时给出所有ICD代码的新列。这就是它的样子:
WhatIWant <- data.frame(PatientNumber=c(11,11,22,22,33,33,44,55),
VisitDate=c("11/03/07", "11/20/07", "12/20/08", "12/30/09", "09/20/12","10/25/07","05/09/08","06/09/13"),
ICD9_1=c(10,10,30,30,25,25,14,40),
ICD9_2= c(15,NA,NA,NA,60,NA,NA,13))
> WhatIWant
PatientNumber VisitDate ICD9_1 ICD9_2
1 11 11/03/07 10 15
2 11 11/20/07 10 NA
3 22 12/20/08 30 NA
4 22 12/30/09 30 NA
5 33 09/20/12 25 60
6 33 10/25/07 25 NA
7 44 05/09/08 14 NA
8 55 06/09/13 40 13
我尝试过重塑,但它似乎在列中添加了所有ICD9代码,如果它们有值或没有值,则在列中添加值(如下所示)。我最终会得到像200列,我只想3(我拥有的数据集中每位患者的最大代码数量,即ICD9_1,ICD9_2,ICD9_3)。
test <- reshape(foo, idvar = c("VisitDate"), timevar = c("PatientNumber"), direction = "wide")
> test
VisitDate ICD9.11 ICD9.22 ICD9.33 ICD9.44 ICD9.55
1 0007-11-03 10 NA NA NA NA
3 0007-11-20 10 NA NA NA NA
4 0008-12-20 NA 30 NA NA NA
5 0009-12-30 NA 30 NA NA NA
6 0012-09-20 NA NA 25 NA NA
8 0007-10-25 NA NA 25 NA NA
9 0008-05-09 NA NA NA 14 NA
10 0013-06-09 NA NA NA NA 40
如果标题不具体,我很抱歉,我不确定如何准确标题我要查找的内容。
提前感谢您的帮助!
答案 0 :(得分:7)
另外,
library(dplyr)
library(tidyr) # See below on how to get tidyr
foo %>%
group_by(PatientNumber, VisitDate) %>%
mutate(n=paste("ICD9",row_number(), sep="_")) %>%
spread(n, ICD9)
#Source: local data frame [8 x 4]
# PatientNumber VisitDate ICD9_1 ICD9_2
#1 11 11/03/07 10 15
#2 11 11/20/07 10 NA
#3 22 12/20/08 30 NA
#4 22 12/30/09 30 NA
#5 33 09/20/12 25 60
#6 33 10/25/07 25 NA
#7 44 05/09/08 14 NA
#8 55 06/09/13 40 13
套件tidyr
尚未在CRAN上提供。像这样安装(见tidyr
git):
# install.packages("devtools")
devtools::install_github("hadley/tidyr")
答案 1 :(得分:6)
reshape
在这种情况下的基本问题是它没有真正的&#34;时间&#34;变量。使用ave
:
foo$time <- with(foo, ave(rep(1, nrow(foo)),
PatientNumber, VisitDate,
FUN = seq_along))
然后,您可以按如下方式使用reshape
:
reshape(foo, direction = "wide",
idvar=c("PatientNumber", "VisitDate"),
timevar="time")
# PatientNumber VisitDate ICD9.1 ICD9.2
# 1 11 11/03/07 10 15
# 3 11 11/20/07 10 NA
# 4 22 12/20/08 30 NA
# 5 22 12/30/09 30 NA
# 6 33 09/20/12 25 60
# 8 33 10/25/07 25 NA
# 9 44 05/09/08 14 NA
# 10 55 06/09/13 40 13
当然,一旦你有了#34;时间&#34;变量,您也可以使用来自&#34; reshape2&#34; {/ p>的dcast
library(reshape2)
dcast(foo, PatientNumber + VisitDate ~ time, value.var="ICD9")
答案 2 :(得分:2)
您可以使用aggregate
:
max_visits = 2
aggregate(ICD9 ~ PatientNumber + VisitDate, foo,
function(x) x[seq_len(max_visits)]) #note that output is 3 columns
# PatientNumber VisitDate ICD9.1 ICD9.2
#1 44 05/09/08 14 NA
#2 55 06/09/13 40 13
#3 33 09/20/12 25 60
#4 33 10/25/07 25 NA
#5 11 11/03/07 10 15
#6 11 11/20/07 10 NA
#7 22 12/20/08 30 NA
#8 22 12/30/09 30 NA
如果您不知道可能的最大访问次数(“max_visits”),您可以:
max_visits = max(ave(foo[["ICD9"]],
foo[["PatientNumber"]], foo[["VisitDate"]],
FUN = length))
max_visits
#[1] 2
编辑:
正如@AnandaMahto在评论中指出的那样,您可以将3列aggregate
d“foo”(比如“aggfoo”)改为4列,例如:
dim(aggfoo)
#[1] 8 3
dim(do.call(data.frame, aggfoo))
#[1] 8 4
dim(data.frame(unclass(aggfoo)))
#[1] 8 4
但这并不是必要的,因为即使有3列,调用每个“ICD9”列仍然很方便:
aggfoo$ICD9[, 1]
和aggfoo$ICD9[, 2]
代替aggfoo$ICD9.1
和aggfoo$ICD9.2
。