使用压缩列从长到宽重塑数据帧

Say I have (fake) patient data from their visits:

假设我的访问中有（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))

Which gives:

这使：

   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

I would like to have a unique row for each patient at a given visit date. If the patient has multiple codes for a date, I would like a new column for all ICD code given at that visit. This is what it would look like:

我希望在给定的访问日期为每位患者设置一个独特的行。如果患者有多个日期代码，我想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

I've tried reshape, but it seems to add all the ICD9 codes in a column and add the value in a column if they have a value or not (as shown below).I will end up with something like 200 columns, I would only like 3 (the max # of codes per patient per visit in the data set I have, ie ICD9_1, ICD9_2, ICD9_3).

我已经尝试过重塑，但它似乎在列中添加了所有ICD9代码，如果它们有值或者没有值，则在列中添加值（如下所示）。我最终会得到类似200列的内容，我会只有3（我拥有的数据集中每位患者每个患者的最大代码数，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

Sorry if the title isn't as specific as it could be, I'm not really sure how to exactly title what I am looking for.

对不起，如果标题不是那么具体，我真的不确定如何准确标题我想要的。

Thanks in advance for your help!

在此先感谢您的帮助！

3 个解决方案

#1

The basic problem for reshape in this case is that it doesn't have a real "time" variable. That's easy to create with ave:

在这种情况下重塑的基本问题是它没有真正的“时间”变量。使用ave很容易创建：

foo$time <- with(foo, ave(rep(1, nrow(foo)), 
                          PatientNumber, VisitDate, 
                          FUN = seq_along))

Then, you can use reshape as follows:

然后，您可以使用reshape如下：

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

---

Of course, once you have that "time" variable, you can also use dcast from "reshape2".

当然，一旦你有了“时间”变量，你也可以使用“reshape2”中的dcast。

library(reshape2)
dcast(foo, PatientNumber + VisitDate ~ time, value.var="ICD9")

#2

Also,

也，

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

Package tidyr is not available on CRAN yet. Install it like this (see tidyr git):

CRAN尚未提供包tidyr。像这样安装（参见tidyr git）：

# install.packages("devtools")
devtools::install_github("hadley/tidyr")

#3

You could use aggregate:

你可以使用聚合：

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

If you don't know the maximum possible visits ("max_visits"), you could:

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

EDIT:

编辑：

As noted by @AnandaMahto in the comments you could turn your 3-column aggregated "foo" (say "aggfoo") to 4 columns with something like:

正如@AnandaMahto在评论中所指出的那样，你可以将你的3列聚合“foo”（比如说“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

That's not necessary, though, as even with 3 columns it's still convenient to call each "ICD9" column: aggfoo$ICD9[, 1] and aggfoo$ICD9[, 2] instead of aggfoo$ICD9.1 and aggfoo$ICD9.2.

这是没有必要的，但是，因为即使有3列，它仍然方便的调用每个 “ICD9” 列：aggfoo $ ICD9 [，1]和aggfoo $ ICD9 [，2]而不是aggfoo $ ICD9.1和aggfoo $ ICD9.2 。