用于比较具有相同 ID 的特定列的行之间的函数

Question

我有一个很大的实验室数据库，一些ID有多个结果，我还创建了另一个关键变量，首字母+年龄+性别变量用于与医院病历的其他匹配目的。但是我注意到有时不同的首字母缩写有相同的医院 ID。我想写一个函数来检测这种不一致。

所以数据库的例子：

df=data.frame(ID=c("5606","5606","5728","5824","5824","5824","5824"),
                     key2=c("TN35M","TN35M","JJ26M","CD47F","CD47F","DG44M","DG44M"),
              date_sample=c("12/03/2012","12/03/2012","19/04/2012","21/05/2012","21/05/2012","19/10/2012","19/10/2012"), service=c("ORTHO","ORTHO","BLOC","VISC","VISC","BLOC","BLOC"), germe=c("Acinetobacter sp","Burkholderia pseudomallei","Stenotrophomonas maltophilia","Staphylococcus haemolyticus"," Enterobacter cloacae","Escherichia  coli","Pseudomonas aeruginosa"))


ID      key2    date_sample service germe

5606    TN35M   12/03/2012  ORTHO   Acinetobacter sp
5606    TN35M   12/03/2012  ORTHO   Burkholderia pseudomallei
5728    JJ26M   19/04/2012  BLOC    Stenotrophomonas maltophilia
5824    CD47F   21/05/2012  VISC    Staphylococcus haemolyticus
5824    CD47F   21/05/2012  VISC    Enterobacter cloacae
5824    DG44M   19/10/2012  BLOC    Escherichia coli
5824    DG44M   19/10/2012  BLOC    Pseudomonas aeruginosa

每个 ID 都应该有一个唯一的 key2 变量。我如何比较相同 "ID" 变量的 "key2" 变量的行，并有一个输出变量来检测我所有不连贯的行，以确保每个 ID 给一个唯一的病人而不是由超过 1 名患者共享？

喜欢：


ID       key2   date_sample service germe                        incoherence

5606    TN35M   12/03/2012  ORTHO   Acinetobacter sp                N
5606    TN35M   12/03/2012  ORTHO   Burkholderia pseudomallei       N
5728    JJ26M   19/04/2012  BLOC    Stenotrophomonas maltophilia    N
5824    CD47F   21/05/2012  VISC    Staphylococcus haemolyticus     Y
5824    CD47F   21/05/2012  VISC    Enterobacter cloacae            Y
5824    DG44M   19/10/2012  BLOC    Escherichia coli                Y
5824    DG44M   19/10/2012  BLOC    Pseudomonas aeruginosa          Y

Answer 1

您可以统计每个组的唯一值。如果大于 1，则 Y（或在本例中为 TRUE），即

!with(df, ave(key2, ID, FUN = function(i) length(unique(i)))) == 1
#[1] FALSE FALSE FALSE  TRUE  TRUE  TRUE  TRUE

注意：确保你的变量是字符，而不是因子

Answer 2

使用dplyr

library(dplyr)
df %>%
 group_by(ID) %>%
 mutate(incoherence = c("N", "Y")[(n_distinct(key2) > 1) +1])

#   ID    key2 incoherence
#  <fct> <fct> <chr>      
#1 5606  TN35M N          
#2 5606  TN35M N          
#3 5728  JJ26M N          
#4 5824  CD47F Y          
#5 5824  CD47F Y          
#6 5824  DG44M Y          
#7 5824  DG44M Y

和data.table

library(data.table)
setDT(df)[, incoherence := c("N", "Y")[(uniqueN(key2) > 1) +1], by = ID]

用于比较具有相同 ID 的特定列的行之间的函数

Function to compare rows among themselves of a specific column with the same ID

compare

r

rows