找出两个表之间的差异

Question

我正在使用SAS/SQL背景中的R,我正在尝试编写代码来获取两个表,比较它们,并提供差异列表.此代码将重复用于许多不同的表集,因此我需要避免硬编码.

我正在使用识别R中两个数据集之间的特定差异,但它并没有让我一路走来.

示例数据,使用LastName/FirstName(唯一)的组合作为键 -

Dataset One --

Last_Name  First_Name  Street_Address   ZIP     VisitCount
Doe        John        1234 Main St     12345   20
Doe        Jane        4321 Tower St    54321   10
Don        Bob         771  North Ave   23232   5
Smith      Mike        732 South Blvd.  77777   3        

Dataset Two --

Last_Name  First_Name  Street_Address   ZIP     VisitCount
Doe        John        1234 Main St     12345   20
Doe        Jane        4111 Tower St    32132   17
Donn       Bob         771  North Ave   11111   5

   Desired Output --

   LastName FirstName VarName         TableOne        TableTwo
   Doe      Jane      StreetAddress   4321 Tower St   4111 Tower St 
   Doe      Jane      Zip             23232           32132
   Doe      Jane      VisitCount      5               17

请注意,此输出忽略两个表中没有相同ID的记录(例如,因为Bob的姓氏在一个表中是"Don",而另一个表中的"Donn",我们完全忽略该记录).

我已经通过在两个数据集上应用融合函数,然后比较它们来探索这一点,但我正在使用的大小数据表明这是不切实际的.在SAS中,我使用了Proc Compare进行这种工作,但是我没有在R中找到一个完全相同的东西.

Answer 1

这是一个基于以下的解决方案data.table:

library(data.table)

# Convert into data.table, melt
setDT(d1)
d1 <- d1[, list(VarName = names(.SD), TableOne = unlist(.SD, use.names = F)),by=c('Last_Name','First_Name')]

setDT(d2)
d2 <- d2[, list(VarName = names(.SD), TableTwo = unlist(.SD, use.names = F)),by=c('Last_Name','First_Name')]

# Set keys for merging
setkey(d1,Last_Name,First_Name,VarName)

# Merge, remove duplicates
d1[d2,nomatch=0][TableOne!=TableTwo]

#     Last_Name First_Name        VarName      TableOne      TableTwo
#     1:       Doe       Jane Street_Address 4321 Tower St 4111 Tower St
#     2:       Doe       Jane            ZIP         54321         32132
#     3:       Doe       Jane     VisitCount            10            17

输入数据集是:

# Input Data Sets
d1 <- structure(list(Last_Name = c("Doe", "Doe", "Don", "Smith"), First_Name = c("John", 
"Jane", "Bob", "Mike"), Street_Address = c("1234 Main St", "4321 Tower St", 
"771  North Ave", "732 South Blvd."), ZIP = c(12345L, 54321L, 
23232L, 77777L), VisitCount = c(20L, 10L, 5L, 3L)), .Names = c("Last_Name", 
"First_Name", "Street_Address", "ZIP", "VisitCount"), class = "data.frame", row.names = c(NA, -4L))                                                                                                               

d2 <- structure(list(Last_Name = c("Doe", "Doe", "Donn"), First_Name = c("John", 
"Jane", "Bob"), Street_Address = c("1234 Main St", "4111 Tower St", 
"771  North Ave"), ZIP = c(12345L, 32132L, 11111L), VisitCount = c(20L, 
17L, 5L)), .Names = c("Last_Name", "First_Name", "Street_Address", 
"ZIP", "VisitCount"), class = "data.frame", row.names = c(NA, -3L))

Answer 2

dplyr而tidyr在这里工作得很好.首先,略微减少的数据集:

dat1 <- data.frame(Last_Name = c('Doe', 'Doe', 'Don', 'Smith'),
                   First_Name = c('John', 'Jane', 'Bob', 'Mike'),
                   ZIP = c(12345, 54321, 23232, 77777),
                   VisitCount = c(20, 10, 5, 3),
                   stringsAsFactors = FALSE)
dat2 <- data.frame(Last_Name = c('Doe', 'Doe', 'Donn'),
                   First_Name = c('John', 'Jane', 'Bob'),
                   ZIP = c(12345, 32132, 11111),
                   VisitCount = c(20, 17, 5),
                   stringsAsFactors = FALSE)

(对不起,我不想全部输入.如果它很重要,请提供一个可重现的示例,其中包含明确定义的数据结构.)

此外,看起来你的"想要的输出"与Jane Doe ZIP和VisitCount.

你想融化它们的想法很好:

library(dplyr)
library(tidyr)
dat1g <- gather(dat1, key, value, -Last_Name, -First_Name)
dat2g <- gather(dat2, key, value, -Last_Name, -First_Name)
head(dat1g)
##   Last_Name First_Name        key value
## 1       Doe       John        ZIP 12345
## 2       Doe       Jane        ZIP 54321
## 3       Don        Bob        ZIP 23232
## 4     Smith       Mike        ZIP 77777
## 5       Doe       John VisitCount    20
## 6       Doe       Jane VisitCount    10

从这里开始,它看似简单:

dat1g %>%
    inner_join(dat2g, by = c('Last_Name', 'First_Name', 'key')) %>%
    filter(value.x != value.y)
##   Last_Name First_Name        key value.x value.y
## 1       Doe       Jane        ZIP   54321   32132
## 2       Doe       Jane VisitCount      10      17